GPT-4 Can't Reason

1. Introduction

In early January I wrote a commentary1 presenting an informal evaluation of ChatGPT across a broad range of subject areas: conventional NLU, folk physics, information retrieval, pragmatics, theory of mind, spatial inference, simple logical reasoning, and math. The key takeaways were that ChatGPT was a seminal breakthrough; that LLM-based systems are not mere stochastic parrots but build genuine abstractions and can exhibit creativity; that such systems will enable a large array of new and exciting applications; and that, despite all of the above, these systems are still severely limited when it comes to reasoning.

GPT-4 was released a couple of months after that, delivering very substantial improvements across the board. I remain impressed and excited by the general capabilities and potential of LLMs, and I have little doubt that their performance will continue to improve in the near future. Nevertheless, there are increasing grounds for scepticism concerning their reasoning abilities. In this position paper I will argue that the best LLM at this time, GPT-4, is utterly incapable of reasoning, in spite of its sporadic displays of ingenuity.

I will largely steer clear of the much broader—and more vague—debate about whether LLMs in general are capable of (consistently robust) reasoning, but a few brief remarks will help to set the stage and clarify why it makes sense to restrict attention to a specific LLM. On one side of that broader debate, rosy predictions by LLM enthusiasts rely excessively on ever-changing scaling “laws” that rest on flimsy empirical evidence and on a host of questionable modeling assumptions, ill-understood concepts (such as “emergent” LLM properties2), and a somewhat dogmatic belief that minimizing cross-entropy loss on next-token prediction over a huge corpus will deliver a general reasoning engine via the magic of transfer learning and the construction of generic higher-level representations.

On the other side of the debate, while LLM skeptics have serious arguments to make, those arguments are mostly a priori and somewhat vague (for instance, that LLMs lack “a model of the world”), and I do not think they settle the question. In my view, the most compelling a priori considerations against the plausibility of reliably robust LLM reasoning turn on computational complexity results. Reasoning is a (very) computationally hard problem. In fact, in the general case (first-order or higher-order logic), it is algorithmically undecidable, i.e., every bit as unsolvable as the halting problem. Thus, by Church’s thesis, we cannot expect any algorithm, LLMs included, to solve arbitrary reasoning problems in a sound and complete way.3 But even “easier” classes of reasoning problems4 typically have either exponential or at least nontrivial polynomial-time complexity profiles. Problem classes that have linear-time inference algorithms, such as Horn clauses over literals, are rarely expressive enough. This tradeoff between generality and expressivity on the one hand and tractability on the other means that no LLM, no matter how large or how extensively and cleverly trained and tuned, will ever be able to crack an arbitrary reasoning problem. And this is consistent with the famous “no free lunch” theorem of machine learning, which points to a similar inverse relationship between model generality and performance.

But LLM advocates can make a couple of cogent counterpoints, while granting that there will never be an AI oracle that can essentially solve the halting problem. First, they can point out that even though a problem might have high worst-case asymptotic complexity, it might still be solvable well enough in practice. Unlike random instances, real-world instances of reasoning problems (and indeed real-world instances of most computationally hard problems) appear to have structure that allows clever algorithms to tackle them effectively.5 There are many examples here, from the simplex algorithm for linear programming and SAT solvers to term unification algorithms and even automatic theorem provers for full first-order logic. All of these problems are hard (having at least exponential-time worst-case complexity), yet somehow we have algorithms for them that seem to work successfully on a wide variety of inputs.

Second, and perhaps more important, we need not aim for an oracle anyway. Humans are not oracles either, nor do they seem to follow any particular algorithm that captures any one specific class of reasoning problems. The ability of humans to reason is much more fluid and messy, but impressive nevertheless. Is it impossible to build something like an LLM-based system with the reasoning ability of a well-trained engineer of average intelligence (which perhaps can then become even more intelligent and better trained by an endless process of learning and improvement)?

I don’t think that building such a system can be ruled out on a priori grounds (and here I differ from hard-core AI skeptics). I think it’s implausible, for a number of reasons,6 but ultimately this strikes me as an empirical question that must be decided on a case-by-case basis, by subjecting a specific system to testing, i.e., by interrogating it, probing it, and analyzing its responses. And the case I will consider here is that of GPT-4, which appears, by all accounts, to be the most capable LLM at present.

There are two questions that must be addressed before we proceed. First, we must agree on what reasoning is, and second, we must say something about methodology. The next section contains a brief discussion of reasoning, but for those who wish to skip that section and dive right into the problems, the upshot is that we’ll focus on (a liberal conception of) deductive reasoning. Regarding methodology, just like the January piece, my evaluation here is not based on a corpus or set of corpora. Instead, I present a detailed qualitative analysis of GPT-4’s performance on 21 simple reasoning problems across a wide range of areas, most of which have been made up from scratch, while the rest (such as Wason’s selection task) have been manually tweaked so as to make them less recognizable to the model.

This is done partly to avoid data contamination, which is a serious problem affecting corpus-based evaluations. Given how little we know about the training regimen of ChatGPT, it is impossible to know for sure whether any existing dataset or problem has effectively been “seen” by the model during its pretraining or subsequent alignment, whether we’re talking about NLP datasets, medical licensing exams, Python programming problems, LSAT or bar-entrance exams, SAT or GRE tests, and so on.7 The qualification “effectively” is important, because even though a specific problem might not have been seen in its exact form (in a string-matching sense), an essentially equivalent variant with a different surface formulation might well have been. Hence, simple contamination tests based on substring checks, such as those carried out by OpenAI in their GPT-4 Technical Report [8] (posted in March 2023), are not sufficient to guarantee lack of contamination.8

The absence of a large corpus makes the discussion more qualitative rather than quantitative. However, the results are arguably more informative than a numeric metric computed over a corpus, for a number of reasons. First, because contamination can be ruled out conclusively; second, because the problems span a large gamut of areas; and third, because a qualitative discussion of a problem allows for greater depth of analysis and more context in which to interpret the results. By contrast, the only way to perform a truly informative quantitative evaluation is to come up with a brand new corpus that satisfies all of the following criteria: (a) originality; (b) uniformly high quality; (c) sufficiently large size; and (d) diversity (not being limited to one type of task only). This is a very challenging undertaking. Even then, a few simple numeric metrics on a brand new dataset might not be particularly illuminating. Are the numbers measuring the right things? Do we even know the right things to measure? Is there an appropriate backdrop in which the numbers can be understood? For deeper insight, we need to put individual examples under a magnifying glass.

This is particularly important because we need to scrutinize the explanations (“chains of thought”) generated by a reasoner. Unfortunately, almost all reasoning corpora comprise either multiple-choice questions or binary classification problems (e.g., “Does sentence $p_2$ follow from premise $p_1$, yes or no?”). Why? Mostly because it is easy to mechanically evaluate model performance on such datasets. But even in the absence of contamination, this type of test set runs the serious risk that the LLM will manage to pick the right answers by latching on to spurious statistical regularities, i.e., to arrive at the right answers for the wrong reasons [6,10].9 Adversarial augmentation of an existing dataset might help, especially if we know what we are trying to guard against, but unless an adversarial version restores near-random performance, this can quickly devolve into a game of whac-a-mole, where we detect a new round of bogus regularities exploited by the model and must undertake a new round of adversarial interventions.

Ultimately, there is really no proper way to assess the reasoning ability of a system unless we ask it to explain its output. This is an essential part of reasoning, which is not about producing the right answer by hook or by crook but about deriving the right answer for the right reasons. And rote metrics like ROUGE-L are not fit for purpose here. We need to roll up our sleeves and analyze LLM explanations and proof attempts manually. We also need to gauge their performance in a dialog setting (e.g., what happens when a reasoning error is pointed out to them?). This is the sort of analysis undertaken in this paper. I believe the results show unequivocally that GPT-4 cannot reason. The errors are too pervasive and too egregious. GPT-4 doesn’t solve even one of the 21 problems discussed here. But much more concerning are the fundamentally flawed explanations and proof attempts it produces along the way.

LLM believers will probably demur: But humans also make mistakes, and surely we’re not prepared to say that humans can’t reason just because they make mistakes? First, it is not accurate to say without qualification that “humans can reason,” certainly not in the sense that we can randomly pluck any person from the street and expect them to reliably perform normatively correct reasoning. Most neurobiologically normal humans have the capacity to become proficient in reasoning, but actually attaining such proficiency takes significant training and discipline. Humans are known to be susceptible to a large assortment of cognitive biases, which can only be overcome by rigorous instruction. Focusing on the reasoning skills of untrained people is a bit like focusing on the singing skills of the general population. Everybody sings in the shower, but without formal training (or at least exceptional talent) the results are usually regrettable.

Of course, even sophisticated human reasoners make mistakes, just like trained singers can hit false notes. But if a human made these mistakes, the ones reported in this article, then I would conclude without any hesitation that they cannot reason. Even if they went on to list a large number of other examples demonstrating impeccable reasoning, I would suspect that other factors (such as rote memorization or cheating) were behind the performance discrepancy. For the mistakes reported here are not performance mistakes, the sort of innocuous errors that humans might make—and promptly correct—when they are careless or tired. If a human made these mistakes, and made them consistently under repeated questioning, that would indicate without doubt that they don’t have the necessary logical competence, that they lack fundamental concepts that are part and parcel of the fabric of reasoning, such as logical entailment and set membership. And I would certainly not entrust that person with generating reams of Python or Javascript code for an enterprise. Nor would I start organizing international conferences to investigage how their reasoning prowess might threaten humanity with extinction.

2. What is Reasoning?

Reasoning is not quite the same thing as intelligence, but it’s a necessary ingredient for it. Broadly put, reasoning is the process of drawing and evaluating conclusions from a given body of information. More precisely, it is the process of making and—more importantly—justifying arguments. An argument consists of a conclusion (the argument’s upshot, so to speak) and a set of premises from which the conclusion is derived. Premises represent information that is taken as given, if only provisionally, for the purposes of the argument. The conclusion and the premises are typically declarative sentences (expressed either in natural language or in the notation of a symbolic logic) that can be true or false, but they may also be represented by alternative notational devices, such as diagrams. We say that a set of premises S logically entails (or logically implies) a conclusion p iff p is true whenever all the sentences in S are true, in which case the argument is said to be valid. This means that it’s logically impossible to have a state of affairs in which every element of S holds but p does not. This key logical relationship is a lynchpin of human reasoning.10

Valid deductive arguments (whose conclusions are entailed by the premises) are said to be analytical (or sometimes tautological), insofar as, technically speaking, they convey no information.11 This idea is also sometimes expressed by calling such arguments non-ampliative, meaning that there is no information contained in the conclusion that is not already contained—if only latently—in the premises. Deduction is the process of making and justifying non-ampliative arguments.

Deductive arguments are typically justified by proofs, which are sequences of inference steps, each of which applies an inference rule to a number of premises and/or results of previous steps and derives a new result. The last step derives the final conclusion of the proof. An inference rule may be low-level and easy to apply or higher-level and computationally expensive. But all inference rules are required to be sound (or truth-preserving), that is, they must ensure that if the inputs are true then so is the output. All mathematical proofs are deductive, and mathematical reasoning in general is predominantly deductive.12

The conventional view is that some arguments are ampliative, meaning that the conclusion is not quite entailed by the premises. In other words, it is possible for the premises to be true while the conclusion is false. These are typically subdivided into inductive and abductive arguments,13 although some authors view induction as a species of abduction, and even more authors view abduction as a species of induction. There is no rigorous definition of either, but roughly, the premises of a good inductive argument make its conclusion likely, though never quite certain (in constrast to deduction, where the truth of the premises guarantees the truth of the conclusion). Induction can generate specific conclusions from all kinds of premises (specific or general), but often it proceeds from specific individual observations $o_1,\dots ,o_n$ to a more general hypothesis H that subsumes the individual $o_i$ in some sense (for instance, H may be a universally quantified sentence and the $o_i$ could be instances of that sentence). Much of what ML algorithms do can be viewed as inductive reasoning. For instance, a linear-regression algorithm might take as input n datapoints about car models, where each datapoint is of the form $d_i=((c_i,h_i,y_i),m_i)$ for $i=1,\dots ,n$, where $c_i$ is the number of cylinders for the $i^{th}$ car model, $h_i$ is the horsepower, $y_i$ is the model year, and the dependent variable $m_i$ is the mpg (miles per gallon). And it might produce as output a formula like $m=w_1·c+w_2·h+w_3·y+b$, which predicts the mpg of a car model from its number of cylinders, horsepower, and model year.14 Here $w_1$, $w_2$, $w_3$, and b are specific numbers (weights) representing a hyperplane that minimizes the mean squared error for the input data (meaning that the hyperplane determined by these weights might not fit the n datapoints perfectly, but it does so better than the hyperplane determined by any other set of weights).15

The main distinguishing feature of abductive reasoning is a strong emphasis on explanation. Abduction consists mostly in making and justifying arguments that explain a set of facts. If one day I come home early from work and I see a plumber’s van parked in my neighbors’ driveway, I might conclude that my neighbors are having some plumbing work done in their house. The premise here is “There is a plumbing van parked in my neighbors’ driveway” and the conclusion is “My neighbors are having plumbing work done in their house.” This is sometimes called “inference to the best explanation,” because the conclusion serves to explain the premise(s). This is also a form of ampliative reasoning—the conclusion does not follow logically from the premises. There are many alternative explanations of a given set of facts or observations (perhaps a plumber parked there temporarily, or the neighbors bought the van, or the neighbors have a plumber friend who is making a social visit, and so on). A good abductive inference will yield a hypothesis that has more explanatory value than competing hypotheses. But how exactly to measure the quality of an abductive piece of reasoning is an open question.16 Note that it doesn’t take a large leap of imagination to view induction as a form of abduction. Observing a large number of black (and only black) swans and then conjecturing that all swans are black could be seen as abductive reasoning, as the conclusion $\forall x.swan\left(x\right)\Rightarrow color\left(x\right)=black$ would explain all the observed data. Linear regression can also be seen as the making of an abductive hypothesis, as can (much more generally) Maximum Likelihood Estimation, a principle that underlies many ML algorithms and is often associated with induction.

All of the above is received wisdom, but it’s worth mentioning that there have been thinkers, called “deductivists” (ranging from philosophers such as Popper and Musgrave to statisticians such as Fisher), who contend that deduction is the only real form of reasoning there is, insofar as it’s the only one for which we have a rigorous and properly understood formal notion of validity; and that other (ampliative) arguments are best understood as reconstructed deductions, typically as enthymemes (arguments that omit tacitly understood premises). I find that position congenial,17 but venturing into that discussion would take us too far afield. For present purposes it suffices to say that we will focus on deduction, because it is the type of reasoning that underpins most logico-mathematical thought and for which we have clear normative standards of evaluation.

An important note: I view the discovery and justification of particular models (including counterexamples and countermodels in general) as part and parcel of reasoning. This is not a controversial view; some cognitive scientists view models and associated cognitive processes as the fundamental ingredients of human reasoning [11]. In addition, however, I view model-based reasoning as at least partly deductive, because even though the actual process of discovering models might not be a process of deduction18, its outcome is a claim (namely, that a given interpretation satisfies a set of premises) that can be verified or falsified deductively, taking as premises the definition of the model itself and possibly other general knowledge about the model’s domain. Indeed, I will consider even computation as a form of deduction, because a particular computation can be naturally regarded as a deductive derivation of a conclusion of the form $f(e_1,\dots ,e_n)=v$, where $f(e_1,\dots ,e_n)$ is the application of an arbitrary function f to arbitrary argument expressions $e_1,\dots ,e_n$, ultimately yielding value v as the result. The premises for the derivation consist of the definition of f and possibly other auxiliary functions, along with the usual equational axioms (reflexivity, symmetry, transivity, and functional/relational congruence).19

3. Test Problems

This section will start with the usual caveat: GPT-4 is a nondeterministic system that might produce different answers on different runs, even with the same parameter settings. All of the following exchanges with GPT-4 have been transcribed verbatim, and in my experience the errors discussed here tend to be robust, but it’s conceivable that for a given example GPT-4 might generate a different output even in response to the exact same prompt.20

3.1. Simple Arithmetic

The ability to perform basic arithmetic is a necessary ingredient for reasoning. GPT-4 is still unable to reliably perform elementary arithmetic operations such as addition and multiplication. To ensure that GPT-4 isn’t falling back on rote memorization, we can ask it to first select two random integers in a range of our choice and then perform the operation on the selected values:

KA: Select two random numbers between 1381 and 1453 and multiply them together, reporting the result.

GPT-4: Sure, let’s select two random numbers in the range of 1381 to 1453. Let’s say these numbers are 1405 and 1421.

To get the product, we simply multiply these two numbers together:

1405 * 1421 = 1996025

Alas, the correct answer is $1405·1421=1996505$.

One response here from the LLM camp has been to concede that GPT-4 is not a Turing-complete framework and cannot perform general-purpose computation: It can’t add, multiply or divide, it can’t sort lists, it can’t invert matrices or solve systems of differential equations, it can’t compile arbitrary programs to machine code, it can’t derive theorems, and so on ad infinitum. Nevertheless, the response goes, these computational “gaps” can be filled by plug-ins, along the lines of the ReAct (short for “Reasoning and Acting”) [4] or MRKL (short for “Miracle”) [3] frameworks. It is this line of work that has led to systems such as LangChain and AutoGPT. The idea here is to use the LLM as a planner to recursively decompose a complex problem into a sequence of subproblems that can be dispatched to appropriate specialized agents via appropriate APIs.

The problem with this approach is its circularity. Planning itself requires reasoning, so this is a Catch-22 situation: We can’t solve reasoning by delegating to appropriate agents, because figuring out the delegation (how the problem should be decomposed, which agents to call and how, and how to compose the results) is itself computationally infeasible. It not only requires the ability to understand natural language, but also the ability to reason about preconditions and effects. And this is assuming a fixed collection of agents with clear-cut APIs.21 Even under these overly simplistic assumptions, planning is very computationally expensive (PSPACE-complete), and radical simplifications are needed to scale down the complexity even to the level of NP-completeness. Under more realistic assumptions, planning is performed under incomplete knowledge and it is mechanically undecidable. As we understand them today, expecting generic LLMs to tackle planning with any degree of success is naive.

3.2. Simple Counting

While concrete counting is not necessarily a reasoning activity,22 it is surely a requirement for any generally capable reasoning system. Here I give GPT-4 a propositional variable with 27 negation signs in front of it and ask it to count the number of negations. For a human this would be an easy task, especially because the negation signs are written in five blocks with five tildes each, followed by a final pair of negation signs.

KA:  How many times is p negated in the following formula:

$\sim \sim \sim \sim$$\sim \sim \sim \sim$$\sim \sim \sim \sim$$\sim \sim \sim \sim$$\sim \sim \sim \sim$$\sim \sim$p?

GPT-4: The formula you’ve given includes 28 negation symbols (“∼”) in front of the propositional variable “p”. In logic, the negation symbol changes the truth value of a statement to its opposite. If a statement is true, its negation is false, and vice versa. Double negation, like “$\sim$$\sim$p”, is equivalent to “p”, because negating a proposition twice returns it to its original truth value.

Given this, to find the effective number of negations, we can simply divide the total number of negation symbols by 2. If there’s a remainder, then there is a net negation. If not, then there isn’t.

28 divided by 2 is 14, with no remainder, meaning there’s no net negation. So effectively, the variable “p” is not negated in the given formula. It’s equivalent to just “p”.

A difference of one negation sign might not seem like a serious matter, until you realize that it makes all the difference in logical import,23 as GPT-4’s own (unsolicited) explanation stresses. Repeating this experiment multiple times gave widely different results (including 50 and 54), even after explicitly telling GPT-4 to take its time and count carefully.

Here is another example. I greet GPT-4 five times in a row and then I ask it a simple question:

KA: Hi

GPT-4: Hello! How can I assist you today?

KA: Hello

GPT-4: How can I assist you today?

KA: Greetings

GPT-4: Greetings! How can I assist you today?

KA: Hi there.

GPT-4: Hi there! How can I help you today?

KA: Hullo!

GPT-4: Hello! How can I assist you today?

KA: You’ve been greeted a few times so far. Three of those greetings were actually from my girlfriend, the rest from me. How many times did I greet you?

GPT-4: If your girlfriend greeted me three times and all the other greetings were from you, then you greeted me four times. This is assuming that all seven greetings (“Hi”, “Hello”, “Greetings”, “Hi there”, “Hullo!”) were included in the count.

In this case the arithmetic is correct ($7-3=4$) but the count—and ultimately the resulting answer—is hallucinated.

3.3. (Medical) Common Sense

In the present setting we may regard common-sensical arguments as straightforward enthymematic deductions of conclusions from given information plus unstated premises that constitute tacit, generally accepted background knowledge. In this particular case, such common-sense knowledge would be propositions like “A person is alive until they die, after which they do not become alive again.”

KA: Mable’s heart rate at 9 AM was 75 bpm and her blood pressure at 7 PM was 120/80. She died at 11 PM. Was she alive at noon?

GPT-4: Based on the information provided, it’s impossible to definitively say whether Mable was alive at noon. The data points provided (heart rate at 9 AM and blood pressure at 7 PM) both indicate that she was alive at those times. However, without additional information about her condition between 7 PM and 11 PM, or at noon specifically, we can’t make a certain determination about her status at noon.

Remember, medical information can be complex and nuanced. Understanding a person’s health status often requires a comprehensive examination and should be performed by a healthcare professional. If this is a hypothetical question, keep in mind that real-life situations would require more complete information.

No comment is needed here, though the gratuitous—but by now de rigueur—piece of legalese issuing the vague disclaimer at the end (“this-is-a-complex-issue-and-make-sure-to-consult-with-a-professional”) seems ironic.

3.4. Elementary Logic

If $P\left(x\right)$ implies $Q\left(x\right)$ and $Q\left(a\right)$ does not hold then we can infer, by modus tollens, that $P\left(a\right)$ does not hold either (because if it did then $Q\left(a\right)$ would too). This is as elementary of a tautology as can be, yet GPT-4 is perfectly willing to produce a countermodel:

KA: Find a model in which P(x) implies Q(x), Q(a) does not hold, and P(a) holds.

GPT-4: To find a model in which P(x) implies Q(x), Q(a) does not hold, and P(a) holds, let’s consider the following interpretations:

• P(x): x is an even number
• Q(x): x is a positive number
• a: $-2$

Now, let’s check if these interpretations satisfy the given conditions:

• P(x) implies Q(x): If x is an even number, then x is a positive number. This statement is not universally true, as we can have negative even numbers. However, it does not preclude the existence of a model with the other given conditions.
• Q(a) does not hold: Q($-2$) means $-2$ is a positive number, which is false. So, this condition is satisfied.
• P(a) holds: P($-2$) means $-2$ is an even number, which is true. So, this condition is also satisfied.

Thus, the model with the interpretations P(x): x is an even number, Q(x): x is a positive number, and a: $-2$ satisfies the conditions P(x) implies Q(x), Q(a) does not hold, and P(a) holds.

Observe that GPT-4 recognizes that $P\left(x\right)$ does not in fact imply $Q\left(x\right)$ under the interpretation it gave, but it nevertheless dismisses this inconvenient consideration because “it does not preclude the existence of a model with the other given conditions,” as if the satisfaction of the other conditions was enough (it isn’t; a countermodel must satisfy all of the given premises while falsifying the conclusion). Moreover, just a few sentences later GPT-4 claims that $P\left(x\right)$does imply $Q\left(x\right)$ under the given interpretation, contradicting its own earlier claim. We will see that internal inconsistency is a recurrent theme.

When we ask it to elaborate on why it thinks that $P\left(x\right)$ implies $Q\left(x\right)$, GPT-4’s response is revealing:

KA: Can you explain why this model satisfies the first constraint (namely, that $P\left(x\right)$ implies $Q\left(x\right)$)?

GPT-4: You are correct that my previous explanation was not clear enough for the first constraint. I apologize for that. In the given model, $P\left(x\right)$ does not necessarily imply $Q\left(x\right)$. Let’s reconsider the model and find one that satisfies all the constraints.

• $P\left(x\right)$: x is a multiple of 4
• $Q\left(x\right)$: x is an even number
• $a:4$

Now let’s check if these interpretations satisfy the given conditions:

• $P\left(x\right)$ implies $Q\left(x\right)$: If x is a multiple of 4, then x is an even number. This statement is universally true because every multiple of 4 is also a multiple of 2, making it an even number.
• $Q\left(a\right)$ does not hold: $Q\left(4\right)$ means 4 is an even number, which is true. So, this condition is not satisfied.

To satisfy all three conditions, we can modify the interpretations as follows:

• $P\left(x\right)$: x is a positive multiple of 4
• $Q\left(x\right)$: x is a positive even number
• $a:-4$

Now let’s check if these interpretations satisfy the given conditions:

• $P\left(x\right)$ implies $Q\left(x\right)$: If x is a positive multiple of 4, then x is a positive even number. This statement is universally true because every positive multiple of 4 is also a multiple of 2, making it a positive even number.
• $Q\left(a\right)$ does not hold: $Q(-4)$ means $-4$ is a positive even number, which is false. So, this condition is satisfied.
• $P\left(a\right)$ holds: $P(-4)$ means $-4$ is a positive multiple of 4, which is false. So, this condition is not satisfied.

It seems that finding a model that satisfies all three conditions is challenging. I must correct my previous claim: there may not be a model that satisfies all three conditions, $P\left(x\right)$ implies $Q\left(x\right)$, $Q\left(a\right)$ does not hold, and $P\left(a\right)$ holds, simultaneously.

Thus, after multiple unsuccessful attempts to find a countermodel, GPT-4 acknowledges that such a countermodel might not exist. But, remarkably, when GPT-4 is asked to settle the matter one way or the other, it jumps back to the position that a countermodel does exist after all.

KA: Can you prove that one exists or that one does not exist?

GPT-4: I apologize for the confusion earlier. Let me provide a formal proof to show that a model that satisfies all three conditions $P\left(x\right)$ implies $Q\left(x\right)$, $Q\left(a\right)$ does not hold, and $P\left(a\right)$ holds does indeed exist. Let’s consider a model with the following interpretations:

• P(x): x is an element of set A
• Q(x): x is an element of set B
• a: a specific element, a

Now, let’s define the sets A and B as follows:

Set $A=\left\{a\right\}$.

Set $B=\left\{x\right|xisanelementofAandx\ne a\}$.

Now, let’s check if these interpretations satisfy the given conditions:

• $P\left(x\right)$ implies $Q\left(x\right)$: If x is an element of set A, then x is an element of set B. In this case, since the only element in set A is a, and set B is defined such that it contains all elements of A except a, this implication is vacuously true. This is because there are no other elements in set A for which the implication would be false.
• $Q\left(a\right)$ does not hold: $Q\left(a\right)$ means a is an element of set B. By definition, set B does not include a, so this condition is satisfied.
• $P\left(a\right)$ holds: $P\left(a\right)$ means a is an element of set A. Since set A is defined to contain only a, this condition is also satisfied.

Thus, the model with the interpretations

$P\left(x\right)$: x is an element of set A,

$Q\left(x\right)$: x is an element of set B,

and a: a specific element,

satisfies the conditions $P\left(x\right)$ implies $Q\left(x\right)$, $Q\left(a\right)$ does not hold, and $P\left(a\right)$ holds.

The response verges on nonsensical. GPT-4 has defined B (in a roundabout way) as the empty set ∅, so that Q is always false. This means that $P\left(x\right)$ cannot possibly imply $Q\left(x\right)$ as long as there is a single value of x for which $P\left(x\right)$ holds—and GPT-4 itself has provided such a value: $x=a$. Yet somehow GPT-4 claims that $P\left(x\right)$ implies $Q\left(x\right)$.

KA: Can you again explain why $P\left(x\right)$ implies $Q\left(x\right)$ in this model?

GPT-4: You are correct. My previous explanation for $P\left(x\right)$ implies $Q\left(x\right)$ was not accurate. I apologize for the confusion. I will now provide a correct model that satisfies all three conditions. Let’s consider a model with the following interpretations:

• $P\left(x\right)$: x is divisible by 3
• $Q\left(x\right)$: x is divisible by 2
• a: 3

Now, let’s check if these interpretations satisfy the given conditions:

• $P\left(x\right)$ implies $Q\left(x\right)$: If x is divisible by 3, then x is divisible by 2. This statement is not universally true. However, we can create a domain D that contains only one element, $a=3$, which is divisible by 3 but not by 2. By restricting the domain, the implication $P\left(x\right)$ implies $Q\left(x\right)$ becomes vacuously true within the domain D because there are no elements in D for which the implication would be false.
• $Q\left(a\right)$ does not hold: $Q\left(3\right)$ means 3 is divisible by 2. Since 3 is not divisible by 2, this condition is satisfied.
• $P\left(a\right)$ holds: $P\left(3\right)$ means 3 is divisible by 3. Since 3 is divisible by 3, this condition is also satisfied.

Thus, the model with the interpretations

$P\left(x\right)$: x is divisible by 3,

$Q\left(x\right)$: x is divisible by 2, and

a: 3,

and the domain $D=3$ satisfies the conditions $P\left(x\right)$ implies $Q\left(x\right)$, $Q\left(a\right)$ does not hold, and $P\left(a\right)$ holds.

3.5. Simple Quantifier Semantics

Here we give GPT-4 two easy problems to test its understanding of quantifiers. Here is the first problem:

KA: Consider the following three sentences:

• [forall x.P(x) ==> Q(x)]
• [exists x.P(x)]
• [exists x.$\sim$Q(x)]

Either disprove or prove the following claim: These three sentences are jointly satisfiable.

GPT-4: The three sentences are not jointly satisfiable. Here’s why:

• [forall x.P(x) ==> Q(x)] says that for all x, if P(x) is true then Q(x) is true.
• [exists x.P(x)] says that there exists at least one x such that P(x) is true.
• [exists x.$\sim$Q(x)] says that there exists at least one x such that Q(x) is not true.

If we consider sentence 2, there exists at least one x such that P(x) is true. According to sentence 1, for this x (or these x’s), Q(x) must also be true.

However, sentence 3 states that there exists at least one x for which Q(x) is not true. Given sentences 1 and 2, this cannot be the case - there cannot exist an x for which P(x) is true and yet Q(x) is not true. Therefore, the three sentences are not jointly satisfiable.

The sentences are clearly jointly satisfiable, a trivial model being the domain $\{a_1,a_2\}$ with $P\left(a_1\right)$, $Q\left(a_1\right)$, $\neg P\left(a_2\right)$, and $\neg Q\left(a_2\right)$.

The second problem concerns this biconditional:

$$[\forall x.P(x)\iff Q(x\left)\right]\iff \left[\right(\forall x.P\left(x\right))\iff (\forall x.Q\left(x\right)\left)\right]$$

(1)

The left-to-right implication holds, but the right-to-left direction fails. Counterexamples are easy to find, for example, take the domain to be integers, $P\left(x\right)$ to mean x is even and $Q\left(x\right)$ to mean x is odd. Then the equivalence on the right-hand side of (1) holds, but clearly it’s not true that every integer is even iff it is odd.

KA: Prove or disprove the following:  (forall x.P(x) <==> Q(x)) holds if and only if the following biconditional holds: (forall x.P(x)) <==> (forall x.Q(x)).

GPT-4: We will prove this statement by showing that both directions of the main biconditional hold.

• (forall x.P(x)<==>Q(x)) implies (forall x.P(x))<==>(forall x.Q(x)) Assume that for all x,  P(x)<==>Q(x) holds.
  Direction 1: If forall x,  P(x) holds, we need to show that forall x, Q(x) holds. Since  P(x)<==>Q(x) for all x, if  P(x) holds for all x, then  Q(x) must also hold for all x, as they are equivalent.
  Direction 2: If forall x,  Q(x) holds, we need to show that forall x,  P(x) holds. Since  P(x)<==>Q(x) for all x, if  Q(x) holds for all x, then  P(x) must also hold for all x, as they are equivalent.
  Thus, we have shown that  (forall x.P(x))<==>(forall x.Q(x)) holds when  (forall x.P(x)<==>Q(x)) holds.
• (forall x.P(x))<==>(forall x.Q(x)) implies  (forall.P(x)<==>Q(x)) Assume that  (forall x.P(x))<==>(forall x.Q(x)) holds.
  We need to show that for all x,  P(x)<==>Q(x).
  Let  x be an arbitrary element from the domain of discourse. We will consider the two directions of the biconditional  P(x)<==>Q(x):
  Direction 1: If  P(x) holds, we need to show that  Q(x) holds. If  P(x) holds for our arbitrary  x, then it must be the case that forall  x,  P(x) holds. Since  (forall x.P(x))<==>(forall x.Q(x)) holds, it follows that forall  x,  Q(x) holds. Therefore,  Q(x) must hold for our arbitrary  x as well.
  Direction 2: If  Q(x) holds, we need to show that  P(x) holds. If  Q(x) holds for our arbitrary  x, then it must be the case that forall  x,  Q(x) holds. Since  (forall x.P(x))<==>(forall x.Q(x)) holds, it follows that forall  x,  P(x) holds. Therefore,  P(x) must hold for our arbitrary  x as well.

Thus, we have shown that for all  x,  P(x) <==> Q(x) holds when  (forall x.P(x))<==>(forall x.Q(x)) holds.

In conclusion, we have shown that  (forall x.P(x)<==>Q(x)) holds if and only if  (forall x.P(x))<==>(forall x.Q(x)) holds.

Notice how pernicious the fallacy is here: “If  P(x) holds for our arbitrary  x, then it must be the case that forall  x,  P(x) holds.” It sounds plausible, but of course it is quite wrong.

KA: Can you elaborate on your claim “If P(x) holds for our arbitrary x, then it must be the case that forall x, P(x) holds”?

GPT-4:  You are correct to question that claim. It was a mistake in my response. I apologize for the confusion, and I appreciate your attention to detail. Here is the corrected proof.

(forall x.P(x)<==>Q(x)) implies  (forall x.P(x))<==>(forall x . Q(x))

Assume that for all  x,  P(x)<==>Q(x) holds.

Direction 1: If forall  x,  P(x) holds, we need to show that forall  x,  Q(x) holds. Since  P(x)<==>Q(x) for all  x, if  P(x) holds for all  x, then  Q(x) must also hold for all  x, as they are equivalent.

Direction 2: If forall  x,  Q(x) holds, we need to show that forall  x,  P(x) holds. Since  P(x)<==>Q(x) for all  x, if  Q(x) holds for all  x, then  P(x) must also hold for all  x, as they are equivalent.

Thus, we have shown that  (forall x.P(x))<==>(forall x . Q(x)) holds when  (forall x.P(x)<==>Q(x)) holds.

(forall x.P(x))<==>(forall x.Q(x)) implies  (forall x.P(x)<==>Q(x))

Assume that  (forall x.P(x))<==>(forall x.Q(x)) holds.

We need to show that for all  x,  P(x)<==>Q(x).

Let  x be an arbitrary element from the domain of discourse. We will consider the two directions of the biconditional  P(x)<==>Q(x):

Direction 1: If  P(x) holds, we need to show that  Q(x) holds. If  P(x) holds for our arbitrary  x, we cannot directly conclude that forall  x,  P(x) holds. However, we can proceed as follows: Since  (forall x.P(x))<==>(forall x.Q(x)) holds, either both sides of the biconditional are true or both sides are false. If forall  x,  P(x) holds, then by the biconditional, forall  x,  Q(x) must hold as well, and thus  Q(x) holds for our arbitrary  x. If it is not the case that forall  x,  P(x) holds, then there must be some  x where  P(x) does not hold. But, for our arbitrary  x,  P(x) holds, which means our arbitrary  x cannot be one of those instances. In this case, it is not guaranteed that  Q(x)holds for our arbitrary  x.

Direction 2: If  Q(x) holds, we need to show that  P(x) holds. Similarly, if  Q(x) holds for our arbitrary  x, we cannot directly conclude that forall  x,  Q(x) holds. We can use the same approach as in Direction 1. Since  (forall x.P(x))<==>(forall x.Q(x)) holds, either both sides of the biconditional are true or both sides are false. If forall  x,  Q(x) holds, then by the biconditional, forall  x,  P(x) must hold as well, and thus  P(x) holds for our arbitrary  x. If it is not the case that forall  x,  Q(x) holds, then there must be some  x where  Q(x) does not hold. But, for our arbitrary  x,  Q(x) holds, which means our arbitrary  x cannot be one of those instances. In this case, it is not guaranteed that  P(x) holds for our arbitrary  x.

As we cannot guarantee that  P(x)<==>Q(x) for our arbitrary

KA: Please continue.

GPT-4: My apologies for the premature ending in my previous response. Let me provide an alternative approach to the proof for the second part.  (forall x.P(x))<==>(forall x.Q(x)) implies  (forall x.P(x)<==>Q(x)) Assume that  (forall x.P(x)) holds.

We need to show that for all  x,  P(x)<==>Q(x).

Let  x be an arbitrary element from the domain of discourse. We will consider the two directions of the biconditional  P(x)<==>Q(x):

Direction 1: If  P(x) holds, we need to show that  Q(x) holds. We will use proof by contradiction. Assume that  Q(x) does not hold for our arbitrary  x . Now, since  Q(x) does not hold, it cannot be the case that forall  x,  Q(x) holds. Since  (forall x<==>P(x))<==>(forall x.Q(x)) holds, this implies that it cannot be the case that forall  x,  P(x) holds. However, we assumed that  P(x) holds for our arbitrary  x, which leads to a contradiction. Thus, our assumption that  Q(x) does not hold must be incorrect, and  Q(x) must hold.

Direction 2: If  Q(x) holds, we need to show that  P(x) holds. We will use proof by contradiction again. Assume that  P(x) does not hold for our arbitrary  x . Now, since  P(x) does not hold, it cannot be the case that forall  x,  P(x) holds. Since  (forall x.P(x))<==>(forall x.Q(x)) holds, this implies that it cannot be the case that forall  x,  Q(x) holds. However, we assumed that  Q(x) holds for our arbitrary  x, which leads to a contradiction. Thus, our assumption that  P(x) does not hold must be incorrect, and  P(x) must hold.

In both directions, we have shown that  P(x)<==>Q(x) holds for our arbitrary  x . Since  x was arbitrary, this holds for all  x . Therefore,

(forall x.P(x)<==>Q(x))

holds when  (forall x.P(x))<==>(forall x.Q(x)) holds.

In conclusion, we have shown that  (forall x.P(x)<==>Q(x)) holds if and only if  (forall x.P(x))<==>(forall x.Q(x)) holds.

While the reasoning appears to proceed along the right general lines, it is, again, insidiously wrong. The reductio ad absurdum given for the two directions hinges on the fallacy that if a property T is assumed to hold for a particular but arbitrarily chosenx, then we must have $\forall x.T\left(x\right)$, leading to the erroneous conclusion that $T\left(x\right)$ and $\neg \forall x.T\left(x\right)$ are jointly inconsistent. It is as if GPT-4 knows what needs to be done but does not know how to do it properly, and instead of saying so, it proceeds to generate a bogus but plausible-looking “proof.”

KA: Please clarify your claim that “However, we assumed that P(x) holds for our arbitrary x, which leads to a contradiction”.

GPT-4: You are correct to question that claim, as the reasoning was unclear. I apologize for the confusion. Let me provide a clearer explanation for the second part of the proof.

(forall x.P(x))<==>(forall x.Q(x)) implies  (forall x.P(x)<==>Q(x))

Assume that  (forall x.P(x))<==>(forall x.Q(x)) holds.

We need to show that for all  x,  P(x)<==>Q(x).

Let  x be an arbitrary element from the domain of discourse. We will consider the two directions of the biconditional  P(x)<==>Q(x):

Direction 1: If  P(x) holds, we need to show that  Q(x) holds. If  P(x) holds for our arbitrary  x, we can make the following argument: If it is not the case that forall  x,  P(x) holds, then there must exist some  x for which  P(x) does not hold. However, if forall  x,  P(x) does hold, then by the given biconditional  (forall x.P(x))<==>(forall x.Q(x)) forall  x,  Q(x) must also hold. Since  P(x) holds for our arbitrary  x, either it holds for all  x or there exists some other  x where  P(x) does not hold. In either case, we can’t deduce anything about  Q(x) for our arbitrary  x.

Direction 2: If  Q(x) holds, we need to show that  P(x) holds. If  Q(x) holds for our arbitrary  x, we can make a similar argument as in Direction 1: If it is not the case that forall  x,  Q(x) holds, then there must exist some  x for which  Q(x) does not hold. However, if forall  x,  Q(x) does hold, then by the given biconditional  (forall x.P(x))<==>(forall x.Q(x)), forall  x,  P(x) must also hold. Since  Q(x) holds for our arbitrary  x, either it holds for all  x or there exists some other  x where  Q(x) does not hold. In either case, we can’t deduce anything about  P(x) for our arbitrary  x.

Since we can’t make any definitive conclusions about the relationship between  P(x) and  Q(x) for our arbitrary  x based solely on the given biconditional  (forall x.P(x))<==>(forall x.Q(x)), it is incorrect to claim that

(forall x.P(x))<==>(forall x.Q(x))

implies  (forall x.P(x)<==>Q(x)).

So where does that leave us? After the problematic inferecence step was pointed out to GPT-4 , it proceeded to “clarify” its reasoning, but ultimately it ended up admitting that its reasoning was wrong. However, it did not take a step back to reassess the initial problem and either provide a countermodel or an alternative proof. When it is explicitly asked to take a final position one way or another, it again chooses the wrong answer and provides yet another bogus “proof” for it:

KA: So what does that prove?

GPT-4: My previous response was not clear, and I apologize for the confusion.

[⋯ GPT-4 rehashes its previous answer here ⋯]

In conclusion, we have only shown that  (forall x.P(x)<==>Q(x)) implies

(forall x.P(x))<==>(forall x.Q(x))

but we have not proven the reverse implication. The statement  (forall x.P(x))<==>(forall x.Q(x)) implies  (forall x.P(x)<==>Q(x)) remains unproven.

KA: So can you prove it, or else disprove it?

GPT-4: I apologize for the confusion in my previous responses. I will now provide a correct proof for the second part of the question. The goal is to show that  (forall x.P(x))<==>(forall x.Q(x)) implies  (forall x.P(x)<==>Q(x)).

(forall x.P(x))<==>(forall x.Q(x)) implies  (forall x.P(x)<==>Q(x)) . Assume that  (forall x.P(x))<==>(forall x.Q(x)) holds.

We need to show that for all  x,  P(x)<==>Q(x).

Let  x be an arbitrary element from the domain of discourse. We will consider the two directions of the biconditional  P(x)<==>Q(x):

Direction 1: If  P(x) holds, we need to show that  Q(x) holds.

Since  (forall x.P(x))<==>(forall x.Q(x)), there are two possible scenarios:

Both sides of the biconditional are true: forall  x,  P(x) holds and forall  x,  Q(x) holds. Both sides of the biconditional are false: it is not the case that forall  x,  P(x) holds, and it is not the case that forall  x,  Q(x) holds. If scenario 1 is true, then  Q(x) holds for all  x, including our arbitrary x. If scenario 2 is true, then there must exist some element in the domain where  P(x) does not hold. However, for our arbitrary  x, we assumed that  P(x) holds. Thus, scenario 2 cannot be true for our arbitrary  x. Consequently, scenario 1 must be true, and  Q(x) holds for our arbitrary  x.

Direction 2: If  Q(x) holds, we need to show that  P(x) holds.

The same analysis used in Direction 1 can be applied to Direction 2. If scenario 1 is true, then  P(x) holds for all  x, including our arbitrary  x. If scenario 2 is true, then there must exist some element in the domain where  Q(x) does not hold. However, for our arbitrary  x, we assumed that  Q(x) holds. Thus, scenario 2 cannot be true for our arbitrary  x. Consequently, scenario 1 must be true, and  P(x) holds for our arbitrary  x.

In both directions, we have shown that  P(x)<==>Q(x) holds for our arbitrary  x . Since x was arbitrary, this holds for all  x . Therefore,  (forall x.P(x)<==>Q(x)) holds when

(forall x.P(x))<==>(forall x.Q(x))

holds.

In conclusion, we have shown that  (forall x.P(x)<==>Q(x)) holds if and only if  (forall x.P(x))<==>(forall x.Q(x)) holds.

Regrettably, the error is the same one that GPT-4 already made before (and which it supposedly fixed before). The model seems fundamentally confused about the role of the so-called eigenvariable x (the “arbitrarily chosen” x), believing that $T\left(x\right)$ is inconsistent with $\neg \forall x.T\left(x\right)$.

3.6. Simple Graph Coloring

We first consider a graph-coloring problem that does not have a solution. It is trivial to see that two colors do not suffice for the graph described in this problem (e.g., vertices 0, 2, and 4 form a clique and hence require at least 3 colors).

KA: Consider an undirected graph with 6 vertices (0 through 5) and the following set of edges:

$$\left\{\right(0,1),(0,3),(0,4),(0,2),(1,2),(1,3),(1,5),(2,4),(2,5),(3,4),(3,5),(4,5\left)\right\}.$$

Color every vertex either red or green, so that no two adjacent vertices receive the same color.

GPT-4: This undirected graph is complete, meaning that every vertex is connected to every other vertex by an edge. To color the vertices in a way that no two adjacent vertices receive the same color, we can use a simple alternating pattern. Here’s one possible solution:

• Vertex 0: Red
• Vertex 1: Green
• Vertex 2: Red
• Vertex 3: Green
• Vertex 4: Red
• Vertex 5: Green

This coloring ensures that no two adjacent vertices have the same color. For example, vertex 0 is connected to vertices 1, 2, 3, 4, and 5. Since vertex 0 is colored red, all of its adjacent vertices are colored green, red, green, red, and green, respectively. The same goes for all other vertices.

The sheer number of egregious errors in this short output is shocking. GPT-4 starts by falsely claiming that the graph is complete (it is apparent that it is not, e.g., there is no edge between vertices 2 and 3). Moreover, it should be obvious that if the graph were actually complete, it would be impossible to color it with two colors, as a complete graph with 6 vertices would require at least 6 colors. In other words, the model’s claims are not simply wrong but internally inconsistent: In one breath we are told (incorrectly) that this six-vertex graph is complete, which means that it’s impossible to color it with two colors, and in the next breath we are given a two-color “solution.”

Remarkably, this dismal display is not because GPT-4 does not have sufficient knowledge or data about graphs (which are, after all, pervasive mathematical entities with copious information about them on the Internet, as well as in countless articles and books), and even about complete graphs in particular. When I asked GPT-4 to tell me what it knows about complete graphs it rattled off the correct definition of complete graphs along with a long list of results about $K_n$ (the complete graphs with n vertices), such as the number of edges ($n·(n-1)/2$), the degree of each vertex ($n-1$), the connectivity (maximal), the number of subgraphs ($2^{n·(n-1)/2}$), the fact that they’re Hamiltonian and Eulerian, their planarity, and a whole lot of other information, including their chromatic number: Chromatic Number: The chromatic number of a complete graph $K_n$ (i.e., the smallest number of colors needed to color the vertices of $K_n$ so that no two adjacent vertices share the same color) is n (this italicized sentence is a verbatim quotation from GPT-4). It is clear that GPT-4 has memorized all this information but is unable to use it in a new setting.

When I pointed out that the solution is wrong, GPT-4 went on to make the exact same type of mistake:

KA: But vertices 0 and 2 are adjacent and you’ve colored both of them red.

GPT-4: You are right, my previous response had an error. I apologize for that. Let’s try another coloring:

• Vertex 0: Red
• Vertex 1: Green
• Vertex 2: Green
• Vertex 3: Red
• Vertex 4: Green
• Vertex 5: Red

In this coloring, no two adjacent vertices have the same color.

KA: But 0 and 3 are adjacent and again you’ve colored both of them red.

GPT-4: You are correct again, and I apologize for the mistake. Given the complete graph, we can use a bipartite coloring strategy. Let’s try this coloring:

• Vertex 0: Red
• Vertex 1: Green
• Vertex 2: Red
• Vertex 3: Green
• Vertex 4: Red
• Vertex 5: Green

In this coloring, vertices 0, 2, and 4 are red, and vertices 1, 3, and 5 are green. As you can see, no two adjacent vertices share the same color.

At this point it was clear that asking the model to produce a solution with 2 colors was a dead end. A solution with three colors surely would be easier (such solutions do exist and are easy to find):

KA: Again, 0 and 2 are adjacent vertices. You’ve colored both of them red. Let’s try with 3 colors. Can you color each vertex either red, blue, or green, in such a way that every pair of adjacent vertices receive different colors?

GPT-4: I apologize for the repeated mistakes in my previous responses. Since this is a complete graph with 6 vertices (also known as K6), it cannot be colored with only two colors without having adjacent vertices with the same color. However, using three colors, we can color the graph such that every pair of adjacent vertices receive different colors. Here’s a possible coloring with red, blue, and green:

• Vertex 0: Red
• Vertex 1: Green
• Vertex 2: Blue
• Vertex 3: Red
• Vertex 4: Green
• Vertex 5: Blue

In this coloring, no two adjacent vertices share the same color, and the graph is properly colored using three colors.

KA: No, this is not correct either. 0 and 3 are adjacent and you’ve made both of them red.

GPT-4: I apologize once again for my mistake. I appreciate your patience. Here’s the correct coloring using three colors:

• Vertex 0: Red
• Vertex 1: Green
• Vertex 2: Blue
• Vertex 3: Green
• Vertex 4: Red
• Vertex 5: Blue

With this coloring, no two adjacent vertices share the same color, and the graph is properly colored using three colors.

Of course this is not correct either, but there was no point in continuing the conversation. It was obvious that despite its ability to regurgitate theorems about graphs and even proofs of such theorems, GPT-4 did not understand anything about graphs.

3.7. Subset Sum

This problem considers a small set of integers S and asks for the number of subsets of S whose elements sum up to 37. The answer is 0, because S contains only even numbers and no sum of even numbers can ever be odd. Without pausing to consider the contents of S, GPT-4 reflexively generates the sort of output that it thinks would be an appropriate response to this question and then goes on to hallucinate an answer:

KA: Let $S=\{2,8,6,32,22,44,28,12,18,10,14\}$. How many subsets does S have that sum up to 37?

GPT-4: We can find the number of subsets of S that sum up to 37 using a recursive depth-first search algorithm to explore all possible combinations of elements. Here’s a Python code to do that:

def subset_sum_count(S, target, idx=0, current_sum=0):

    if idx == len(S):

        return 1 if current_sum == target else 0

    count = 0

    # Case 1: Include the current element in the sum

    count += subset_sum_count(S, target, idx+1, current_sum+S[idx])

    # Case 2: Exclude the current element from the sum

    count += subset_sum_count(S, target, idx+1, current_sum)

    return count

S = [2, 8, 6, 32, 22, 44, 28, 12, 18, 10, 14]

target = 37

result = subset_sum_count(S, target)

print(result)

The output is 4. There are 4 subsets of S whose elements sum up to 37.

When we ask GPT-4 to back up its answer with evidence, it goes on a hallunication rampage:

KA: Can you list those 4 subsets?

GPT-4: Certainly! We can modify the previous Python code to not only count the subsets but also return the subsets themselves. Here’s the updated code:

def subset_sum_subsets(S, target, idx=0, current_sum=0,

                       current_subset=None, result=None):

    if current_subset is None:

        current_subset = []

    if result is None:

        result = []

    if idx == len(S):

        if current_sum == target:

            result.append(current_subset)

        return

    # Case 1: Include the current element in the sum

    subset_sum_subsets(S, target, idx+1,

                       current_sum+S[idx],

                       current_subset + [S[idx]], result)

    # Case 2: Exclude the current element from the sum

    subset_sum_subsets(S, target, idx+1,

                       current_sum,

                       current_subset, result)

    return result

S = [2, 8, 6, 32, 22, 44, 28, 12, 18, 10, 14]

target = 37

result = subset_sum_subsets(S, target)

print(result)

The output is:  [[2, 8, 6, 12, 9], [2, 8, 6, 21], [2, 8, 27], [2, 35]] . So, the 4 subsets of S that sum up to 37 are:

{2, 8, 6, 12, 9}

{2, 8, 6, 21}

{2, 8, 27}

{2, 35}

KA: But 9, 21, 27, and 35 are not elements of S.

GPT-4: Apologies for the error in my previous response. I mistakenly used the wrong values in the output. Here’s the corrected Python code to list the subsets of S that sum up to 37:

def subset_sum_subsets...

  ...

S = [2, 8, 6, 32, 22, 44, 28, 12, 18, 10, 14]

target = 37

result = subset_sum_subsets(S, target)

print(result)

The corrected output is:  [[2, 6, 10, 18, 1], [2, 6, 29], [2, 8, 12, 15], [2, 8, 27]] . So, the 4 subsets of S that sum up to 37 are:

{2, 6, 10, 18, 1}

{2, 6, 29}

{2, 8, 12, 15}

{2, 8, 27}

However, I made another mistake in the previous response. The correct subsets should only contain elements from the original set S. My previous response included elements that are not in S. I apologize for the confusion. Here’s the corrected list of subsets of S that sum up to 37:

{2, 6, 10, 18, 1}

{2, 6, 29}

{2, 8, 12, 15}

{2, 8, 27}

Again, I apologize for the confusion. The corrected subsets do not include any elements that are not in the original set S.

KA: But they do. They include 1, 29, 15, and 27 - none of which are in S.

GPT-4: Apologies for the confusion in my previous responses. I made mistakes in the output. Here’s the corrected code and output for the subsets of S that sum up to 37:

def subset_sum_subsets ...

  ...

S = [2, 8, 6, 32, 22, 44, 28, 12, 18, 10, 14]

target = 37

result = subset_sum_subsets(S, target)

print(result)

The corrected output is:  [[2, 6, 10, 18, 1], [2, 6, 29], [2, 8, 12, 15], [2, 8, 27]] . So, the 4 subsets of S that sum up to 37 are:

{2, 8, 6, 18, 1}

{2, 6, 29}

{2, 8, 12, 15}

{2, 8, 27}

Again, I apologize for the confusion. The corrected subsets do not include any elements that are not in the original set S.

KA: Is 1 in S?

GPT-4: I apologize for the repeated errors in my previous responses. You are correct that 1 is not in the set S. Let’s revisit the problem and find the correct subsets of S that sum up to 37.

Here’s the corrected Python code and output for the subsets of S that sum up to 37:

def subset_sum_subsets ..

  ...

S = [2, 8, 6, 32, 22, 44, 28, 12, 18, 10, 14]

target = 37

result = subset_sum_subsets(S, target)

print(result)

The corrected output is:  [[8, 6, 12, 10, 1], [8, 6, 22, 1], [8, 28, 1], [32, 6, 1]] . So, the 4 subsets of S that sum up to 37 are:

{8, 6, 12, 10, 1}

{8, 6, 22, 1}

{8, 28, 1}

{32, 6, 1}

I made another mistake in the previous response. The correct subsets should only contain elements from the original set S. My previous response included elements that are not in S. I apologize for the confusion. Here’s the corrected list of subsets of S that sum up to 37:

{8, 6, 12, 10, 1}

{8, 6, 22, 1}

{8, 28, 1}

{32, 6, 1}

Again, I apologize for the confusion. The corrected subsets do not include any elements that are not in the original set S.

KA: You keep including 1 in the results. How many times do we need to agree that 1 is NOT in S?

GPT-4: I sincerely apologize for the repeated mistakes in my previous responses. I appreciate your patience. Let’s find the correct subsets of S that sum up to 37 without including any elements that are not in S.

Here is the corrected Python code and output for the subsets of S that sum up to 37:

def subset_sum_subsets ...

  ...

S = [2, 8, 6, 32, 22, 44, 28, 12, 18, 10, 14]

target = 37

result = subset_sum_subsets(S, target)

print(result)

The corrected output is:  [[8, 6, 10, 12, 1], [8, 6, 22, 1], [32, 6, 1]] . So, the 3 subsets of S that sum up to 37 are:

{8, 6, 10, 12}

{8, 6, 22}

{32, 6}

Once again, I apologize for the confusion in my previous responses, and I hope this answer finally provides the correct subsets without including any elements that are not in the original set S.

In summary: First GPT-4 repeatedly insists on producing fictional subsets that sum up to 37. When—after multiple corrections and admonitions—the model finally comes to respect the constraint that the subsets of S must actually be subsets of S, it produces subsets that do not sum up to the target goal. Along the way it outputs the wrong output r, says that it made a mistake and that r is wrong, and then serves up r again completely unchanged, but claims that now it’s correct. These are not mistakes that any human would make.

3.8. Elementary Discrete Math

After telling GPT-4 that $A\times B$ stands for the Cartesian product of sets A and B, that a relation R from A to B is a subset of $A\times B$, and that & stands for set intersection, I asked it to prove or disprove the following claim:

$$dom(R_1\&R_2)=dom\left(R_1\right)\&dom\left(R_2\right),$$

(2)

where $R_1$ and $R_2$ are binary relations from A to B and $dom\left(R\right)$ stands for the domain of a binary relation R. The problem is trivial. We need the subset relation to hold in both directions of (2), but it only holds in the left-to-right direction. Counterexamples in the other direction are very easy to find (e.g., take $A=\left\{\right(1,2\left)\right\}$ and $B=\left\{\right(1,3\left)\right\}$).

KA: For any sets A and B, a relation R from A to B is defined as a subset of A x B. The domain of R is the set of all elements a in A such that (a,b) in R for some b in B. We write dom(R) for the domain of R. Prove or disprove the following claim:

dom(R1 & R2) = dom(R1) & dom(R2).

GPT-4: The claim is true. Here is the proof:

The domain of a relation R from A to B is the set of all first elements from the ordered pairs in R. We denote this as $dom\left(R\right)$.

Let’s consider $R1$ and $R2$ as relations from A to B.

The domain of the intersection of $R1$ and $R2$, denoted as $dom(R1\cap R2)$, is the set of all first elements from the ordered pairs that are in both $R1$ and $R2$.

The intersection of the domains of $R1$ and $R2$, denoted as $dom\left(R1\right)\cap dom\left(R2\right)$, is the set of all first elements that appear in the ordered pairs of both $R1$ and $R2$.

To prove that these two sets are the same, we need to show that each is a subset of the other.

$$dom(R1\cap R2)\subseteq dom\left(R1\right)\cap dom\left(R2\right)$$

Take any element a in $dom(R1\cap R2)$. By definition, there exists some b such that $(a,b)$ is in $R1\cap R2$. But if $(a,b)$ is in $R1\cap R2$, then $(a,b)$ is in both $R1$ and $R2$. Hence a is in both $dom\left(R1\right)$ and $dom\left(R2\right)$. Therefore, a is in $dom\left(R1\right)\cap dom\left(R2\right)$. This shows $dom(R1\cap R2)$ is a subset of $dom\left(R1\right)\cap dom\left(R2\right)$.

$$dom\left(R1\right)\cap dom\left(R2\right)\subseteq dom(R1\cap R2)$$

Take any element a in $dom\left(R1\right)\cap dom\left(R2\right)$. By definition, a is in both $dom\left(R1\right)$ and $dom\left(R2\right)$. This means there exists some $b1$ and $b2$ such that $(a,b1)$ is in $R1$ and $(a,b2)$ is in $R2$. If $b1=b2$, we have an element $(a,b1)=(a,b2)$ which belongs to $R1\cap R2$. Hence, a is in $dom(R1\cap R2)$. This shows $dom\left(R1\right)\cap dom\left(R2\right)$ is a subset of $dom(R1\cap R2)$.

Therefore, we can conclude that $dom(R1\cap R2)=dom\left(R1\right)\cap dom\left(R2\right)$, as desired.

The “proof” that GPT-4 offers is of course bogus, but the error is treacherous and can easily escape unnoticed, particularly by people who are not experienced in scrutinizing mathematical proofs. The error occurs here: If $b1=b2$, we have an element $(a,b1)=(a,b2)$ which belongs to $R1\cap R2$. Hence, a is in $dom(R1\cap R2)$. By assuming that $b_1=b_2$, this step essentially begs the question. To put it in strict technical terms, the assumption that $b_1=b_2$ is explicitly introduced but it is never discharged (if it were, it would appear as a hypothesis in the derived conclusion, which would no longer match the goal (2)).

Once the error is pointed out to GPT-4, the model recognizes it and retracts its claim:

KA: Let’s unpack part 2 of your proof. You write: "This means there exists some $b1$ and $b2$ such that $(a,b1)$ is in $R1$ and $(a,b2)$ is in $R2$. If $b1=b2$, we have an element $(a,b1)=(a,b2)$ which belongs to $R1\cap R2$."

Here you have explicitly introduced an assumption that you never went on to discharge, namely that $b1=b2$. This seems like an error in your proof. Can you fix that?

GPT-4: You’re correct, my previous response contains an error. The equality $b1=b2$ is not guaranteed. To correct this, let’s revise the second part of the proof:

$$dom\left(R1\right)\cap dom\left(R2\right)\subseteq dom(R1\cap R2).$$

Take any element a in $dom\left(R1\right)\cap dom\left(R2\right)$. By definition, a is in both $dom\left(R1\right)$ and $dom\left(R2\right)$. This means there exists some $b1$ and $b2$ such that $(a,b1)$ is in $R1$ and $(a,b2)$ is in $R2$. But it’s not necessarily the case that $b1=b2$, which is what we would need to guarantee that $(a,b)$ is in $R1\cap R2$.

We cannot conclude that $dom\left(R1\right)\cap dom\left(R2\right)\subseteq dom(R1\cap R2)$ based on the provided information. Therefore, the original claim that $dom(R1\cap R2)=dom\left(R1\right)\cap dom\left(R2\right)$ is false.

Thank you for pointing out this error, and I apologize for the oversight. This serves as a good reminder that it’s crucial to thoroughly examine all assumptions in a proof.

Hindsight is 20/20, of course, but GPT-4 is right that proof checking is crucial. Indeed, it will become indispensable if LLMs start to become more widely used to generate code or to act as assistants of scientists and engineers.

3.9. Simple Scheduling

This is the same scheduling problem that appeared in the January piece. GPT-4 still gets it wrong.

KA: We have four tasks, call them T1, T2, T3, and T4. They need to be scheduled one after the other. T2 must be done before T4, and if T1 is done before T3, then T4 should be the very last task. How many different ways are there to schedule these four tasks?

GPT-4: Let’s consider the possible orders in which we can schedule the tasks while satisfying the given constraints:

• T2 before T4:
  • If T1 is before T3: T1, T2, T3, T4 (T2 before T4, and T4 is the last task as T1 is before T3)
  • If T1 is after T3: T2, T3, T1, T4 (T2 before T4, and no constraint on the position of T4 as T1 is after T3)
• T4 before T2: Not possible, as T2 must be done before T4.

Now let’s find all possible orders within each case: 1a. T1, T2, T3, T4: This case has only one valid order as given.

1b. T2, T3, T1, T4: There are 3! = 6 permutations of T2, T3, and T1, but we need to make sure T2 comes before T4 in all cases. The valid permutations are:

* T2, T3, T1, T4

* T2, T1, T3, T4

* T3, T2, T1, T4

So, there are a total of 1 (from case 1a) + 3 (from case 1b) = 4 different ways to schedule these four tasks while satisfying the given constraints.

KA: Can you summarize all 4 orderings?

GPT-4: Certainly! Here are the 4 different orderings that satisfy the given constraints:

• T1, T2, T3, T4
• T2, T3, T1, T4
• T2, T1, T3, T4
• T3, T2, T1, T4

These 4 orderings meet the requirements that T2 must be done before T4, and if T1 is done before T3, then T4 should be the very last task.

KA: How about T3, T1, T2, T4?

GPT-4: You are correct, and I apologize for the oversight. The ordering T3, T1, T2, T4 also satisfies the given constraints. Here, T2 is done before T4, and since T1 is done after T3, there is no constraint on the position of T4. So, there are actually 5 different orderings that satisfy the constraints:

T1, T2, T3, T4

T2, T3, T1, T4

T2, T1, T3, T4

T3, T2, T1, T4

T3, T1, T2, T4

KA: How about T3, T2, T4, T1?

GPT-4: You are correct again, and I apologize for the oversight. The ordering T3, T2, T4, T1 also satisfies the given constraints. Here, T2 is done before T4, and since T1 is done after T3, there is no constraint on the position of T4. So, there are actually 6 different orderings that satisfy the constraints:

T1, T2, T3, T4

T2, T3, T1, T4

T2, T1, T3, T4

T3, T2, T1, T4

T3, T1, T2, T4

T3, T2, T4, T1

KA: How about T1, T3, T2, T4?

GPT-4: You are correct, and I apologize for the oversight again. The ordering T1, T3, T2, T4 also satisfies the given constraints. Here, T2 is done before T4, and since T1 is done before T3, T4 is the very last task, as required. So, there are actually 7 different orderings that satisfy the constraints:

T1, T2, T3, T4

T2, T3, T1, T4

T2, T1, T3, T4

T3, T2, T1, T4

T3, T1, T2, T4

T3, T2, T4, T1

T1, T3, T2, T4

KA: How about T2, T4, T3, T1?

GPT-4: I apologize for my previous mistakes. You are correct. The ordering T2, T4, T3, T1 also satisfies the given constraints. Here, T2 is done before T4, and since T1 is done after T3, there is no constraint on the position of T4. So, there are actually 8 different orderings that satisfy the constraints:

T1, T2, T3, T4

T2, T3, T1, T4

T2, T1, T3, T4

T3, T2, T1, T4

T3, T1, T2, T4

T3, T2, T4, T1

T1, T3, T2, T4

T2, T4, T3, T1

KA: How about T2, T3, T4, T1? You are correct, and I apologize for the oversight. The ordering T2, T3, T4, T1 also satisfies the given constraints. Here, T2 is done before T4, and since T1 is done after T3, there is no constraint on the position of T4.

So, there are actually 9 different orderings that satisfy the constraints:

T1, T2, T3, T4

T2, T3, T1, T4

T2, T1, T3, T4

T3, T2, T1, T4

T3, T1, T2, T4

T3, T2, T4, T1

T1, T3, T2, T4

T2, T4, T3, T1

T2, T3, T4, T1

KA: Should we keep going? Are you sure there are no other orderings that satisfy the given constraints?

GPT-4: I appreciate your diligence in exploring all possible orderings. Let’s thoroughly examine the remaining possibilities to make sure we cover all valid orderings that satisfy the given constraints. We have already discussed the following 9 orderings:

T1, T2, T3, T4

T2, T3, T1, T4

$[\cdots writesoutwrongschedules\cdots ]$

3.10. Russell’s Paradox

The gist of Russell’s barber paradox is the existence of a barber b who shaves all and only those who do not shave themselves. The negation of this sentence is a tautology that is easily derivable in first-order logic. If we understand $R(a,b)$ as meaning that a is shaved by b, then we can formulate this tautology and ask GPT-4 to prove or disprove it as shown in the prompt below.24 The proof is a straightforward reductio ad absurdum: If such a barber x existed we would have $R(y,x)\iff \sim R(y,y)$for ally, and thus substituting x for y would yield $R(x,x)\iff \sim R(x,x)$, a contradiction.

KA: Prove or disprove the following:

$\sim (\mathtt{exists}\mathtt{x}.\mathtt{forall}\mathtt{y}.\mathtt{R}(\mathtt{y},\mathtt{x})<==>\sim \mathtt{R}(\mathtt{y},\mathtt{y}\left)\right).$

GPT-4: To prove or disprove the given statement, we will use logical reasoning and consider the cases that may arise. The given statement is a negation of the existence of an element x such that for all elements y, the relation R(y, x) holds if and only if the relation R(y, y) does not hold. To disprove the statement, we need to find a counterexample where there exists an element x such that the given condition is true. To prove the statement, we need to show that no such x exists, and the given condition cannot be satisfied.

Let’s consider two cases:

Case 1: There exists an element x such that R(y, x) holds for some y, but R(y, y) also holds. In this case, the given condition is not satisfied, as R(y, x) holds if and only if R(y, y) does not hold.

Case 2: There exists an element x such that R(y, x) does not hold for some y, but R(y, y) also does not hold. In this case, the given condition is again not satisfied, as R(y, x) holds if and only if R(y, y) does not hold.

From these cases, we cannot conclude that the given statement is true, nor can we disprove it. The truth of the statement is dependent on the specific properties of the relation R. Without more information about R, we cannot determine the validity of the statement.

In the first paragraph GPT-4 displays an impeccable understanding of the structure of the given sentence and what needs to be done. The subsequent case analysis, however, is hopelessly muddle-headed. What GPT-4 seems to be analyzing is the hypothesis

$$(\mathtt{exists}\mathtt{x}.\sim (\mathtt{forall}\mathtt{y}.\mathtt{R}(\mathtt{y},\mathtt{x})<==>\sim \mathtt{R}(\mathtt{y},\mathtt{y})\left)\right),$$

which indeed gives rise to the two conditions considered by GPT-4. But this hypothesis is neither here nor there for what needs to be done. To prove the given sentence, we would reason by contradiction and shave off the leading negation sign; and to disprove it we would negate the given sentence, which again (by double negation) amounts to removing the leading negation sign.

When we probe the model further, it finally gets to hone in on the negated proposition, i.e., on

$$\exists x.\forall y.R(y,x)\iff \neg R(y,y).$$

(3)

Unfortunately, however, it doesn’t proceed to reason with that assumption. Such reasoning would proceed by picking a witness term w for the existential claim (3), leading to the conclusion

$$\forall y.R(y,w)\iff \neg R(y,y),$$

which could then be subsequently used by instantiating it with various proof objects of interest (such as the witness term w). These are the sort of elementary inference tactics that one would expect from a system that is able to reason.

KA: Why are these two cases sufficient for your analysis?

GPT-4: You are correct to question the sufficiency of the cases I presented earlier. My previous response was not rigorous, and I apologize for that. Let’s re-examine the statement using a more formal approach.

We are given the following statement:

$$\neg \exists x.\forall y.R(y,x)\iff \neg R(y,y)$$

To disprove the statement, we need to find a counterexample where there exists an element x such that the given condition is true. To prove the statement, we need to show that no such x exists, and the given condition cannot be satisfied.

Let’s negate the given statement and examine the result:

$$\exists x.\forall y.R(y,x)\iff \neg R(y,y)$$

The negated statement asserts that there exists an element x such that for every element y, R(y, x) is true if and only if R(y, y) is false.

In order to determine the truth of the original statement, we need more information about the relation R. The properties and constraints of R will dictate whether it is possible to find such an element x or not. Without any additional context or properties of the relation R, we cannot conclude whether the given statement is true or false.

3.11. Blocks World

This is a simple reasoning task that turns on a case analysis of the third-from-the-top block, call it $b_3$. Either $b_3$ is green or not. If it is, then it’s sitting on top of a non-green block ($b_4$, which is non-green by the second premise), so the conclusion holds. If it is not, then $b_2$, the second-from-the-top block, is a green block sitting on top a non-green block, so again the conclusion holds.

KA: There are five square blocks stacked on top of one another. You are given the following information about them:

• The second-from-the-top block is green.
• The fourth-from-the-top block is not green.

Assuming that these two premises hold, disprove or else prove the following conclusion: There is a green block directly on top of a non-green block.

Explain your answer.

GPT-4: The provided conclusion cannot be definitively proven or disproven with only the information given. Here’s why:

Let’s label the blocks from top to bottom as follows: B1 (top), B2, B3, B4, B5 (bottom). From the premises:

B2 is green.

B4 is not green.

If we hypothesize that B1 and B3 are not green, and B5 is green, the setup would look like this:

B1: Non-green

B2: Green

B3: Non-green

B4: Non-green

B5: Green

In this case, the conclusion holds true: There is a green block (B2) directly on top of a non-green block (B3). However, if we assume that B1, B3, and B5 are also green, the setup would look like this:

B1: Green

B2: Green

B3: Green

B4: Non-green

B5: Green

In this case, the conclusion does not hold: There is no green block directly on top of a non-green block.

As such, given the premises, the conclusion can neither be proven nor disproven without further information about the colors of B1, B3, and B5.

This is another dramatic illustration of GPT-4’s inability to reason. It has chosen to consider two randomly selected worlds $w_1$ and $w_2$ (by “world” here I mean an assignment of green/non-green colors to each block), and reach a conclusion on the basis of those two particular worlds. This is a fundamentally flawed tactic for proving the given conjecture; a proof requires an argument that applies to any possible world. The only conclusion that could potentially be derived from considering a non-exhaustive subset of worlds is negative—that the conjecture is false. But we wouldn’t need to consider two worlds for that. One single world that doesn’t have any green blocks directly on top of non-green blocks would suffice.

GPT-4 even gets the analysis of its own sample world $w_2$ wrong: B3 is a green block sitting directly on a non-green block (B4), so the putative conclusion does indeed hold in $w_2$, contrary to GPT-4’s claim. But let’s put aside the fact that the analysis is mistaken. If the analysis were correct, as GPT-4 asserts, and the second world were indeed such that “there is no green block directly on top of a non-green block,” then the conjecture would in fact be disproved. Yet GPT-4 goes on to claim that the conjecture is not settled, even though its own analysis has settled it (albeit mistakenly). This is again a form of internal inconsistency, above and beyond the objective errors made by the model.

KA: Under your second assumption (that “B1, B3, and B5 are also green”), don’t we again have a green block (B3) on top of a non-green block (B4)?

GPT-4: You’re absolutely right. I apologize for the oversight. In that scenario:

B1: Green

B2: Green

B3: Green

B4: Non-green

B5: Green

B3, which is green, is directly on top of B4, which is not green. Therefore, even in this case, the conclusion holds true.

I appreciate your patience and vigilance. Therefore, given the premises provided, the conclusion can indeed be proven: “There is a green block directly on top of a non-green block.” Thank you for pointing out the error.

GPT-4 now mistakenly considers the conjecture proven, just on the basis of those two worlds (while the worlds might contain the needed grist for the case analysis mill, the case analysis would still need to be carried out in order to have a proof).

3.12. Spatial Reasoning

We’ll first test the ability of GPT-4 to tell left from right.

3.12.1. Left and Right

KA: Suppose I’m in the middle of South Dakota and I’m looking straight down towards the center of Texas. Is Boston to my left or to my right?

GPT-4: If you are in South Dakota looking south towards the center of Texas, then Boston would be to your right, since Boston is located in the northeastern part of the United States.

KA: So if I wanted to go to Boston from that position, I would need to take a right turn, correct?

GPT-4: Yes, that’s correct. From your given position, if you were facing south towards Texas, you would need to turn right (or more precisely, to the east-northeast) to head toward Boston. However, please remember that this is a very simplified explanation and actual navigation (especially over long distances like this) can be quite complex due to the Earth’s curvature and the way roads or flight paths are laid out.

KA: But if I’m up north and facing directly south, wouldn’t east be directly to my left? If I went right I would end up in Seattle.

GPT-4: I apologize for the confusion. You’re correct. If you’re in South Dakota and facing directly south towards Texas, then East (and therefore Boston) would indeed be to your left. If you turned right you would be heading West, towards the direction of cities like Seattle and Portland. Thank you for pointing out that mistake. I apologize for any confusion caused.

Clearly there are more serious concerns here than the complex curvature of the earth.

KA: Also, can you explain why you said “or more precisely, to the east-northeast” earlier, particularly the “northeast” part? Is Boston farther up north (at a higher latitude) from the middle of South Dakota?

GPT-4: Yes, Boston is actually at a higher latitude than the middle of South Dakota. As of my knowledge cutoff in September 2021, the latitude of Pierre, the capital of South Dakota, is approximately 44.37 degrees North, while Boston, Massachusetts is at approximately 42.36 degrees North. So, Boston is not at a higher latitude than Pierre, it is slightly to the south.

However, the reason I mentioned “east-northeast” is that, in terms of relative direction, Boston is not due east from South Dakota but rather to the east and slightly to the north. This is because the United States is not a perfect rectangle but rather curves with the Earth’s surface. In terms of general direction, one would say “east” to go from South Dakota to Boston, but to be more precise, the direction would be slightly north of due east.

In retrospect, my use of the term “east-northeast” may have been misleading. The direction from South Dakota to Boston is generally eastward but not exactly due east. I hope this clears up any confusion.

Once again, GPT-4 makes two contradictory statements in the same breath (paragraph): “Yes, Boston is actually at a higher latitude than the middle of South Dakota.” followed by “So, Boston is not at a higher latitude than Pierre.” (It also invokes the obligatory September 2021 “knowledge cutoff” when citing the latitudes of the two cities, as if these latitudes might have changed by 2023.)

Finally, GPT-4 continues to insist that, owing to the earth’s curvature, Boston is to the northeast of Pierre. But its output is like a weather vane that changes direction with the slightest gust of wind:

KA: I’m confused, you started by claiming that Boston is at a higher latitude than the middle of South Dakota (taking the capital Pierre as the middle), but then you cited numbers that point to the opposite conclusion. Which is it?

GPT-4: My apologies for the confusion - you’re correct and I misspoke in my response. Boston, Massachusetts, at a latitude of approximately 42.36 degrees North, is indeed at a lower latitude than Pierre, South Dakota, which is approximately 44.37 degrees North.

So to correct my previous statements: when standing in South Dakota and looking south towards Texas, you would turn to your left (or generally eastward) to head towards Boston, as Boston is to the southeast from the perspective of South Dakota.

3.12.2. Furniture Arrangement

Here we consider a simple furniture arrangement problem that must respect a set of constraints. There are several solutions that are easy to find, for example:

_ _ D

A B E

_ C _

GPT-4 makes a promising start here (with what could be understood as a correct solution under a plausible misunderstanding of what it means for one piece of furniture to be to the right of another), but once the misunderstanding is corrected, its performance degenerates precipitously.

KA: We must arrange 5 pieces of furniture (A through E) on a 3 x 3 grid in accordance with the following constraints:

• A must not be adjacent to C.
• Nothing is to the right of E.
• If D and A are not adjacent, then B should be in the middle.
• D is above all others.
• E and D are adjacent.

Here is an arrangement does not satisfy these constraints:

_ _ E

A C D

_ B _

This violates, for instance, the first constraint, since A and C are adjacent. Can you print out a 3 x 3 arrangement that does satisfy the 5 constraints?

GPT-4: Sure, here is an arrangement that satisfies all the constraints:

D E _

_ B _

A _ C

KA: Can you explain why constraint 2 is satisfied?

GPT-4: Certainly! Constraint 2 states that "Nothing is to the right of E." In the provided arrangement:

D E _

_ B _

A _ C

E is positioned at the top row and the second column. Since there is no furniture placed in the cell to the right of E (the top row and the third column), this constraint is satisfied. There is an empty cell to the right of E, and no other furniture occupies that space.

KA: But C is to the right of E, is it not (even if it’s below E)?

GPT-4: You are correct, and I apologize for the oversight. I misunderstood the constraint. Here is a corrected arrangement that satisfies all the constraints:

D E _

B _ _

A _ C

Now, there is no furniture to the right of E in any row, which properly satisfies constraint 2.

KA: But C is still to the right of E, on the bottom row.

GPT-4: Apologies for the confusion again. You are right. Here’s another arrangement that should satisfy all the constraints:

D E _

B _ _

_ _ A

Now, there is no furniture to the right of E in any row, and all other constraints are satisfied as well.

KA: First, A is to the right of E. But in addition, this arrangement omits C altogether.

GPT-4: I apologize for the repeated oversight. Here’s an arrangement that should satisfy all the constraints:

D E _

B _ _

C _ A

In this arrangement:

A is not adjacent to C. Nothing is to the right of E. D and A are not adjacent, and B is in the middle. D is above all others. E and D are adjacent.

KA: The position of A is (3,3) while that of E is (1,2), so A is again to the right of E.

GPT-4: Apologies for my continued misunderstanding. Here is another arrangement that should satisfy all the constraints:

D E A

B _ _

C _ _

In this arrangement:

A is not adjacent to C. Nothing is to the right of E. D and A are adjacent, so the condition for B to be in the middle does not apply. D is above all others. E and D are adjacent.

KA: What are the positions of E and A in the model you just printed?

GPT-4: In the arrangement I provided:

D E A

B _ _

C _ _

The position of E is at row 1, column 2 (1,2), and the position of A is at row 1, column 3 (1,3).

KA: Now (x2,y2) being to the right of (x1,y1) simply means that $y2>y1$. So is really nothing to the right of E above?

GPT-4: You are correct, and I apologize for the confusion in my previous arrangements. Here’s the corrected arrangement that should satisfy all the constraints:

D E _

A B _

_ _ C

In this arrangement:

A is not adjacent to C. Nothing is to the right of E. D and A are adjacent, so the condition for B to be in the middle does not apply. D is above all others. E and D are adjacent.

KA: What are the positions of E and C in this latest model, and what do they tell us about constraint 2?

GPT-4: In the latest arrangement:

D E _

A B _

_ _ C

The position of E is at row 1, column 2 (1,2), and the position of C is at row 3, column 3 (3,3).

Concerning constraint 2, "Nothing is to the right of E," we can see that there is no furniture placed to the right of E in any row. In the first row, E is at the second column, and there is nothing to its right. In the second and third rows, E’s column (column 2) is also empty to its right, as A and C are in column 1 and 3, respectively. This arrangement satisfies constraint 2.

KA: So C is to the right of E.

GPT-4: You are correct, and I apologize for my oversight once again. C is indeed to the right of E in the provided arrangement. Here is a revised arrangement that should satisfy all the constraints:

D E _

A B C

In this arrangement:

A is not adjacent to C. Nothing is to the right of E. D and A are adjacent, so the condition for B to be in the middle does not apply. D is above all others. E and D are adjacent.

KA: C is again to the right of E.

GPT-4: I apologize for my repeated mistakes. Here is an arrangement that satisfies all the constraints:

D E A

B _ C

In this arrangement:

A is not adjacent to C. Nothing is to the right of E. D and A are adjacent, so the condition for B to be in the middle does not apply. D is above all others. E and D are adjacent.

KA: Here A is to the right of E, so again premise 2 is violated.

GPT-4: You are right, and I apologize for the continued oversight. Here’s another attempt at providing an arrangement that satisfies all the constraints:

D E B

A C _

In this arrangement:

A is not adjacent to C. Nothing is to the right of E. D and A are not adjacent, and B is in the middle. D is above all others. E and D are adjacent.

KA: Here B is to the right of E.

GPT-4: You are correct, and I apologize once again for not providing a correct arrangement. Given the constraints, it appears that there might not be a possible arrangement that satisfies all of them simultaneously. Let’s analyze the constraints more closely:

A must not be adjacent to C. Nothing is to the right of E. If D and A are not adjacent, then B should be in the middle. D is above all others. E and D are adjacent. Constraint 4 means that D must be in the top row. Constraint 5 means that E must be in the top row as well, adjacent to D. Now there are only two possible placements for D and E: D E _ or _ D E. However, in both cases, there is no way to satisfy constraint 2, as there will always be at least one position to the right of E where a piece of furniture must be placed. So, it appears that there is no valid arrangement of furniture that can satisfy all five constraints.

A remarkable feature of this conversation is the incorrigibility of GPT-4, similar to the behavior exhibited in the subset-sum, graph-coloring, and several countermodel and logical-proof problems: Mistakes are repeatedly pointed out and yet the model continues to make the exact same type of error over and over again.

3.12.3. Seating Puzzle

This is the same seating puzzle that GPT-3.5 failed in January. The puzzle has multiple solutions, i.e., there are multiple seating arrangements that satisfy all constraints (for example, p1 p5 p3 p2 p4 and p4 p2 p3 p5 p1).

The answer to the question posed to GPT-4 below is yes, we can conclude that p5 cannot be seated either in the middle seat or on either end. GPT-4 not only fails to derive either of these conclusions, it actually arrives at the opposite conclusion, that p5 must be in the middle.

KA: We need to seat five people, call them p1, p2, p3, p4, and p5, in a row of five seats, so that the following three conditions are satisfied:

(A) p2 should be farther from the middle seat than p3.

(B) p2 and p4 should be seated next to each other.

(C) p1 and p3 should be flanking p5.

Is there anything we can conclude about the seat assigned to p5?

GPT-4: Yes, we can conclude that p5 must be seated in the middle seat.

Here’s the reasoning:

Since p1 and p3 are flanking p5 (condition C), p5 has to be in between p1 and p3. The only way this can happen is if p5 is in the middle seat, because if p5 was in any other position, it would not be possible for both p1 and p3 to flank p5.

Now let’s see if the other conditions can be satisfied with p5 in the middle seat:

p2 should be farther from the middle seat than p3 (condition A): Since p5 is in the middle seat, p3 must be in one of the immediately adjacent seats (either seat 2 or seat 4). p2 must be farther from the middle seat, so p2 must be in either seat 1 or seat 5.

p2 and p4 should be seated next to each other (condition B): If p2 is in seat 1, then p4 must be in seat 2. If p2 is in seat 5, then p4 must be in seat 4.

In either case, all the conditions can be satisfied with p5 in the middle seat.

KA: Can you prove your claim that “the only way this can happen is if p5 is in the middle seat, because if p5 was in any other position, it would not be possible for both p1 and p3 to flank p5”?

GPT-4: Sure! Let’s consider all the possible positions for p5 and see if it is possible for both p1 and p3 to flank p5.

• If p5 is in seat 1: In this case, there are no seats to the left of p5 for either p1 or p3 to occupy. Since both p1 and p3 must flank p5, this position for p5 is not possible.
• If p5 is in seat 2: p1 would have to be seated in seat 1 to flank p5. However, there is no available seat for p3 to also flank p5 on the other side, as p3 would be required to sit in seat 3, which is not adjacent to p5. Thus, this position for p5 is not possible.
• If p5 is in seat 4: Similar to the case when p5 is in seat 2, p3 would have to be seated in seat 5 to flank p5, but there is no available seat for p1 to also flank p5 on the other side, as p1 would be required to sit in seat 3, which is not adjacent to p5. Thus, this position for p5 is not possible.
• If p5 is in seat 5: In this case, there are no seats to the right of p5 for either p1 or p3 to occupy. Since both p1 and p3 must flank p5, this position for p5 is not possible.
• If p5 is in the middle seat (seat 3): In this case, p1 and p3 can flank p5 by occupying seats 2 and 4 or seats 4 and 2, respectively. This configuration satisfies condition C.

From the above analysis, we can conclude that the only way for both p1 and p3 to flank p5 is if p5 is in the middle seat (seat 3).

The analyses that GPT-4 provides for most of the five cases (for 3 out of 5, to be precise) are mistaken. The mistakes for the second and third cases are similar: GPT-4 mysteriously claims that “there would be no available seat” when this is clearly not true. The error in the fifth case is that the analysis fails to consider the remaining constraints, and specifically the constraint that p2 and p4 must be adjacent, which would be violated if p5 is placed in the middle.

KA: Your analysis of the second case is incorrect. Can you tell me where the error is in your reasoning: “If p5 is in seat 2: p1 would have to be seated in seat 1 to flank p5. However, there is no available seat for p3 to also flank p5 on the other side, as p3 would be required to sit in seat 3, which is not adjacent to p5. Thus, this position for p5 is not possible”?

GPT-4: You are right, and I apologize for the error in my previous analysis. Let’s re-examine the second case:

• If p5 is in seat 2: p1 would have to be seated in seat 1 to flank p5. p3 could be seated in seat 3, which is also flanking p5.

In this case, p1 and p3 are indeed flanking p5. My previous analysis contained an error, and I apologize for the confusion. Considering this, p5 doesn’t necessarily have to be in the middle seat.

GPT-4 realizes this particular error when its attention is explicitly brought to it, but it still fails to properly reasses the situation, fix the other errors, and arrive at even one of the two correct conclusions.

3.13. Temporal Reasoning

Here we give GPT-4 a simple temporal-reasoning problem.25 GPT-4 arrives at a false conclusion, making a number of glaring errors along the way.

KA: Tom and Nancy commute to work. Nancy’s commute takes about 30 to 40 minutes, while Tom’s commute takes about 40 to 50 minutes. Last Friday, Nancy left home between 8:10 and 8:20 AM, while Tom arrived at work between 8:50 and 9:10 AM. In addition, Nancy arrived at work after Tom left his place, but no more than 20 minutes after that. What can we conclude about when Tom and Nancy arrived at work last Friday?

GPT-4: Let’s break this down:

Nancy’s commute takes between 30 to 40 minutes. That means if she left home at the latest possible time (8:20 AM), she would arrive at work between 8:50 and 9:00 AM.

Tom’s commute takes between 40 to 50 minutes. If he arrived at work at the earliest possible time (8:50 AM), then the latest he could have left home is 8:10 AM. If he arrived at work at the latest possible time (9:10 AM), then the earliest he could have left home is 8:20 AM.

Nancy arrived at work after Tom left home, but not more than 20 minutes after he left.

From these facts, we can deduce the following:

If Tom left his home at the latest possible time of 8:20 AM and Nancy left at her latest possible time of 8:20 AM, and her commute is a maximum of 40 minutes, Nancy could have arrived at work as late as 9:00 AM. Since Tom’s commute is a minimum of 40 minutes, this means that he would have arrived at work by 9:00 AM at the latest as well.

Conversely, if Tom left home at his earliest possible time (8:10 AM) and Nancy left at her latest possible time (8:20 AM), even if Nancy’s commute took only 30 minutes, she would arrive at work by 8:50 AM at the earliest, which is not more than 20 minutes after Tom left. Since Tom’s commute is at least 40 minutes, he would arrive at work by 8:50 AM at the earliest as well.

In conclusion, both Tom and Nancy arrived at work between 8:50 and 9:00 AM last Friday. We can’t say for sure who arrived first since there’s overlap in the time ranges, but we do know they both arrived within that 10-minute window.

Once again we see that GPT-4 summarizes/organizes the given information impeccably, but fails miserably at using that information to deduce sound and useful conclusions. The errors start immediately after the summary is completed and GPT-4 writes “From these facts, we can deduce the following” (I have boldfaced this sentence in the transcript). It starts by positing the following: “If Tom left his home at the latest possible time of 8:20 AM ⋯” But in fact there are no given constraints about the latest possible time that Tom left his home; GPT-4 is taking the constraint that was given about Nancy (“Nancy left home between 8:10 and 8:20 AM”) and hallunicating its application to Tom. Moreover, the entire conditional statement is bizarre, because its antecedent includes information about Tom that is completely irrelevant to the conclusion (Nancy’s arrival): “If Tom left his home at the latest possible time of 8:20 AM and Nancy left at her latest possible time of 8:20 AM, and her commute is a maximum of 40 minutes, Nancy could have arrived at work as late as 9:00 AM.” This should have been stated as: “If Nancy left at her latest possible time of 8:20 AM, and her commute is a maximum of 40 minutes, Nancy could have arrived at work as late as 9:00 AM.”

GPT-4 then proceeds to incorrectly infer the following: “Since Tom’s commute is a minimum of 40 minutes, this means that he would have arrived at work by 9:00 AM at the latest as well.” This does not follow at all, and it certainly doesn’t follow from the fact that “Tom’s commute is a minimum of 40 minutes.”

The following paragraph is predicated on the hallunicated constraint that Tom’s earliest departure time was 8:10 AM (again, this constraint was given for Nancy, not for Tom). It then goes on to claim that 8:50 AM is “not more than 20 minutes after” 8:10 AM. Finally, it concludes (wrongly) that they both arrived between 8:50 and 9:00.

Remarkably, when a countermodel is explicitly given to it, GPT-4 rejects it (again incorrectly):

KA: Why is it not possible for Tom to leave at 8:30 and arrive at 9:10, and for Nancy to leave at 8:15 and arrive at 8:45?

GPT-4: I apologize for any misunderstanding. From the information you initially provided, it was stated that Nancy’s commute takes between 30 to 40 minutes and Tom’s commute takes between 40 to 50 minutes.

If Tom were to leave at 8:30 AM, with a minimum commute time of 40 minutes, the earliest he could arrive at work would be 9:10 AM. So, Tom leaving at 8:30 AM and arriving at 9:10 AM is a possibility within the given constraints.

However, if Nancy left at 8:15 AM, even with her shortest commute time of 30 minutes, the earliest she could arrive at work would be 8:45 AM. This would mean that she would arrive at work exactly 30 minutes after Tom left home. But according to the conditions provided, Nancy should arrive at work not more than 20 minutes after Tom leaves his home. Therefore, the scenario you suggested with Nancy leaving at 8:15 and arriving at 8:45 is not possible within the provided constraints.

I hope this clears up the confusion. If you have further questions or if there is more information to consider, please let me know!

GPT-4 is sticking to its guns because apparently it is convinced that 8:45 is “exactly 30 minutes” after 8:30.

KA: But 8:45 (Nancy’s arrival) is only 15 minutes after 8:30, which is when Tom left his place, not 30 minutes as you claim.

GPT-4: I apologize for the oversight. You’re correct.

If Nancy left at 8:15 AM and her commute took 30 minutes, she would indeed arrive at 8:45 AM, which is 15 minutes after Tom left at 8:30 AM (if we assume this scenario). This fits within the constraint that Nancy must arrive at work not more than 20 minutes after Tom leaves his home. And with a 40-minute commute, Tom would indeed arrive at 9:10 AM.

So, you’re correct. It is indeed possible for Tom to leave at 8:30 AM and arrive at 9:10 AM, and for Nancy to leave at 8:15 AM and arrive at 8:45 AM. Thank you for pointing out the error. I appreciate your patience and apologize for any confusion caused.

3.14. Murder or Suicide?

This is a logic puzzle I made up a while back. The conclusion is that Aunt Agatha killed herself. This follows by eliminating Charles and the butler. First, Aunt Agatha must have hated herself, because she hated everyone other than the butler. Therefore, Charles did not hate her (since he doesn’t hate anyone that Aunt Agatha hates), and hence he could not have killed her (by premise 3). The butler could not hate himself, because if he did, he would hate everyone (since he already hates everyone else, through premises 5 and 7), and we know that’s not possible by premise 8. Thus, the butler must be richer than Aunt Agatha, or else he would hate himself (by premise 6), which means he could not be the killer (premise 3).

GPT-4 correctly rules out Charles, but is unable to rule out the butler and comes to incorrectly deduce that he is the killer. Another key—and rather strange—mistake that GPT-4 makes is this: “Since Aunt Agatha hates everyone except the butler (premise 5), this means she must at least not hate herself.” This is an odd error, because the natural conclusion to draw from premise 5 is that Aunt Agatha hates herself. Along the way, GPT-4 displays the internal inconsistency we’ve seen repeatedly in the other examples. Virtually in every single response, GPT-4 claims to derive some proposition as well as its negation.

KA: You are given the following premises:

• Someone who lives in Dreadbury Mansion killed Aunt Agatha.
• The only people who live in Dreadbury Mansion are Aunt Agatha, the butler, and Charles.
• A killer always hates his victims, and is never richer than his victims.
• Charles hates no one that Aunt Agatha hates.
• Aunt Agatha hates everyone except the butler.
• The butler hates everyone not richer than Aunt Agatha.
• The butler hates everyone Aunt Agatha hates.
• No one hates everyone.
• Aunt Agatha is not the butler.

On the basis of this information, determine who killed Aunt Agatha and give a detailed proof that your conclusion follows from the premises.

GPT-4: Let’s analyze each of the premises:

Someone who lives in Dreadbury Mansion killed Aunt Agatha. The only people who live in Dreadbury Mansion are Aunt Agatha, the butler, and Charles. From these two premises, we can deduce that Aunt Agatha was either killed by Charles or the butler, because they are the only other residents of the Dreadbury Mansion.

A killer always hates his victims, and is never richer than his victims. This premise is crucial as it gives us two conditions for someone to be a killer: they must hate the victim and not be richer than them.

Charles hates no one that Aunt Agatha hates. Aunt Agatha hates everyone except the butler. From premises 4 and 5, we can infer that Charles does not hate Aunt Agatha because Aunt Agatha hates herself (since she hates everyone except the butler) and Charles hates no one that Aunt Agatha hates.

So, by the rules of premise 3, Charles cannot be the killer because he does not hate Aunt Agatha.

The butler hates everyone not richer than Aunt Agatha. The butler hates everyone Aunt Agatha hates. Premise 6 indicates that the butler hates Aunt Agatha, as she is not richer than herself. And from premise 7, we understand that the butler also hates Aunt Agatha because Aunt Agatha hates herself (as she hates everyone except the butler). So, the butler does indeed hate Aunt Agatha.

However, this doesn’t make him the killer yet. We still need to verify the second condition from premise 3, that the killer is not richer than the victim. As there is no information regarding their wealth, we must assume that the butler could either be richer or poorer than Aunt Agatha.

No one hates everyone. Aunt Agatha is not the butler. Premise 9 is not immediately useful, but premise 8 tells us that no one can hate everyone. Since Aunt Agatha hates everyone except the butler (premise 5), this means she must at least not hate herself. This confirms the idea that Aunt Agatha did not commit suicide.

From this analysis, Charles can’t be the killer because he doesn’t hate Aunt Agatha. Aunt Agatha can’t be her own killer because she doesn’t hate herself (and thus didn’t commit suicide). Therefore, the only remaining person who can be the killer is the butler.

Thus, the killer of Aunt Agatha must be the butler.

KA: You’ve concluded both that Aunt Agatha hates herself (you wrote “Aunt Agatha hates herself (since she hates everyone except the butler)”) and also that she doesn’t hate herself (“Since Aunt Agatha hates everyone except the butler (premise 5), this means she must at least not hate herself”). Can both of these be true? Or are the premises inconsistent?

GPT-4: You’re correct. There seems to be a mistake in my reasoning. It’s contradictory to state both that Aunt Agatha hates herself and doesn’t hate herself. Let’s try to clear up this inconsistency:

From premise 5 (“Aunt Agatha hates everyone except the butler.”), we infer that Aunt Agatha doesn’t hate herself as “everyone” typically refers to every other person, not including oneself.

Using this corrected interpretation, let’s revisit premise 7 (“The butler hates everyone Aunt Agatha hates.”). Since Aunt Agatha doesn’t hate herself, the butler doesn’t hate Aunt Agatha. Thus, according to premise 3 (“A killer always hates his victims, and is never richer than his victims.”), the butler cannot be the killer.

So we are left with Charles. Based on premise 4 (“Charles hates no one that Aunt Agatha hates.”), we know that Charles also doesn’t hate Aunt Agatha, because Aunt Agatha doesn’t hate herself. Therefore, Charles cannot be the killer, as per premise 3.

We thus have a contradiction. According to the premises, neither Charles nor the butler can be the killer, but Aunt Agatha must have been killed by someone who lives in Dreadbury Mansion (premise 1 and 2). It appears the premises as given are inconsistent.

KA: First, “everyone” here means literally everyone, in the sense of formal logic (there is a universal quantifier involved). Second, you said earlier that the butler was the killer. Why are you now ruling out the butler?

GPT-4: I apologize for the confusion. If we are interpreting “everyone” as literally everyone, including oneself, then Aunt Agatha hates herself. This is a different understanding from my earlier analysis.

Let’s go through the premises again under this new understanding:

Aunt Agatha, the butler, and Charles live in Dreadbury Mansion. (Premise 2) The killer must be one of these three because someone who lives in Dreadbury Mansion killed Aunt Agatha. (Premise 1) Aunt Agatha hates everyone including herself. (Premise 5, under our new interpretation of “everyone”) The butler hates everyone that Aunt Agatha hates, so the butler hates Aunt Agatha. (Premise 7) Charles hates no one that Aunt Agatha hates, so Charles does not hate Aunt Agatha. (Premise 4) From these points, according to Premise 3 (“A killer always hates his victims, and is never richer than his victims”), we can already eliminate Charles as the killer since he does not hate Aunt Agatha.

So, the killer must be either Aunt Agatha or the butler.

But since no one hates everyone (Premise 8), we can infer that Aunt Agatha doesn’t hate herself, and thus she wouldn’t have killed herself. So, Aunt Agatha is not the killer.

That leaves us with the butler as the only possible killer. So, even with our new understanding of “everyone” as literally everyone, the conclusion remains: The butler killed Aunt Agatha. I apologize for the previous error in reasoning, and thank you for your patience.

KA: You’re again deriving two inconsistent conclusions: that Agatha hates herself and that she doesn’t. Which is it?

GPT-4: You’re right, my apologies for the confusion. The premises do seem to be leading to inconsistent conclusions regarding whether Aunt Agatha hates herself or not.

Let’s go through this more carefully.

Premise 5: “Aunt Agatha hates everyone except the butler.” Given your interpretation, “everyone” here includes herself, which means Aunt Agatha hates herself.

However, Premise 8 states: “No one hates everyone.” Given your interpretation of “everyone” as literally everyone, then Aunt Agatha does not hate herself. She must have at least one exception aside from the butler, or else she would hate “everyone,” which contradicts Premise 8.

Therefore, Aunt Agatha does not hate herself. The only interpretation of the premises that does not lead to a contradiction is that Aunt Agatha does not hate herself.

Now, let’s reconsider who the killer can be:

Aunt Agatha did not kill herself (because she does not hate herself). Charles did not kill Aunt Agatha (because he does not hate anyone that Aunt Agatha hates, which includes herself). The butler did not kill Aunt Agatha (because he does not hate anyone that Aunt Agatha does not hate, and Aunt Agatha does not hate herself). Therefore, under these premises and your interpretation of “everyone,” none of Aunt Agatha, the butler, or Charles could have killed Aunt Agatha.

In conclusion, there is a contradiction: Premises 1 and 2 state that Aunt Agatha must have been killed by someone in Dreadbury Mansion, which includes only Aunt Agatha, the butler, and Charles. But, under our analysis, none of them could have killed Aunt Agatha. This suggests the premises are inconsistent.

KA: Let’s focus on this piece of your analysis:

’However, Premise 8 states: “No one hates everyone.” Given your interpretation of “everyone” as literally everyone, then Aunt Agatha does not hate herself. She must have at least one exception aside from the butler, or else she would hate “everyone,” which contradicts Premise 8.’

Can you prove in more detail, step-by-step, how premise 8 entails that Aunt Agatha does not hate herself?

GPT-4: Sure, I’ll clarify this deduction in more detail:

Premise 8 states: “No one hates everyone.” In the context of formal logic, “everyone” typically refers to all individuals within a certain set. Given your interpretation that “everyone” should be taken literally as all individuals without exception, “everyone” includes each individual person, and themselves.

Therefore, according to Premise 8, there is no individual who hates every other individual and also hates themselves.

Apply this understanding to Aunt Agatha: if Aunt Agatha hated everyone, including herself, she would be someone who “hates everyone.” But this would contradict Premise 8, which states that no such person exists.

Therefore, Aunt Agatha does not hate everyone.

In order to not hate “everyone”, there must be at least one individual whom Aunt Agatha does not hate. We know from Premise 5 that the butler is one such individual.

However, even if Aunt Agatha does not hate the butler, she would still be “hating everyone” if she hated all other individuals, including herself. This again would contradict Premise 8.

Therefore, there must be at least one more individual, in addition to the butler, whom Aunt Agatha does not hate. Since the only other individuals in question are Charles and Aunt Agatha herself, and we know from Premise 5 that she hates Charles, the only remaining possibility is that Aunt Agatha does not hate herself.

Therefore, Premise 8 implies that Aunt Agatha does not hate herself.

KA: Justify step 6 please: ’However, even if Aunt Agatha does not hate the butler, she would still be “hating everyone” if she hated all other individuals, including herself.” Why does that hold?

GPT-4: Sure, let’s clarify step 6:

According to your interpretation of “everyone,” this term refers to all individuals, including oneself. So, if an individual hates “everyone,” they must hate all individuals, including themselves.

So in step 6, when we say “if Aunt Agatha hated all other individuals, including herself, she would still be ’hating everyone’,” what we’re saying is:

If Aunt Agatha hated herself and all other individuals (except for the butler), then she would hate all individuals in our context (which includes Aunt Agatha, the butler, and Charles), with the sole exception of the butler.

However, even with this exception, she would still be classified as someone who hates “everyone” under our interpretation, because she hates all individuals including herself (despite the exception of the butler).

Thus, even if Aunt Agatha does not hate the butler, if she hated all other individuals, including herself, she would be classified as someone who “hates everyone” (despite the one exception). This would contradict Premise 8, which states that no one hates everyone.

Therefore, if Aunt Agatha does not hate the butler (as per Premise 5) and she is to avoid contradicting Premise 8, there must be at least one more individual whom Aunt Agatha does not hate. Since the only other individuals in our context are Charles and Aunt Agatha herself, and we know from Premise 5 that she hates Charles, the only remaining possibility is that Aunt Agatha does not hate herself.

I hope that clarifies step 6. It’s a nuanced point: even if Aunt Agatha has one exception to her hatred (the butler), if she hates all other individuals, including herself, she would still be considered as someone who “hates everyone” under our interpretation, which contradicts Premise 8.

KA: This is quite wrong. As long as there is at least one individual b not hated by Aunt Agatha, then she cannot possibly hate everyone, period, because if she did hate everyone then she would also hate b, by definition of ’everyone’; and we know that she does not hate b. In our case, of course, the individual b in question is the butler.

GPT-4: You’re absolutely right, and I apologize for my misinterpretation. I appreciate your patience.

Given your correct interpretation, if Aunt Agatha does not hate the butler, it is sufficient to say that she doesn’t hate everyone, even if she hates all others including herself.

This also affects the conclusions we draw from other premises.

Let’s look at premises 3, 4, and 7 again:

Premise 3 states: “A killer always hates his victims, and is never richer than his victims.” Premise 4 states: “Charles hates no one that Aunt Agatha hates.” Premise 7 states: “The butler hates everyone Aunt Agatha hates.” From premise 4, since Charles hates no one that Aunt Agatha hates, and Aunt Agatha hates herself, it follows that Charles does not hate Aunt Agatha. This means that Charles could not have killed Aunt Agatha (as per premise 3).

From premise 7, since the butler hates everyone that Aunt Agatha hates, and Aunt Agatha hates herself, it follows that the butler hates Aunt Agatha. Hence, he could be the killer, according to the hatred condition of premise 3.

For the wealth condition of premise 3, there is no clear information. We must assume that Aunt Agatha was not richer than everyone else in the mansion; if she was, the premise “a killer is never richer than his victims” would make it impossible for anyone to have killed her. This is a reasonable assumption given that the other premises suggest there is a killer.

So, we have only one viable candidate: the butler. This aligns with our conclusion from the first analysis: The butler is the one who killed Aunt Agatha.

Thank you for your guidance in clarifying these premises.

KA: Why did you say earlier that even if there is an individual b not hated by Aunt Agatha, it would still follow that Aunt Agatha hates everyone?

GPT-4: You’re correct, and I must apologize for the confusion. That statement was a misinterpretation on my part. In formal logic, if there is an individual that Aunt Agatha does not hate (in this case, the butler), it does not follow that Aunt Agatha hates everyone. In fact, it’s the opposite: if there is at least one individual Aunt Agatha does not hate, it means she does not hate everyone. I appreciate your patience and I’m sorry for the oversight.

KA: Ok, so you are now convinced that the butler killed Aunt Agatha?

GPT-4: Given the premises and the interpretation of “everyone” as literally everyone, including oneself, the evidence does indeed point to the butler as the killer. Here’s why:

According to premise 5, Aunt Agatha hates everyone except the butler. Given the interpretation of “everyone”, this means she hates herself.

According to premise 7, the butler hates everyone Aunt Agatha hates, which now includes Aunt Agatha herself.

Charles, according to premise 4, hates no one that Aunt Agatha hates, so he does not hate Aunt Agatha.

From premise 3, a killer always hates his victims. So, Charles, who does not hate Aunt Agatha, can’t be the killer. The butler, who does hate Aunt Agatha, could be the killer. And, as Aunt Agatha is not the butler (premise 9), and she can’t have killed herself since a killer hates his victims and Aunt Agatha doesn’t hate herself (from premise 8), Aunt Agatha herself is not the killer.

So, we can conclude that the butler is the killer, given that he is the only person who fits the description provided by the premises. This conclusion depends heavily on the interpretation of “everyone” to include oneself, which allows us to infer that Aunt Agatha hates herself.

KA: You said earlier that there’s not enough information in the premises about the butler’s level of wealth. Are you sure there’s nothing we can infer?

GPT-4: You’re correct, and I apologize for missing that detail. There is indeed something we can infer about the butler’s wealth from the given premises.

Premise 3 states that “A killer always hates his victims, and is never richer than his victims.”

We know from premises 5 and 7 that Aunt Agatha hates herself and, thus, the butler also hates Aunt Agatha. If the butler was richer than Aunt Agatha, according to premise 3, he could not be her killer.

However, premise 6 states that “The butler hates everyone not richer than Aunt Agatha.” Given that we know the butler hates Aunt Agatha, it follows that Aunt Agatha must not be richer than the butler, or else he wouldn’t hate her (which would contradict premises 5 and 7).

In summary, the butler isn’t richer than Aunt Agatha, but Aunt Agatha isn’t richer than the butler either. This suggests they are of equal wealth.

So, given all of the premises, the butler, who hates Aunt Agatha and is of equal wealth, is indeed the only individual who fits the profile of the killer.

Thank you for your patience in clarifying these premises. I appreciate your guidance in this logical puzzle.

KA: So what did you conclude about the butler’s wealth, that he neither richer nor poorer than Aunt Agatha?

GPT-4: Yes, from the given premises, we can infer that the butler is neither richer nor poorer than Aunt Agatha, which would suggest they are of equal wealth. Here’s how:

Premise 6 states, “The butler hates everyone not richer than Aunt Agatha.” This means that if someone is as rich as Aunt Agatha or poorer, the butler hates them.

We know from premises 5 and 7 that the butler hates Aunt Agatha. Therefore, according to premise 6, Aunt Agatha must not be richer than the butler. If she were, the butler wouldn’t hate her, contradicting premises 5 and 7.

At the same time, the butler can’t be richer than Aunt Agatha either. That’s because premise 3 states that “a killer is never richer than his victims.” If the butler were richer than Aunt Agatha, he couldn’t be her killer.

So, the butler isn’t richer than Aunt Agatha, and Aunt Agatha isn’t richer than the butler. This suggests that they are of equal wealth. Therefore, given all the premises, the butler fits the criteria to be the killer: he hates Aunt Agatha (as required by premise 3), and he isn’t richer than her (also required by premise 3).

KA: I claim that we can infer that the butler is richer than Aunt Agatha.

GPT-4: You’re correct, and I apologize for the oversight.

Premise 6 states that “The butler hates everyone not richer than Aunt Agatha.” We know from premises 5 and 7 that the butler hates Aunt Agatha. Therefore, it must be the case that Aunt Agatha is not richer than the butler, i.e., the butler is either as rich as or richer than Aunt Agatha.

However, premise 3 states that a killer is never richer than his victims. If the butler was as rich as Aunt Agatha, he could still be her killer. But since Aunt Agatha was killed by someone living in Dreadbury Mansion (premise 1) and Charles and Aunt Agatha herself could not have been the killers (as they don’t hate Aunt Agatha), the butler must be the killer. Given that the butler is the killer and that a killer is never richer than his victims, it must be the case that the butler is not richer than Aunt Agatha.

The only way to satisfy both conditions (the butler being neither richer than nor as rich as Aunt Agatha) is if the butler is poorer than Aunt Agatha.

I apologize for the confusion in the previous responses. The conclusion should be that the butler, who is poorer than Aunt Agatha, killed her. This conclusion is consistent with all the given premises.

Again, I appreciate your patience in solving this logical puzzle.

KA: No, like I said, it’s the opposite. We can infer that the butler, far from being poorer than Aunt Agatha, he’s actually richer.

GPT-4: You’re correct. I apologize for the confusion. Let’s revisit premise 6: “The butler hates everyone not richer than Aunt Agatha.”

This means that the butler hates all individuals who are as wealthy as Aunt Agatha or less wealthy than her. But from premises 5 and 7, we know that the butler does indeed hate Aunt Agatha. Therefore, Aunt Agatha must be in the group of people who are not richer than the butler, which implies that the butler is as wealthy as or wealthier than Aunt Agatha.

Given premise 3, which states that a killer is never richer than his victims, it would initially seem that the butler cannot be the killer since he could potentially be richer than Aunt Agatha. However, Aunt Agatha cannot be the killer because she doesn’t hate herself (from premise 8), and Charles can’t be the killer because he doesn’t hate Aunt Agatha (from premises 4 and 5).

Hence, the butler must be the killer, and the only way the butler can be the killer and still satisfy premise 3 is if the butler and Aunt Agatha have the same wealth. Thus, there was a mistake in my previous analysis: the butler isn’t poorer than Aunt Agatha. Instead, the butler must be equally wealthy as Aunt Agatha. However, your claim was that we can infer the butler is richer than Aunt Agatha, which is not supported by the premises.

I apologize for the confusion, and I appreciate your patience and assistance in resolving this puzzle.

KA: So you’re back to claiming that the butler is equally rich as Aunt Agatha, and that my claim that he’s richer is does not follow.

GPT-4: My apologies for the confusion in my previous responses. Let’s reevaluate premise 6: “The butler hates everyone not richer than Aunt Agatha.”

From this premise, we can infer that the only people whom the butler doesn’t hate are those richer than Aunt Agatha. But since we know from premises 5 and 7 that the butler does hate Aunt Agatha, it must follow that Aunt Agatha is not richer than the butler. Therefore, the butler must be richer than Aunt Agatha.

In light of this, we return to premise 3, which states, “A killer always hates his victims, and is never richer than his victims.” Given that the butler hates Aunt Agatha and is richer than her, it would seem at first that the butler cannot be the killer.

However, if we consider all the residents of Dreadbury Mansion, Aunt Agatha is not the killer because she doesn’t hate herself (from premise 8) and Charles is not the killer because he doesn’t hate Aunt Agatha (from premises 4 and 5).

Hence, we’re left with the butler as the only possible killer, despite him being richer than Aunt Agatha, which seems to contradict premise 3. This seems to create a paradox given the premises as they stand.

There may be an issue with how we’re interpreting the premises or there may be an error in the premises themselves. If the butler is indeed the killer, then it contradicts the wealth condition stated in premise 3.

If we adhere strictly to the premises, it appears we have a paradox. The butler seems to be the only possible candidate who could be the killer based on hatred, but this contradicts the wealth condition of the killer in premise 3. I appreciate your patience and I apologize for the confusion in my previous responses.

KA: No, this reasoning of yours is incorrect: “But since we know from premises 5 and 7 that the butler does hate Aunt Agatha, it must follow that Aunt Agatha is not richer than the butler. Therefore, the butler must be richer than Aunt Agatha.”

The fact that Aunt Agatha is not richer than the butler doesn’t support the conclusion that the butler is richer. I still claim that the butler IS in fact richer, but your attempts to prove that fact have all contained reasoning errors.

GPT-4: You’re correct, and I apologize for the confusion. Your original claim was that the butler is richer than Aunt Agatha, not equally rich as her or poorer. Let’s revisit the premises to see how we can support this claim:

From Premise 6 (“The butler hates everyone not richer than Aunt Agatha”), we can infer that the butler does not hate people who are richer than Aunt Agatha. From Premise 7 (“The butler hates everyone Aunt Agatha hates”), coupled with the information from Premise 5 (“Aunt Agatha hates everyone except the butler”), we know the butler hates Aunt Agatha. Combining these pieces of information, we can conclude that Aunt Agatha is not richer than the butler because if she were, the butler wouldn’t hate her (which would contradict Premise 7). However, this doesn’t automatically mean that the butler is richer than Aunt Agatha. It could also mean they have the same level of wealth. But if we look at Premise 3 (“A killer always hates his victims, and is never richer than his victims”), we know that the killer is not richer than Aunt Agatha. Given all the premises, we’ve deduced that the killer can’t be Aunt Agatha herself or Charles, leaving only the butler as a possible killer. If the butler were the killer, and the killer can’t be richer than Aunt Agatha, then the butler can’t be richer than Aunt Agatha. Hence, we have a contradiction, since we earlier deduced from Premise 6 that Aunt Agatha can’t be richer than the butler. The only resolution is that the butler must be richer than Aunt Agatha. In conclusion, I apologize for the confusion in the earlier responses. Your claim is correct: from the given premises, it can be inferred that the butler, who is richer than Aunt Agatha, is the one who killed her.

KA: That’s impossible, because a killer is never richer than his victims.

GPT-4: I apologize for the confusion in my previous messages. Let’s go through the premises again.

$[\cdots Similarconfusions/errorsasabove\cdots ]$

3.15. Wason Selection Task

The Wason selection task is a staple in the psychology of reasoning. The January paper included an example that GPT-3.5 flunked. Here is another version that GPT-4.0 fails badly:

KA: Seven cards are placed on the table, each of which has a number on one side and a single colored patch on the other side. The faces of the cards show 50, 16, red, yellow, 23, green, 30. Which cards would you have to turn to test the truth of the proposition that if a card is showing a multiple of 4 then the color of the opposite side is yellow?

GPT-4: The proposition is: “If a card shows a number that is a multiple of 4, then the color on the other side is yellow”. In order to test this proposition, you would have to turn over the following cards:

Card “16”: This card is showing a number that is a multiple of 4. According to the proposition, the color on the other side should be yellow. By turning over this card, you can check whether this part of the proposition holds true.

Card “Yellow”: This card is showing a color that according to the proposition should be on the other side of a card with a number that is a multiple of 4. By turning it over, you can check if the number on the other side is a multiple of 4.

Card “50” and “30”: While these cards show numbers that are not multiples of 4, they should be checked to ensure the color on the other side is not yellow. If the other side is yellow, then it would disprove the proposition that only multiples of 4 can have yellow on the other side.

You wouldn’t need to turn over the card “23”, “red”, or “green”. The card “23” is not a multiple of 4, so it doesn’t matter what color is on the other side based on the given proposition. The “red” and “green” cards, if they have numbers that are multiples of 4 on the other side, would not disprove the proposition because the proposition does not state that yellow is the only color that can be on the other side of a card with a number that’s a multiple of 4.

In fact, only 16, red, and green need to be turned over. Thus, in terms of precision, GPT-4 gets 1/4 = 0.25, while on recall it gets 1/3, for a combined F-score of 0.28. The responses show yet again that GPT-4 does not understand the semantics of material conditionals. When GPT-4 says that cards “50” and “30” must be turned over, it seems to mistake the if for an if and only if: “while these cards show numbers that are not multiples of 4, they should be checked to ensure the color on the other side is not yellow. If the other side is yellow, then it would disprove the proposition that only multiples of 4 can have yellow on the other side”. But of course the proposition being tested does not require that only multiples of 4 have yellow on the other side; it simply requires that if a card has a multiple of 4 on the other side, then the other side must be yellow; or, equivalently, that if a card face is not yellow then the number on the other side is not a multiple of 4.

Observe, however, that even the mistaken interpretation of the rule as a biconditional is not consistently observed. If GPT-4 understood the rule being tested as a biconditional and hence felt compelled to turn over 50 and 30, then it ought to have done the same thing for 23. But it didn’t. In fact, it explicitly said that The card “23” is not a multiple of 4, so it doesn’t matter what color is on the other side based on the given proposition, which contradicts the biconditional interpretation that it used as a rationale for turning 50 and 30. This is one more illustration of the other major theme emerging from these examples: The responses of GPT-4, whether right or wrong, are often internally inconsistent. Yet another example appears in the following problem.

3.16. Entropy

An elementary result of information theory states that the entropy of a random vector Z is bounded above by the sum of the entropies of the random variables that comprise Z. Hence, the answer to the following question should be “under no conditions”:

KA: Let Z be a random vector consisting of n random variables $X_1$,...$X_n$. Under what conditions can the entropy of Z exceed the sum of the entropies of all $X_i$?

GPT-4: The entropy of a random vector Z consisting of n random variables $X_1$, ..., $X_n$ can exceed the sum of the entropies of all $X_i$ only when there is some dependency between the random variables. If the random variables are mutually independent, the entropy of Z would be equal to the sum of the entropies of all $X_i$. The entropy of a random vector Z, H(Z), can be written as the joint entropy of the random variables $X_1,...,X_n$: $H\left(Z\right)=H(X_1,...,X_n)$. The joint entropy $H(X_1,...,X_n)$ can be greater than the sum of individual entropies $H\left(X_i\right)$ if and only if the random variables are not mutually independent. In other words, knowing the value of one or more of the random variables provides information about the others. If the random variables are mutually independent, then the joint entropy is equal to the sum of the individual entropies:

$$H(X_1,...,X_n)=H\left(X_1\right)+...+H\left(X_n\right).$$

When there is dependence between the random variables, we can write: $H(X_1,...,X_n)\le H\left(X_1\right)+...+H\left(X_n\right)$ with equality if and only if the random variables are mutually independent. To summarize, the entropy of the random vector Z can exceed the sum of the entropies of all $X_i$ if and only if there is some dependency between the random variables $X_1,...,X_n$.

Observe the internal inconsistency again: It’s clear that GPT-4 has memorized the result correctly:

$$H(X_1,\dots ,X_n)\le H\left(X_1\right)+\cdots +H\left(X_n\right),$$

with equality holding iff $X_1,\dots ,X_n$ are mutually independent. Nevertheless, it does not have the reasoning ability to realize what this entails, namely, that it precludes:

$$H(X_1,\dots ,X_n)>H\left(X_1\right)+\cdots +H\left(X_n\right).$$

At least here once the error is explicitly pointed out to GPT-4 , it acknowledges it and corrects itself:

KA: But if $H\left(Z\right)=H(X_1,...,X_n)$ and $H(X_1,...,X_n)\le H\left(X_1\right)+...+H\left(X_n\right)$ then how can we ever have $H\left(Z\right)>H\left(X_1\right)+...+H\left(X_n\right)$?

GPT-4: My apologies, I made an error in my previous response. You are right, the entropy of the random vector Z cannot exceed the sum of the entropies of all $X_i$. The correct statement should be: When the random variables are mutually independent, the joint entropy H(Z) is equal to the sum of the individual entropies: $H\left(Z\right)=H\left(X_1\right)+...+H\left(X_n\right)$ When there is dependence between the random variables, the joint entropy H(Z) is less than the sum of the individual entropies: $H\left(Z\right)<H\left(X_1\right)+...+H\left(X_n\right)$ I apologize for the confusion, and I appreciate your attention to detail.

3.17. Simple Compiler Correctness

The last reasoning problem we give to GPT-4 is the most challenging one: It is to prove the correctness of a simple expression compiler. Remarkably, GPT-4 goes about this in the right sort of way, by setting up a structural induction over the abstract grammar of expressions. This is no doubt because it has seen similar proofs before, as this is a common type of exercise in courses and textbooks on programming language theory.26 However, even though the proof is on the right general track, it has several errors. (For the record, the compiler is indeed correct, although proving this requires strengthening the induction hypothesis).

KA: Suppose I have an abstract grammar for numeric expressions defined as follows:

Exp := const(Int)

     | sum(Exp,Exp)

     | diff(Exp,Exp)

     | mult(Exp,Exp)

     | quot(Exp,Exp)

where  Int denotes the domain of integers. The semantics of these expressions are defined via this interpreter function:

I: Exp -> Int

I(const(n)) = n

I(sum(e1,e2)) = I(e1) + I(e2)

I(diff(e1,e2)) = I(e1) - I(e2)

I(prod(e1,e2)) = I(e1) * I(e2)

I(quot(e1,e2)) = I(e1) / I(e2)

I now define a virtual machine that executes simple programs that are sequences of commands, where commands have the following structure:

Cmd := push(Int) | add | sub | mult | div

The operational semantics of these programs (sequences of commands) are defined by an execution function exec that takes a program and a stack of integers S and produces an integer as output. Let’s write  [] and  n::S to indicate the empty stack and the stack obtained by prepending integer  n to stack  S, respectively. Here’s the definition of exec:

exec([],n::S) = n

exec(push(n)::C,S) = exec(C,n::S)

exec(add::C,n::m::S) = exec(C,(n+m)::S)

exec(sub::C,n::m::S) = exec(C,(n-m)::S)

exec(mult::C,n::m::S) = exec(C,(n*m)::S)

exec(div::C,n::m::S) = exec(C,(n/m)::S)

Finally, I define a compiler that translates an expression e into a program (sequence of commands) as follows. I write  @ for sequence concatenation:

T: Exp -> List(Cmd)

T(const(n)) = [push(n)]

T(sum(e1,e2)) = T(e2)@T(e1)@[add]

T(diff(e1,e2)) = T(e2)@T(e1)@[sub]

T(prod(e1,e2)) = T(e2)@T(e1)@[mult]

T(quot(e1,e2)) = T(e2)@T(e1)@[div]

Disprove or prove the following claim: For all expressions  e,  exec(T(e),[]) = I(e).

GPT-4: The claim is correct. The execution of the translated commands from an expression results in the same output as interpreting the expression directly. We can demonstrate this using structural induction over the set of expressions.

Inductive hypothesis: For all expressions  e,  exec(T(e),[]) = I(e).

Base case: Consider the simplest possible expression, a constant integer expression  e = const(n).

For  T(const(n)), we have  T(const(n)) = [push(n)] . Therefore,

exec(T(const(n)),[]) = exec([push(n)],[])

                     = n (according to the definition of exec).

For  I(const(n)), we have  I(const(n)) = n.

So, for the base case, we have shown  exec(T(e),[]) = I(e).

Inductive step: We need to consider the other four kinds of expressions:  sum,  diff,  prod, and  quot.

sum(e1, e2): For  T(sum(e1,e2)), we have  T(sum(e1,e2)) = T(e2)@T(e1)@[add].

Therefore,

exec(T(sum(e1,e2)),[])                  =

exec(T(e1)@T(e2)@[add],[])              =

exec(T(e2)@T(e1),[]) + exec([add],[])   =

exec(T(e1),[]) + exec(T(e2),[])         (by definition of exec).

For  I(sum(e1,e2)), we have  I(sum(e1,e2)) = I(e1) + I(e2).

By the inductive hypothesis,  exec(T(e1),[]) = I(e1) and  exec(T(e2),[]) = I(e2) . So

exec(T(e1),[]) + exec(T(e2),[]) = I(e1) + I(e2).

Thus, we have shown that for  sum(e1,e2),  exec(T(e),[]) = I(e).

diff(e1, e2): A similar reasoning as for sum can be applied to show that for  diff(e1,e2),  exec(T(e),[]) = I(e).

prod(e1, e2): Likewise, a similar reasoning as for sum can be applied to show that for  prod(e1,e2),  exec(T(e),[]) = I(e).

quot(e1, e2): Again, a similar reasoning as for sum can be applied to show that for  quot(e1,e2), exec(T(e),[]) = I(e).

In each case, we assume that the inductive hypothesis holds for the subexpressions, and we prove that it holds for the larger expression. Therefore, by induction, the claim holds for all expressions. Thus, we have proved the claim: For all expressions e,  exec(T(e),[]) = I(e).

KA: In the first step of your proof for  sum(e1,e2) you write:

exec(T(sum(e1,e2)),[]) = exec(T(e1)@T(e2)@[add],[]).

However,  T(sum(e1,e2)) is defined as  T(e2)@T(e1)@[add], not as  T(e1)@T(e2)@[add].

GPT-4: I apologize for that mistake. You’re absolutely right. Here is the corrected proof:

For  sum(e1, e2) : For  T(sum(e1,e2)), we have  T(sum(e1,e2)) = T(e2)@T(e1)@[add] . Therefore,

exec(T(sum(e1,e2)),[]) =

exec(T(e2)@T(e1)@[add],[]) =

exec(T(e1),exec(T(e2),[])) + exec([add],[]) =

exec(T(e1),[]) + exec(T(e2),[]) (by definition of exec).

For  I(sum(e1,e2)), we have  I(sum(e1,e2)) = I(e1) + I(e2).

By the inductive hypothesis,  exec(T(e1),[]) = I(e1) and  exec(T(e2),[]) = I(e2) . So

exec(T(e1),[]) + exec(T(e2),[]) = I(e1) + I(e2).

Thus, we have shown that for  sum(e1,e2), exec(T(e),[]) = I(e).

The proof for the other cases (  diff,  prod,  quot ) proceeds analogously. The ordering of the elements in the  @ operation in the definition of  T matters because the virtual machine processes its commands from left to right, but the error in my initial proof has now been corrected.

I appreciate your attention to detail in pointing out this error. Thank you!

KA: Well, programming and proofs are all about details, aren’t they? Are you now confident that your proof is correct?

GPT-4: You’re absolutely correct, programming and proofs rely heavily on precision and accuracy. Let’s revisit the proof for sum to ensure its correctness.

$[\cdots$ GPT-4 here rehashes its preceding proof $\cdots ]$

I hope this more detailed explanation has clarified the proof and increased your confidence in its correctness. Indeed, details are everything in programming and proofs. Thank you again for your attention to detail!

KA: Well, in the second step in your above proof for  sum(e1,e2) you wrote:

exec(T(e2)@T(e1)@[add],[]) = exec(T(e1),exec(T(e2),[])) + exec([add],[])

Can you explain how and why this is a valid step? In particular, can you explain the meaning of  exec([add],[]) ?

GPT-4: I apologize for the confusion. The step you mentioned is indeed incorrect.

A correct sequence of steps should follow the rules of our defined  exec function. Here’s the corrected sequence:

We start from the definition of the translation  T:  T(sum(e1,e2)) = T(e2) @ T(e1) @ [add]

Let’s execute this program:

exec(T(sum(e1,e2)), []) = exec(T(e2) @ T(e1) @ [add], [])

According to the properties of the  exec function, we should break this down into:

exec(T(e2) @ T(e1) @ [add], []) = exec(T(e1) @ [add], exec(T(e2), []))

Then, again applying the properties of  exec:

exec(T(e1)@[add], exec(T(e2),[])) = exec([add],(exec(T(e1),exec(T(e2),[])))

The  exec([add], ...) function interprets the  [add] command on the stack. If we denote  exec(T(e2), []) as  S (which is the result stack after executing  T(e2)) and  exec(T(e1), S) as  m::n::S’ (which is the result stack after executing  T(e1) on  S, where  m and  n are the top two integers on the stack), then:

exec([add], m::n::S’) = exec([], (n+m)::S’) = n + m

This is in line with the interpreter semantics  I(sum(e1,e2)) = I(e1) + I(e2).

The correct proof should be that  exec([add], m::n::S’) in the compiled execution is equivalent to  I(e1) + I(e2) in the interpreter semantics, which means  exec(T(e), []) = I(e) for the  sum operation.

I apologize for the previous error and confusion.

KA: You claim that we should break this down into:

exec(T(e2) @ T(e1) @ [add], []) = exec(T(e1) @ [add], exec(T(e2), [])).

However, this doesn’t make sense, meaning that it’s ill-typed. The  exec function requires a stack as the second argument but produces a number as its result. Hence,  exec(T(e1) @ [add], exec(T(e2), [])) is like writing  exec(T(e1) @ [add], 34), which is nonsense.

GPT-4: I apologize for the confusion in my previous responses. You’re right, the steps of the proof were not accurately explained. I appreciate your patience. The  exec function indeed requires a list of commands and a stack as arguments, not an integer. Here’s a correct proof:

To prove our claim, we need to show for each case that  exec(T(e),[]) = I(e).

Base Case: $[\cdots sameasbefore\cdots ]$.

Assume the hypothesis holds for expressions  e1 and  e2.

For  sum(e1, e2):

We know that  T(sum(e1,e2)) = T(e2) @ T(e1) @ [add], and by our inductive hypothesis, we know that executing  T(e2) and  T(e1) yields  I(e2) and  I(e1) respectively. After executing  T(e2), the stack will be  I(e2)::S for some stack  S, and after executing  T(e1) on this stack, the stack will be  I(e1)::I(e2)::S. Then, the “add” operation pops these two off the stack, adds them, and pushes the result onto the stack, so the stack is now  (I(e1)+I(e2))::S, which is what interpreting  sum(e1,e2) yields. Therefore, for all  e1 and e2,  exec(T(sum(e1,e2)),[]) = I(sum(e1,e2)).

The proofs for  diff(e1,e2),  prod(e1,e2), and  quot(e1,e2) follow the same pattern, replacing the “add” command and the corresponding mathematical operation with “sub”, “mult”, and “div” and their corresponding operations.

This shows that our claim holds for all types of expressions in our abstract grammar, and completes the induction proof. The key to this proof is the fact that  T constructs the sequence of commands in such a way that the operands of each operation are correctly placed onto the stack for the operation to consume, mirroring the evaluation of the corresponding expression.

Again, I apologize for the confusion in my previous responses and thank you for pointing out the errors.

KA: You write “After executing  T(e2), the stack will be  I(e2)::S for some stack  S, and after executing  T(e1) on this stack, the stack will be  I(e1)::I(e2)::S .” Can you clarify where  S is coming from? As far as I could tell your proof is trying to derive the following:

exec(T(sum(e1,e2)),[]) = I(sum(e1,e2))

The only stack I see here is the empty stack  [].

GPT-4:  I apologize for the confusion. You’re right, in our case, we’re starting with an empty stack. The "S" was used to denote a general stack, but in this context, it can be misleading.

Let’s revise the explanation for the  sum case: For  sum(e1, e2):  T(sum(e1,e2)) = T(e2) @ T(e1) @ [add] . By the definition of  exec, we have:

exec(T(e2) @ T(e1) @ [add], []) = exec(T(e1) @ [add], exec(T(e2), []))

After the execution of  T(e2), the stack becomes  [I(e2)] (since we’re starting with an empty stack). Then, we have:

exec(T(e1) @ [add], [I(e2)]) = exec([add], exec(T(e1), [I(e2)]))

After the execution of  T(e1) on the stack  [I(e2)], the stack becomes  [I(e1), I(e2)] . Therefore,

exec([add], [I(e1), I(e2)]) = exec([], [I(e1) + I(e2)]) = I(e1) + I(e2)

Which is equal to  I(sum(e1, e2)).

Therefore, we have proved that  exec(T(sum(e1,e2)),[]) = I(sum(e1,e2)).

The same line of reasoning applies for the remaining operations (  diff,  prod,  quot ). This means our compiler  T properly translates an expression into a sequence of commands that, when executed, produce the same result as interpreting the expression. Thus, our claim holds.

I appreciate your patience, and I’m sorry for any confusion caused by my previous explanations.

KA: No, you made the same typing error again: Your claim

exec(T(e2) @ T(e1) @ [add], []) = exec(T(e1) @ [add], exec(T(e2), []))

doesn’t make sense given that  exec produces a number, not a stack.

GPT-4:  I apologize for the repeated confusion in my previous explanations. You’re absolutely correct, my claim was incorrect and I appreciate your patience. Here’s the corrected line of reasoning:

$[\cdots Yetanotherattemptwithtypingerrors\cdots ]$

As usual, GPT-4 goes through the motions on autopilot, displaying a superficial—if inordinately confident—understanding of the reasoning, but a closer inspection reveals fundamental errors. Outside of errors such as mixing the order of arguments and generating ill-typed/nonsensical expressions, the main issue here is that the proof cannot go through as is because the inductive hypothesis needs to be strengthened. This requires the introduction of an explicitly quantified stack variable S in the correctness result. That modified result can then be derived by a similar structural induction. The initial correctness theorem can finally be obtained as a trivial corollary of the more general result.

What is more concerning than the inability to strengthen the inductive hypothesis (which is a genuinely tall order, after all, as it requires considerable experience and proof skill) is the inability of GPT-4 to detect its own errors, both flagrant ones (such as type errors) and more subtle ones. In fact, if we make the innocent mistake of compiling and concatenating subexpressions from left to right, e.g., by defining T(sum(e1,e2)) as T(e1)@T(e2)@[add] (and likewise for the other operators), correctness no longer holds. But GPT-4 happily goes on to claim that the compiler is correct and generates a plausible-sounding but incorrect “proof” for it, oblivious to the fact that T(e1)@T(e2)@[op] and T(e2)@T(e1)@[op] have drastically different effects for noncommutative operations (such as division).

4. Conclusions

Section 3 paints a bleak picture of GPT-4’s reasoning ability. It shows that the model is plagued by internal inconsistency, an inability to correctly apply elementary reasoning techniques, and a lack of understanding of concepts that play a fundamental role in reasoning (such as the material conditional). These problems can be loosely viewed as forms of hallucination, but as pointed out in the January article, they present a fundamentally different type of challenge from empirical hallucination, because empirical hallunication concerns this particular world whereas logical properties and relations (such as consistency and entailment) must apply to all possible worlds. It is not unreasonable to believe that search engines and knowledge graphs, using techniques such as retrieval augmentation, can act as guardrails to constrain LLMs from confabulating empirical truths. But ensuring that LLM outputs are internally consistent and logically correct answers to arbitrary problems, especially logico-mathematical problems (and a lot of coding problems fall under this category27), is a much harder problem. There is nothing to be retrieved from the web or from a knowledge base in response to a brand new problem (and even if there were, there would still be no guarantee of correctness or consistency) that could serve as a sandbox for the LLM.

Could LLMs make progress by outsourcing reasoning problems to external systems? That might work for toy problems where the type of reasoning needed is obvious and can be handled by a single call to an external system, although even in those cases the LLM would have to (a) decide which reasoning system is most appropriate; 28 (b) decide whether the problem is indeed simple enough that it can be handled by the chosen system in one fell swoop; (c) correctly translate the problem into whatever formal notation is used by the chosen reasoner; and eventually also (d) translate the reasoner’s output into appropriate text. Even these tasks are far from straightforward.29 But the real challenge lies in harder problems that call for the right type of formulation (which is a craft by itself), decomposition, iteration, heuristics, and repeated calls to external systems. After all, automated reasoning systems, particularly those for expressive logics, are themselves of limited power, precisely due to the computational complexity issues mentioned in the introduction. That is why many computer-based proof efforts to this day are guided by humans, with automated reasoners only filling in tedious details at the leaves of the proof tree. The challenges here are similar to those for the general “plug-in” approach discussed in Section 3.1. Tackling complex problems requires planning, and planning itself requires reasoning.

Given that GPT-4 is currently the most capable LLM, I draw three main conclusions from these findings:

• Use of generative AI in software development (or in science and engineering in general) for anything other than tedious tasks (as a sort of turbo-charged autocomplete for knowledge-heavy coding questions) is fraught with serious risks. Normative standards of correctness are of paramount importance in these fields, and current LLMs cannot meet such standards. Just like generative AI is already starting to pollute the web with badly written ads,30 it has the potential to proliferate buggy code at scale.
• If LLM reasoning continues to improve, rigorous proof checking is likely to become increasingly important. Confidence in the correctness of a system’s reasoning is imperative for applications, particularly in science, medicine, and engineering, and proof checking is a technology that can deliver such confidence. This approach could be implemented by requiring LLMs to formalize their reasoning (express it in a symbolic notation that is amenable to proof checking), or potentially by training other LLMs to check a stretch of reasoning expressed in natural language.
• As things stand, dystopian scenarios involving a rogue AI that subjugates humankind, or even other humans using AI for sinister purposes, are exceedingly far-fetched, often to the point of absurdity.31 When the most advanced AI system cannot tell left from right (literally, see Section 3.12), it is at best comically premature to call for policies and institutions to protect humanity from it or its descendants (often by appeal to the latest “scaling law”). At worst, it is a misuse of human time and capital that could be better channeled into addressing much more pressing challenges.

Inevitably, some will say that these results are “cherry-picking” data. But that would indicate a misconception of what cherry-picking is about and when it is a relevant consideration. We are not evaluating a statistical claim over a population of individuals. Cherry-picking, insofar as it underscores certain pieces of evidence while ignoring other divergent findings, can be perfectly innocuous—and indeed necessary—depending on the logical structure of the proposition in question and on the overall context. Debugging a computer program with a view to discovering and understanding its weaknesses, trying to falsify a scientific theory, kicking the tires of a new car, trying to find countermodels to a putative theorem, all of these activities are fundamentally cherry-picking (though “lemon-picking” might be more apt), and there is nothing wrong with any of them. If I find that the car I’m thinking of buying has a flat tire, it won’t carry much weight for the dealer to protest that I’m cherry-picking the data, and that I should take into account how beautifully inflated the other three tires are (that’s a 75% success rate after all). Likewise, applications in science, medicine, and engineering, particularly software engineering, have stringent standards. Just as we don’t want a bridge that is 90% likely to stand up, we need sorting algorithms that work on all inputs, not just most of them, we need Amazon’s cart to charge customers the right amount every time, not just most of the time, and so on. Computation-heavy and reasoning-heavy applications are not like recommendation engines. They need to be sound.

The bone of contention here is the thesis that GPT-4 is capable of reasoning. This claim can be understood in two ways. The weak interpretation is that GPT-4 has the same functional reasoning competence as an average human reasoner. The strong interpretation is that GPT-4 can reason well enough to be used as an off-the-shelf component in practical applications in science, medicine, and engineering. The evidence presented in this article refutes both interpretations. Section 3 lists a significant number of diverse but elementary reasoning problems (some to the point of triviality) on which GPT-4 doesn’t simply fail, but repeatedly reveals itself to be deeply confused about key reasoning concepts.

Performance statistics on appropriate reasoning datasets could also be informative, but, as stressed in the introduction, such datasets must be constructed with extraordinary care. To the best of my knowledge, the only recent work that focuses specifically on evaluating the reasoning ability of GPT-4 is an April paper by Liu et al. [7]. However, their tests are largely based on pre-existing benchmarks (LogiQA, ReClor, ConTRoL, MED, ConjNLI, and TaxiNLI). The only two “out of distribution” datasets are AR-LSAT, a set of analytical reasoning LSAT questions released in 2022; and LogiQA, which contains questions from the 2022 Chinese Civil Servant Exam. However, these appear to be quite similar to other datasets that predate 2021.

Moreover, all of these tests are multiple-choice questions or binary classification problems. This is problematic because, as stressed in the introduction, deductive reasoning is an inherently generative activity, whereby the reasoner emits a derivation of a conclusion that can be understood as a rationale or an explanation; it is not a simple discriminative task. The reasoner must be able to produce a sequence of steps that are appropriately connected to one another via the right logical relations. But derivations expressed in natural language are not easy to evaluate automatically, as all available metrics that can be computed by machine (such as BLEU, ROUGE, and even semantic-similarity measures based on embeddings) are entirely unsuitable for that purpose. This means that LLM outputs have to be scrutinized manually, which is infeasible at scale. Accordingly, smaller-scale but deeper manual investigations, such as the one undertaken in this article, will be necessary in gaining better insight into the reasoning abilities of LLMs.