Logic on Planet Zorko

A group of Zorkan mathematicians are sitting around having a conversation in a language that you are unfamiliar with. You are listening in with a translator. This translator is an expert in formal logic, and has decided to play the following game with you. He says:

“After listening to the full conversation, I will translate all the sentences that were said for you. But I won’t translate them into English; I want something more universal. Instead, I will choose a formal language that captures the mathematical content of all the sentences said, while leaving out the vagaries and subtleties of the Zorkan language. I will describe to you the semantics of the formal language I choose, if you don’t already know it.”

“Furthermore,” (says the translator) “I happen to be intimately familiar with Zorkan society and culture. The Zorkans are having a discussion about one particular mathematical structure, and I know which one that is. The mathematicians are all fantastically precise reasoners, such that none of them ever says a sentence that is false of the structure that they are discussing.”

(So for instance if they are talking about the natural numbers, then no mathematician will say “0 = 1”, and if they are talking about abelian groups, then no mathematician will say “∃x∃y (xy ≠ yx)”. But they could say “∃x∃y (xy ≠ yx)” if they are talking about non-abelian groups.)

You know nothing about Zorkan psychology, besides that the Zorkan way of life is so utterly foreign to you that you cannot reliably assume that the mathematical structures that come most naturally to you will also come naturally to them. It might be, for instance, that nonstandard models of arithmetic are much more intuitive to them than the natural numbers. You cannot assume that the structure they are discussing is the one that you think is “most natural”; you can only conclude this if one of them says a sentence that is true of that model and no others.

The conversation finishes, and you are tasked with answering the following two questions:

(1) What structure are they talking about?
(2) Can you come up with a verification procedure for the mathematicians’ sentences (including possible future sentences they might say on the topic)?

So, that’s the setup. Now, the question I want you to consider is the following: Suppose that the structure that the mathematicians have in mind is actually the natural numbers. Is there some conversation, any conversation at all (even allowing infinitely long conversations, and uncomputable conversations – conversations which cannot be produced as the output of any Turing machine), that the mathematicians could have, and some translation of this conversation, such that you can successfully answer both (1) and (2)? If so, what is that conversation? And if not, then why not?

✯✯✯

Let’s work out some simple examples.

Example 1

Suppose the conversation is translated into a propositional language with three atomic propositions {P, Q, R}.

Mathematician A: “P ∨ Q”
Mathematician B: “(Q ∨ R) → (¬P)”
Mathematician C: “R”

From this conversation, you can deduce that the model they are talking about is the one that assigns “False” to P, “True” to Q, and “True” to R.

M: {P is false, Q is true, R is true}

This is the answer to the question 1!

As for the second question, we want to know if there’s some general procedure that produces all the future statements the mathematicians could make. For instance, the set generated by our procedure should include (Q ∧ R) but not (Q ∧ P).

It turns out that such a procedure does exist, and is not too difficult to write out and implement.

Example 2

Take the above conversation and modify it slightly:

Mathematician A: “P ∨ Q”
Mathematician B: “(Q ∨ R) → (¬P)”
Mathematician C: “¬R”

If you work it out, you’ll see that question 1 can no longer be answered unambiguously. The problem is that there are multiple models of the sentences that the mathematicians are saying:

M1: {P is false, Q is true, R is false}
M2: {P is true, Q is false, R is false}

So even though they have one particular structure in mind, you don’t have enough information from their conversation to figure out exactly what that structure is.

Now let’s think about the answer to question 2. We don’t know whether the mathematicians are thinking about M1 or M2, and M1 and M2 differ in what truth value they assign the proposition P. So we can’t construct an algorithm that will generate the set of all their possible future statements, as this would require us to know, in particular, whether P is true or false in the model that they have in mind.

We might suspect that this holds true generally: if you can’t answer question 1, then you won’t be able to answer question 2 either. But we might also wonder: if we can answer question 1, then can we also always answer question 2?

The answer is no, as the next example will show.

Example 3

For this conversation, the translation is in second-order logic. This will allow us to talk about more interesting mathematical structures than before; namely, structures that have a domain of objects on which functions and predicates can act. In particular, we’re in a second-order language with one constant symbol “c” and one function symbol “f”. Here’s the conversation:

Mathematician A: ¬∃x (f(x) = c)
Mathematician B: ¬∃x∃y ((f(x) = f(y)) ∧ ¬(x = y))
Mathematician C: ∀R (R(c) ∧ ∀x(R(x) → R(f(x))) → ∀x R(x))

Notice that the only truly second-order sentence is the third one, in which we quantify over a predicate variable R rather than an individual variable x, y, z, …. But the second-order status of this sentence it makes it that the translator could not have possibly translated this conversation into a first-order language, much less a propositional language.

This time, questions 1 and 2 are much harder to answer than before. But if you work it out, you’ll see that there is exactly one mathematical structure that satisfies all three of the mathematicians’ statements. And that structure is the natural numbers!

So, we know exactly what structure the mathematicians have in mind. But can we also answer question 2 in the positive? Can we produce some verification procedure that will allow us to generate all the future possible sentences the mathematicians could say? Unfortunately, the answer is no. There is no sound and complete proof system for second-order logic, so in particular, we have no general algorithm for producing all the truths in this second order language. So sad.

Example 4

Now let’s move to first-order logic for our final example. The language of translation will be a first order language with a constant symbol for every natural number {0,1,2,3,…}, function symbols for ordinary arithmetic {+, ×}, and relation symbols for orders {≥}

Imagine that the conversation consists of literally all the first-order sentences in the language that are true of the natural numbers. Anything which you can say in the language, and which is true as a statement about ℕ, will be said at some point. This will obviously be a very long conversation, and in fact infinitely long, but that’s fine. It will include sentences like “0 ≠ 1”, “0 ≠ 2”, “0 ≠ 3”, and so on.  (These Zorkans are extremely thorough.)

Given this conversation, can we answer (1) and (2)? Take a guess; the answer may surprise you!

It turns out that even though we can answer (2) positively – we can actually produce an algorithm that will generate one-by-one all the possible future statements of the mathematicians (which really means all the sentences in the language that are true of the natural numbers), we cannot answer (1) positively! There are multiple distinct mathematical structures that are compatible with the entirety of true statements about natural numbers in the language. Earlier we hypothesized that any time we have a negative answer to (1), we will also have a negative answer to (2). But this is not true! We can verify all the true statements about natural numbers in the language… without even knowing that we’re actually talking about the natural numbers! This is an important and unintuitive consequence of the expressive limitations (and in particular, of the compactness) of first-order logic.

The Takeaway

We had an example where we could answer both (1) and (2) for a simple mathematical structure (a model of propositional logic). And we saw examples for natural numbers where we could answer (1) but not (2), as well as examples where we could answer (2) but not (1). But we haven’t yet seen an example for natural numbers where we had both (1) and (2). This is no coincidence!

It is actually a consequence of the theorem I proved and discussed in my last post that no such such conversation can exist. When structures at least as complicated as the natural numbers are being discussed in some language (call it L), you cannot simultaneously (1) know for sure what structure is being talked about and (2) have an algorithmic verification system for L-sentences about the structure.

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A Gödelian Logic Puzzle

There’s an island on which there lives exactly two types of people: truthers and liars. Truthers always say true statements, and liars always say false statements. One day a brilliant logician comes to visit the island. The logician knows all of the above-stated facts about the island. It also happens that the logician is a perfectly sound reasoner – he never proves anything that is false.

The logician encounters an individual named ‘Jal’ that lives on the island. The logician knows that Jal lives on the island, and so is either a truther or a liar. Now, Jal makes a statement from which it logically follows that Jal is a truther. But the logician could never possibly prove that Jal is a truther! (Remember, we never asserted that the logician proves all true things, just that the logician proves only true things). What type of statement could accomplish this?

This puzzle is from a paper by Raymond Smullyan on mathematical logic. Try to answer it for yourself before reading on!

(…)

Alright, so here’s one possible answer. Jal could say to the logician: “You will never prove that I am a truther.” I claim that this sentence logically entails that Jal is a truther, and yet the logician cannot possibly prove it.

First of all, why does it entail that Jal is a truther? Let’s prove it by contradiction. Suppose that Jal is not a truther. Then, since Jal is either a truther or a liar, Jal must be a liar. That means that every statement Jal makes must be false. So in particular, Jal’s statement that “you will never prove that I am a truther” must be false. This entails that the logician must eventually prove that Jal is a truther. But we assumed that Jal isn’t a truther! So the logician must eventually prove a falsehood. But remember, we assumed that our logician’s proofs were always sound, so that he will never prove a falsehood. So we have a contradiction.

Therefore, Jal is a truther.

Now, why can the logician not prove that Jal is a truther? This can be seen very straightforwardly: we just proved that Jal is a truther, which means that all of Jal’s statements must be true. So in particular, Jal’s statement that “you will never prove that I am a truther” must be true. So in other words, it’s true that the logician will never prove that Jal is a truther!

So there you have it, a statement that appears to satisfy both of the criteria!

But now the next question I have for you is a bit trickier. It appears from the line of reasoning above that we have just proven that Jal is a truther. So why couldn’t the logician just run through that exact same line of reasoning? It appears to be perfectly valid, and to use nothing more advanced than basic predicate logic.

But if the logician does go through that line of reasoning, then he will conclude that Jal is a truther, which will make Jal’s statement false, which is a contradiction! So we’ve gone from something which was maybe just unintuitive to an actual paradox. Can you see how to resolve this paradox? (Again, see if you can figure it out yourself before reading on!)

(…)

Okay, so here’s the resolution. If we say that the logician can go through the same line of reasoning as us, then we reach a contradiction (that a truther tells a false statement). So we must deny that the logician can go through the same line of reasoning as us. But why not? As I said above, the reasoning is nothing more complicated than basic predicate logic. So it’s not that we’re using some magical supernatural rules of inference that no mortal logician could get his hands on. It must be that one of the assumptions we used in the argument is an assumption that the logician cannot use.

So look back through the argument, and carefully consider each of the assumptions we used:

First of all, why does it entail that Jal is a truther? Let’s prove it by contradiction. Suppose that Jal is not a truther. Then, since Jal is either a truther or a liar, Jal must be a liar. That means that every statement Jal makes must be false. So in particular, Jal’s statement that “you will never prove that I am a truther” must be false. This entails that the logician must eventually prove that Jal is a truther. But we assumed that Jal isn’t a truther! So the logician must eventually prove a falsehood. But remember, we assumed that our logician’s proofs were always sound, so that he will never prove a falsehood. So we have a contradiction.

In order, we made use of the assumptions that (1) Jal is either a truther or a liar, (2) every statement made by a liar is false, and (3) the logician is a sound reasoner.

I told you at the beginning that facts (1) through (2) are all known to the logician, but I did not say the same of (3)! The logician can only run through this argument if he knows that he is a sound reasoner (that he only proves true things). And this is the problem assumption, which must be rejected.

It’s not that no logician can actually ever be sound (a logician who only ever reasons in first order logic and nothing more fancy would be sound). It’s that the logician, though he really is sound, cannot know himself to be sound. In other words, no sound system can prove its own soundness!

This is very similar to Gödel’s second incompleteness theorem. The only proof system which can assert its own consistency is an inconsistent proof system, and the only type of logician that can prove his own soundness will end up being unsound. Here’s the argument that the logician might make if they believe in their own soundness:

Supposing Jal is a liar, then his statement is false, so I could eventually prove that he is a truther. But then I’d have proven something false, which I know I can never do, so Jal must not be a liar. So he must be a truther. 

Since the logician has now produced a proof that Jal is a truther, Jal’s statement is false. This means that Jal cannot be a truther, so the logician has proven a false statement!

Crazy conditionals

It’s well known that the material implication → of propositional logic does not do a perfect job of capturing what we mean when we make “if… then…” statements in English. The usual examples of failure rest on the fact that any material conditional with a false antecedent is vacuously true (so “if 2 is odd then 2 is even” turns out to be true). But over time, philosophers have come up with a whole lot of different ways in which → can catch us by surprise.

Here’s a list of some such cases. In each case, I will present an argument using if…then… statements that is clearly invalid, but which is actually valid in propositional logic if the if…then… statements are translated as the material conditional!

1. Harper

If I put sugar into my coffee, it will taste fine.
Therefore, if I put sugar and motor oil into my coffee, it will taste fine.

S → F
(S ∧ M) → F

2. Distributivity

If I pull both switch A and switch B, the engine will start.
Therefore, either the engine will start if I pull switch A or the engine will start if I pull switch B.

(A ∧ B) → S
(A → S) ∨ (B → S)

3. Transitivity

If Biden dies before the election, Trump will win.
If Trump wins the election, Biden will retire to his home.
Therefore, if Biden dies before the election, Biden will retire to his home.

B → T
T → R
B → R

4. Principle of Explosion

Either zombies will rise from the ground if I bury a chicken head in my backyard, or zombies will rise from the ground if I don’t bury a chicken head in my backyard.

(B → D) ∨ (¬B → D) is a tautology

5. Contraposition

If I buy a car, I won’t buy a Pontiac.
Therefore, if I buy a Pontiac, I won’t buy a car.

C → ¬P
P → ¬C

6. Simplification

If John is in London then he’s in England, and if he’s in Paris then he’s in France.
Therefore, either (1) if John’s in London he’s in France or (2) if John’s in Paris then he’s in England.

(L → E) ∧ (P → F)
(L → F) ∨ (P → E)

7. False Antecedent

It’s not the case that if God exists then human life is a product of random chance.
Therefore, God exists.

¬(G → C)
G

8. True Consequent

If I will have eternal life if I believe in God, then God must exist.
I do not believe in God.
Therefore, God exists.

(B → E) → G
~B
G

You can check for yourself that each of these is logically valid! Can you figure out what’s going wrong in each case?

A Dice Puzzle

Today I have a wonderfully counterintuitive puzzle to share!

You and a friend each throw a dice. Each of you can see how your own die landed, but not how your friend’s die landed. Each of you is to guess how the other’s die landed. If you both guess correctly, then you each get a reward. But if only one of you guesses correctly, neither of you get anything.

The two die rolls are independent and you are not allowed to communicate with your friend after the dice have been thrown, though you can coordinate beforehand. Given this, you would expect that you each have a 1 in 6 chance of guessing the other’s roll correctly, coming out to a total chance of 1 in 36 of getting the reward.

The question is: Is it possible to do any better?

Answer below, but only read on after thinking about it for yourself!

 

(…)

 

(…)

 

(Spoiler space)

 

(…)

 

(…)

 

The answer is that remarkably, yes, you can do better! In fact, you can get your chance of getting the reward as high as 1 in 6. This should seem totally crazy. You and your friend each have zero information about how the other die roll turned out. So certainly each of you has a 1 in 6 chance of guessing correctly. The only way for the chance of both guessing correctly to drop below 1 in 36 would be if the two guesses being correct were somehow dependent on each other. But the two die rolls are independent of one another, and no communication of any kind is allowed once the dice have been rolled! So from where does the dependence come? Sure you can coordinate beforehand, but it’s hard to imagine how this could help out.

It turns out that the coordination beforehand does in fact make a huge difference. Here’s the strategy that both can adopt in order to get a 1 in 6 chance of getting the reward: Each guesses that the others’ die lands the same way that their own die landed. So if my die lands 3, I guess that my friend’s die landed 3 as well. This strategy succeeds whenever the dice actually do land the same way. And what’s the chance of this? 6 out of 36, or 1 out of 6!

1 1       2 1       3 1       4 1       5 1       6 1
1 2       2 2       3 2       4 2       5 2       6 2
1 3       2 3       3 3       4 3       5 3       6 3
1 4       2 4       3 4       4 4       5 4       6 4
1 5       2 5       3 5       4 5       5 5       6 5
1 6       2 6       3 6       4 6       5 6       6 6

Sum and Product Puzzle

X and Y are two integers.

X < Y
X > 1
Y > 1
X + Y < 100

S and P are two perfect logicians. S knows X + Y, and P knows X × Y.

Everything I’ve just said is common knowledge. S and P have the following conversation:

S: “P, you don’t know X and Y”
P: “Now I do know X and Y!”
S: “And now so do I!”

What are X and Y?

Once you figure out that, here’s a question: If instead of saying that X + Y < 100, we say X + Y < N, then what’s the range of values of N for which this puzzle has a unique solution?

The full solution to the dog puzzle

A couple of days ago I posted A logic puzzle about dogs. Read that post first and try solving it before reading on!

Below is the full explanation of the puzzle. Heavy spoilers, obviously.

The dog society is structured identically to the Robinson axioms for natural number arithmetic. Dogs are numbers, Spot is 0, the alpha of n is n + 1, the referee for n and m is n + m, the counselor for n and m is n × m, and the strength relation is the < relation. This means that “the marriage counselor for Spot’s alpha and the referee of a fight between Spot’s alpha and Spot’s alpha” is translated to 1 × (1 + 1) = 2, which is Spot’s alpha’s alpha. In Robinson arithmetic, you can also prove that ∀n (n < n + 1).

As for the question of if it’s possible for a dog to be stronger than Spot, Spot’s alpha, Spot’s alpha’s alpha, and so on: The primary difference between Robinson arithmetic and Peano arithmetic (the usual axiomatization of natural numbers) is that the Robinson axioms have no induction axiom (which would be something like “If Spot has a property, and if the alpha of any dog with the property also has the property, then all dogs have the property”). The induction axiom serves to rule out many models of the axioms that are not actually the natural numbers.

If the induction axiom is stated using second-order logic, then the axiom system uniquely pins down (ℕ,+,×,>) and there are no dogs besides those in Spot’s hierarchy. But the induction axiom cannot be stated as a single axiom in first order logic, since it involves quantifying over all properties. For first-order Peano arithmetic, we instead have an infinite axiom schema, one for each property that is definable within the first-order language. This turns out to be strictly weaker than the single second-order axiom, as there are some properties of numbers that cannot be described in a first-order language (like being larger than a finite number of numbers).

What this amounts to is that first-order Peano arithmetic with its infinite axiom schema is too weak to pin down the natural numbers as a unique model. There are what’s called nonstandard models of first order PA, which contains the ordinary numbers but also an infinity of weird extra numbers.(In fact, there exist models of first-order PA with every infinite cardinality!) Some of these numbers have the property that they are bigger than all the regular natural numbers.And since Robinson arithmetic is strictly weaker than first-order PA (lacking an induction axiom as it does), this means that Robinson arithmetic is also not strong enough to rule out numbers greater than all elements of ℕ. Which means that we cannot prove that there are no dogs stronger than every dog in Spot’s hierarchy!

I made this puzzle to illustrate three things: First, how the same axioms, and even the same models of the same axioms, can have wildly different interpretations, and that a shift in how you think about these axioms can make seemingly impossible tasks trivial. Second, how axiom systems for structures like ℕ can (and inevitably do) fail to capture important and intuitively obvious features of the structure. And third, how logic is so interesting! Just trying to create a simple rule system to describe one of the most natural and ordinary mathematical structures that we ALL use ALL THE TIME for reasoning about the world, turns out to be so nontrivial, and in fact impossible to do perfectly!

Three paradoxes of self-reference

I’ve spent a lot of time in the past describing the paradoxes that arise when we try to involve infinities into our reasoning. Another way to get paradoxes aplenty is by invoking self-reference. Below are three of the best paradoxes of self-reference for you to sort out.

In each case, I want you to avoid the temptation to just say “Ah, it’s just self-reference that’s causing the problem” and feel that the paradox is thus resolved. After all, there are plenty of benign cases of self-reference. Self-modifying code, flip-flops in computing hardware, breaking the fourth wall in movies, and feeding a function as input to itself are all examples. Self-referential definitions in mathematics are also often unobjectionable: as an example, we can define the min function by saying y = min(X) iff y is in X and for all elements x of X, y ≤ x (the definition quantifies over a group of objects that includes the thing being defined). So if we accept that self-reference is not necessarily paradoxical (just as infinity is sometimes entirely benign), then we must do more to resolve the below paradoxes than just say “self-reference.”

1. Berry’s Paradox

Consider the set of integers definable in an English sentence of under eighty letters. This set is finite, because there are only a finite number of possible strings of English characters of under eighty letters. So since this set is finite and there are an infinity of integers, there must be a smallest integer that’s not in the set.

But hold on: “The smallest positive integer not definable in under eighty letters” appears to now define this integer, and it does so with only 67 letters! So now it appears that there is no smallest positive integer not definable in under eighty letters. And that means that our set cannot be finite! But of course, the cardinality of the set of strings cannot be less than the cardinality of the set of numbers those strings describe. So what’s going on here?

2. Curry’s paradox

“If this sentence is true, then time is infinite.”

Curry’s paradox tells us that just from the existence of this sentence (assuming nothing about its truth value), we can prove that time is infinite.

Proof 1

Let’s call this sentence P. We can then rewrite P as “If P is true, then time is infinite.” Now, let’s suppose that the sentence P is true. That means the following:

Under the supposition that P is true, it’s true that “If P is true, then time is infinite.”

And under the supposition that P is true, P is true.

So under the supposition that P is true, time is infinite (by modus ponens within the supposition).

But this conclusion we’ve just reached is just the same thing as P itself! So we’ve proven that P is true.

And therefore, since P is true and “If P is true, then time is infinite” is true, time must be infinite!

If you’re suspicious of this proof, here’s another:

Proof 2

If P is false, then it’s false that “If P is true then time is infinite.” But the only way that sentence can be false is if the antecedent is true and the consequent false, i.e. P is true and time is finite. So from P’s falsity, we’ve concluded P’s truth. Contradiction, so P must be true.

Now, if P is true, then it’s true that “If P is true, then time is infinite”. But then by modus ponens, time must be infinite.

Nothing in our argument relied on time being infinite or finite, so we could just as easily substitute “every number is prime” for “time is infinite”, or anything we wanted. And so it appears that we’ve found a way to prove the truth any sentence! Importantly, our conclusion doesn’t rest on the assumption of the truth of the sentence we started with! All it requires is the *existence of the sentence*. Is this a proof of the inconsistency of propositional logic? And if not, then where exactly have we gone wrong?

3. Curry’s paradox, Variant

Consider the following two sentences:

1) At least one of these two sentences is false.
2) Not all numbers are prime.

Suppose that (1) is false. Well then at least one of the two sentences is false, which makes (1) true! This is a contradiction, so (1) must be true.

Since (1) is true, at least one of the two sentences must be false. But since we already know that (1) is true, (2) must be false. Which means that all numbers are prime!

Just like last time, the structure of the argument is identical no matter what we put in place of premise 2, so we’ve found a way to disprove any statement! And again, we didn’t need to start out by assuming anything about the truth values of sentences (1) and (2), besides that they have truth values.

Perhaps the right thing to say, then, is that we cannot always be sure that self-referential statements actually have truth values. But then we have to answer the question of how we are to distinguish between self-referential statements that are truth-apt and those that are not! And that seems very non-trivial. Consider the following slight modification:

1) Both of these two sentences are true.
2) Not all numbers are prime.

Now we can just say that both (1) and (2) are true, and there’s no problem! And this seems quite reasonable; (1) is certainly a meaningful sentence, and it seems clear what the conditions for its truth would be. So what’s the difference in the case of our original example?

A logic puzzle about dogs

Spot is a dog. Every dog has one alpha (also a dog), and no two dogs have the same alpha. But Spot, alone amongst dogs, isn’t anybody’s alpha. 🙁

For any two dogs, there is an assigned referee in case they get into a fight and an assigned marriage counselor in case they get married. The dogs have set up the following rules for deciding who will be the referees and counselors for who:

If Spot fights with any dog, the other dog gets to be the referee. The referee of a fight between dog 1’s alpha and dog 2 has to be the alpha of the referee of a fight between dogs 1 and 2.

Spot has to be his own marriage counselor, no matter who he marries. The marriage counselor for dog 1’s alpha and dog 2 has to referee any fight between dog 2 and the marriage counselor for dog 1 and dog 2.

Finally, dog 1 is stronger then dog 2 if and only if dog 1 is the referee for dog 2 and some other dog. Strength is transitive, and no dog is stronger than itself.

Question 1: Who’s the marriage counselor for Spot’s alpha and the referee of a fight between Spot’s alpha and Spot’s alpha?

Question 2: How many dogs are there?

Question 3: Is the referee for dog 1’s alpha and dog 2 always the same as the referee for dog 2’s alpha and dog 1? What if we asked the same question about marriage counselors?

Question 4: Is any dog stronger than their own alpha?

Bonus Question: Is it possible for there to be a dog that’s stronger than Spot, Spot’s alpha, Spot’s alpha’s alpha, and so on?

A challenge: Try to figure out a trick that allows you to figure out the above questions in your head. I promise, it’s possible!

A suspicious proof of God’s existence

Consider the following argument:

  1. If I will have eternal life if I believe in God, then God must exist.
  2. I do not believe in God.
  3. Therefore, God exists.

Intuitively, it seems possible for (1) and (2) to be true and yet (3) to be false.

But now let’s formalize the argument.

B = “I believe in God”
E = “I will get eternal life”
G = “God exists”

  1. (B → E) → G
  2. ~B
  3. Assume ~G
  4. ~(B → E), modus tollens (1,3)
  5. B & ~E, (4)
  6. B, (5)
  7. B & -B, (6,2)
  8. G, proof by contradiction (2 through 7)

This argument is definitely logically valid, so were our initial intuitions mistaken? And if not, then what’s going on here?

Six Case Studies in Consequentialist Reasoning

Consequentialism is a family of moral theories that say that an act is moral or immoral based on its consequences. If an act has overall good consequences then it is moral, and if it has bad consequences then it is immoral. What precisely counts as a “good” or “bad” consequence is what distinguishes one consequentialist theory from another. For instance, act utilitarians say that the only morally relevant feature of the consequences of our actions is the aggregate happiness and suffering produced, while preference utilitarians say that the relevant feature of the consequences is the number and strength of desires satisfied. Another form of consequentialism might strike a balance between aggregate happiness and social equality.

What all these different consequentialist theories have in common is that the ultimate criteria being used to evaluate the moral status of an action is only a function of the consequences of that action, as opposed to, say, the intentions behind the action, or whether the action is an instance of a universalizable Kantian rule.

In this essay, we’ll explore some puzzles in consequentialist theories that force us to take a more nuanced and subtle view of consequentialism. These puzzles are all adapted from Derek Parfit’s Reasons and Persons, with very minor changes.

First, we’ll consider a simple puzzle regarding how exactly to evaluate the consequences of one’s actions, when one is part of a collective that jointly accomplishes some good.

Case 1: There are 100 miners stuck in a mineshaft with flood waters rising. These men can be brought to the surface in a lift raised by weights on long levers. The leverage is such that just four people can stand on a platform and provide sufficient weight to raise the lift and save the lives of the hundred men. But if any fewer than four people stand on the platform, it will not be enough to raise the lift. As it happens, you and three other people happen to be standing there. The four of you stand on the platform, raising the lift and saving the lives of the hundred men.

The question for us to consider is, how many lives did you save by standing on the platform? The answer to this question matters, because to be a good consequentialist, each individual needs to be able to compare their contribution here to the contribution they might make by going elsewhere. As a first thought, we might say that you saved 100 lives by standing on the platform. But the other three people were in the same position as you, and it seems a little strange to say that all four of you saved 100 lives each (since there weren’t 400 lives saved total). So perhaps we want to say that each of you saved one quarter of the total: 25 lives each.

Parfit calls this the Share-of-the-Total View. We can characterize this view as saying that in general, if you are part of a collective of N people who jointly save M lives, then your share of lives saved is M/N.

There are some big problems with this view. To see this, let’s amend Case 1 slightly by adding an opportunity cost.

Case 2: Just as before, there are 100 miners stuck in a mineshaft with flood waters rising, and they can be saved by four or more people standing on a platform. This time though, you and four other people happen to be standing there. The other four are going to stand on the platform no matter what you do. Your choice is either to stand on the platform, or to go elsewhere to save 10 lives. What should you do?

The correct answer here is obviously that you should leave to save the 10 lives. The 100 miners will be saved whether you stay or leave, and the 10 lives will be lost if you stick around. But let’s consider what the Share-of-the-Total View says. According to this view, if you stand on the platform, your share of the lives saved is 100/5 = 20. And if you leave to go elsewhere, you only save 10 lives. So you save more lives by staying and standing on the platform!

This is a reductio of the Share-of-the-Total View. We must revise this view to get a sensible consequentialist theory. Parfit’s suggestion is that we say that when you join others who are doing good, the good that you do is not just your own share of the total benefit. You should also add to your share the change that you caused in the shares of the benefits produced by each other by joining. On their own, the four would each have a share of 25 lives. So by joining, you have a share of 20 lives, minus the 5 lives that have been reduced from the share of each of the other four. In other words, by joining, you have saved 20 – 5(4) lives, in other words, 0 lives. And of course, this is the right answer, because you have done nothing at all by stepping onto the platform!

Applying our revised view to Case 1, we see that if you hadn’t stepped onto the platform, zero lives would be saved. By stepping onto the platform, 100 lives are saved. So your share of those lives is 25, plus 25 lives for each of the others that would have had zero without you. So your share is actually 100 lives! The same applies to the others, so in our revised view, each of the four is responsible for saving all 100 lives. Perhaps on reflection this is not so unintuitive; after all, it’s true for each of them that if they change their behavior, 100 lives are lost.

Case 3: Just as in Case 2, there are 100 miners stuck in a mineshaft. You and four others are standing on the platform while the miners are slowly being raised up. Each of you know of an opportunity to save 10 lives elsewhere (a different 10 lives for each of you), but to successfully save the lives you have to leave immediately, before the miners are rescued. The five of you have to make your decision right away, without communicating with each other.

We might think that if each of the five of you reasons as before, each of you will go off and save the other 10 lives (as by staying, they see that they are saving zero lives). In the end, 50 lives will be saved and 100 lost. This is not good! But in fact, it’s not totally clear that this is the fault of our revised view. The problem here is lack of information. If each of the five knew what the other four planned on doing, then they would make the best decision (if all four planned to stay then the fifth would leave, and if one of the other four planned to leave then the fifth would stay). As things stand, perhaps the best outcome would be that all five stay on the platform (losing the opportunity to save 10 extra lives, but ensuring the safety of the 100). If they can use a randomized strategy, then the optimal strategy is to each stay on the platform with probability 97.2848% (saving an expected 100.66 lives)

Miners Consequentialism

Let’s move on to another type of scenario.

Case 4: X and Y simultaneously shoot and kill me. Either shot, by itself, would have killed.

The consequence of X’s action is not that I die, because if X had not shot, I would have died by Y’s bullet. And the same goes for Y. So if we’re evaluating the morality of X or Y’s action based on its consequences, it seems that we have to say that neither one did anything immoral. But of course, the two of them collectively did do something immoral by killing me. What this tells us that the consequentialist’s creed cannot be “an act is immoral if its consequences are bad”, as an act can also be immoral if it is part of a set of acts whose collective consequences are bad.

Inheriting immorality from group membership has some problems, though. X and Y collectively did something immoral. But what about the group X, Y, and Barack Obama, who was napping at home when this happened? The collective consequences of their actions were bad as well. So did Obama do something immoral too? No. We need to restrict our claim to the following:

“When some group together harm or benefit other people, this group is the smallest group of whom it is true that, if they had all acted differently, the other people would not have been harmed, or benefited.” -Parfit

A final scenario involves the morality of actions that produce imperceptible consequences.

Case 5: One million torturers stand in front of one million buttons. Each button, if pushed, induces a tiny stretch in each of a million racks, each of which has a victim on it. The stretch induced by a single press of the button is so minuscule that it is imperceptible. But the stretch induced by a million button presses produces terrible pain in all the victims.

Clearly we want to say that each torturer is acting immorally. But the problem is that the consequences of each individual torturer’s action are imperceptible! It’s only when enough of the torturers press the button that the consequence becomes perceptible. So what we seem to be saying is that it’s possible to act immorally, even though your action produces no perceptible change in anybody’s conscious experience, if your action is part of a collection of actions that together produce negative changes in conscious experiences.

This is already unintuitive. But we can make it even worse.

Case 6: Consider the final torturer of the million. At the time that he pushes his button, the victims are all in terrible agony, and his press doesn’t make their pain any perceptibly worse. Now, imagine that instead of there being 999,999 other torturers, there are zero. There is just the one torturer, and the victims have all awoken this morning in immense pain, caused by nobody in particular. The torturer presses the button, causing no perceptible change in the victims’ conditions. Has the torturer done something wrong?

It seems like we have to say the same thing about the torturer in Case 6 as we did in Case 5. The only change is that Nature has done the rest of the harm instead of other human beings, but this can’t matter for the morality of the torturer’s action. But if we believe this, then the scope of our moral concerns is greatly expanded, to a point that seems nonsensical. My temptation here is to say “all the worse for consequentialism, then!” and move to a theory that inherently values intentions, but I am curious if there is a way to make a consequentialist theory workable in light of these problems.