On philosophical progress

A big question in the philosophy of philosophy is whether philosophers make progress over time. One relevant piece of evidence that gets brought up in these discussions is the lack of consensus on age old questions like free will, normative ethics, and the mind body problem. If a discipline is progressing steadily towards truth with time, the argument goes, then we should expect that questions that have been discussed for thousands of years should be more or less settled by now. After all, that is what we see in the hard sciences; there are no lingering disputes over the validity of vitalism or the realm of applicability of Newtonian mechanics.

There are a few immediate responses to this line of argument. It might be that the age old questions of philosophy are simply harder than the questions that get addressed by physicists or biologists. “Harder” doesn’t mean “requires more advanced mathematics to grapple with” here, but something more like “it’s unclear what even would count as a conclusive argument for one position or another, and therefore much less clear how to go about building consensus.” Try to imagine what sort of argument would convince you of the nonexistence of libertarian free will with the same sort of finality as a demonstration of time dilation convinces you of the inadequacy of nonrelativistic mechanics.

A possible rejoinder at this point would be to take after the logical positivists and deny the meaningfulness or at least truth-aptness of the big questions of philosophy as a whole. This may go too far; it may well be that a query is meaningful but, due to certain epistemic limitations of ours, forever beyond our ability to decide. (We know for sure that such queries can exist, due to Gödelian discoveries in mathematics. For instance, we know of the existence of a series of numbers that are perfectly well defined, but for which no algorithm can exist to enumerate all of them. The later numbers in this sequence will forever be a mystery to us, and not for lack of meaningfulness.)

I think that the roughly correct position to take is that science is largely about examining empirical facts-of-the-matter, whereas philosophy is largely about analyzing and refining our conceptual framework. While we have a fairly clear set of standards for how to update theories about the empirical world, we are not in possession of such a set of standards for evaluating different conceptual frameworks. The question of “what really are the laws governing the behavior of stuff out there” has much clearer truth conditions than a question like “what is the best way to think about the concepts of right and wrong”; i.e. It’s clearer what counts as a good answer and what counts as a bad answer.

When we’re trying to refine our concepts, we are taking into account our pre-theoretical intuitions (e.g. any good theory of the concept of justice must have something to do with our basic intuitive conception of justice). But we’re not just satisfied to describe the concept solely as the messy inconsistent bundle of intuitions that constitute our starting position on it. We also aim to describe the concept simply, by developing a “theory of justice” that relies on a small set of axioms and from which (the hope is) the rest of our conclusions about justice follow. We want our elaboration of the concept to be consistent, in that we shouldn’t simultaneously affirm that A is an instance of the concept and that A is not an instance of the concept. Often we also want our theory to be precise, even when the concept itself has vague boundaries.

Maybe there are other standards besides these, intuitiveness, simplicity, consistency, and precision. And the application of these standards is very rarely made explicit. But one thing that’s certain is that different philosophers have different mixes of these values. One philosopher might value simplicity more or less than another, and it’s not clear that one of them is doing something wrong by having different standards. Put another way, I’m not convinced that there is one unique right set of standards for conceptual refinement.

We may want to be subjectivists to some degree about philosophy, and say that there are a range of rationally permissible standards for conceptual refinement, none better than any other. This would have the result that on some philosophical questions, multiple distinct answers may be acceptable but some crazy enough answers are not. Maybe compatibilism and nihilism are acceptable stances on free will but libertarianism is not. Maybe dualism and physicalism are okay but not epiphenomenalism. And so on.

This view allows for a certain type of philosophical progress, namely the gradual ruling out of some philosophical positions as TOO weird. It also allows for formation of consensus, through the discovery of philosophical positions that are the best according to all or most of the admissible sets of standards. I think that one example of this would be the relatively recent rise of Bayesian epistemology in philosophy of science, and in particular the Bayesian view of scientific evidence as being quantified by the Bayes factor. In brief, what does it mean to say that an observation O gives evidence for a hypothesis H? The Bayesian not only has an answer to this, but to the more detailed question of to what degree O gives evidence for H. The quantity is cr(O | H) / cr(O), where cr(.) is a credence function encoding somebody’s beliefs before observing O. If this quantity is equal to 1, then O is no evidence for H. If it is greater than 1, then O is evidence for H. And if it’s less than 1, then O is evidence against H.

Not everything in Bayesian epistemology is perfectly uncontroversial, but I would argue that on this particular issue – the issue of how to best formalize the notion of scientific evidence – the Bayesian definition survives all its challenges unscathed. What are some other philosophical questions on which you think there has been definite progress?

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)

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

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

Proving the Completeness of Propositional Logic

The completeness of a logic is a really nice property to establish. For a logic to be complete, it must be that every semantic entailment is also syntactically entailed. Said more simply, it must be that every truth in the language is provable. Gödel’s incompleteness theorems showed us that we cannot have such high hopes for mathematics in general, but we can still establish completeness for some simple logics, such as propositional and first order logic.

I want to post a proof of the completeness of propositional logic here in full for future reference. Roughly the first half of what’s below is just establishing some necessary background, so that this post is fairly self-contained and doesn’t reference lemmas that are proved elsewhere.

The only note I’ll make before diving in is that the notation I(A,P) is a way to denote the smallest set that contains A and is closed under the operations in P. It’s a handy way to inductively define sets that would be enormously complicated to define otherwise. With that out of the way, here we go!

First we define the proof system for propositional logic.

Axiom 1: α→(β→α)
Axiom 2: (α→(β→γ)) → ((α→β)→(α→γ))
Axiom 3: ((¬α)→(¬β)) → (β→α)

The α, β, and γ symbols in these axioms are meant to stand for any well-formed formula. What this means is that we actually have a countable infinity of axioms that fall into the three categories above. For simplicity, I’ll keep calling them “Axioms 1, 2, and 3”, and assume you don’t find it too confusing.

You might also notice that the axioms only involve the symbols → and ¬, but neglect ∧ and ∨. This is okay because → and ¬ are adequate connectives for the semantics of propositional logic (which is to say that any truth function can be expressed in terms of them).

Axioms = {Axiom 1, Axiom 2, Axiom 3}
Deduction rule = Modus Ponens (MP)
….. MP(α, α→β) = β

The set of all provable sentences is just the set of all sentences that includes the axioms and is closed under modus ponens.

Theorems = I(Axioms, MP)

We can also easily talk about the set of sentences that can be proven from assumptions in a set Σ:

Th(Σ) = I(Axioms ∪ Σ, MP)
Notation: Σ ⊢ α iff α ∈ Th(Σ)

With that out of the way, let’s establish some basic but important results about the propositional proof system.

Monotonicity: If Σ ⊆ Σ’, then Th(Σ) ⊆ Th(Σ’).

Strong monotonicity: If Σ ⊢ Σ’, then Th(Σ’) ⊆ Th(Σ).

Intuitively, monotonicity says that if you expand the set of assumptions, you never shrink the set of theorems. Strong monotonicity says that if Σ can prove everything in Σ’, then Σ’ cannot be stronger than Σ. Both of these follow pretty directly from the definition of Th(Σ).

Soundness: If ⊢ α, then ⊨ α.
Proof by structural induction
….. Each axiom is a tautology.
….. Tautology is closed under MP (if ⊨ α and ⊨ (α→β), then ⊨ β).

Extended Soundness: If Σ ⊢ α, then Σ ⊨ α.
Proof by structural induction
….. Σ ⊨ α ∈ Axioms and Σ ⊨ α ∈ Σ.
….. If Σ ⊨ α and Σ ⊨ (α→β), then Σ ⊨ β.

Law of identity: ⊢ (α→α)
….. α→((α→α)→α), Axiom 1
….. (α→((α→α)→α))→((α→(α→α))→(α→α)), Axiom 2
….. (α→(α→α))→(α→α), MP
….. α→(α→α), Axiom 1
….. α→α, MP

Principle of Explosion: If Σ ⊢ α and Σ ⊢ (¬α), then Σ ⊢ β.
….. By strong monotonicity, it suffices to show that Σ ∪ {α} ∪ {¬α} ⊢ β.
……….. (¬α)→((¬β)→(¬α)), Axiom 1
……….. ¬α, Assumption
……….. (¬β)→(¬α), MP
……….. ((¬β)→(¬α))→(α→β), Axiom 3
.………. α→β, MP
……….. α, Assumption
……….. β, MP

And finally, our most important background theorem:

Deduction Theorem: Σ ⊢ (α→β) iff Σ ∪ {α} ⊢ β.
Proof =>
….. Suppose Σ ⊢ (α→β).
….. By monotonicity, Σ ∪ {α} ⊢ (α→β).
….. Also, clearly Σ ∪ {α} ⊢ α.
….. So Σ ∪ {α} ⊢ β.
Proof <=
….. Suppose Σ ∪ {α} ⊢ β.
….. Base cases
………. β ∈ Axioms. (β, β→(α→β), α→β).
………. β ∈ Σ. (β, β→(α→β), α→β).
………. β = α. (⊢ (α→α), so by monotonicity Σ ⊢ (α→α)).
….. Inductive step
………. Suppose Σ ⊢ (α→γ) and Σ ⊢ (α→(γ→𝛿)).
………. By strong monotonicity, suffices to show Σ ∪ {α→γ, α→(γ→𝛿)} ⊢ (α→𝛿)
……………. (α→(γ→𝛿)) → ((α→γ)→(α→𝛿)), Axiom 2
……………. α→(γ→𝛿), Assumption
……………. (α→γ)→(α→𝛿), MP
……………. (α→γ), Assumption
……………. (α→𝛿). MP

Now, let’s go into the main body of the proof. The structure of the proof is actually quite similar to the proof of the compactness theorem I gave previously. First we show that every consistent set of sentences Σ has a maximally consistent extension Σ’. Then show that Σ’ is satisfiable. Now since Σ’ is satisfiable and it’s an extension of Σ, Σ must also be satisfiable. From there it’s a simple matter to show that the logic is complete.

So, let’s define some of the terms I just used.

Σ is consistent iff for no α does Σ ⊢ α and Σ ⊢ (¬α)
….. Equivalently: iff for some α, Σ ⊬ α

Σ is maximally consistent iff Σ is consistent and for every α, either Σ ∪ {α} is inconsistent or Σ ⊢ α.

One final preliminary result regarding consistency before diving into the main section of the proof:

If Σ is satisfiable, then Σ is consistent.
….. Suppose Σ is inconsistent.
….. Then there’s an α such that Σ ⊢ α and Σ ⊢ (¬α).
….. By extended soundness, Σ ⊨ α and Σ ⊨ (¬α).
….. So Σ is not satisfiable.

This is the converse of the result we actually want, but it’ll come in handy. Now, let’s begin to construct our extension!

Any consistent Σ can be extended to a maximally consistent Σ’
….. Choose any ordering {αn} of well-formed-formulas.
….. Define Σ0 = Σ.
….. Σn+1 = Σn if Σn ⊢ (¬αn+1), and Σn ∪ {αn+1} otherwise.
….. For each n, (i) Σn is consistent and (ii) either Σn ⊢ αn or Σn ⊢ (¬αn)
……….. Base case: Σ0 is consistent by assumption, and (ii) doesn’t apply.
……….. Inductive step: Suppose Σn satisfies (i) and (ii). Two cases:
…………….. If Σn ⊢ (¬αn+1), then Σn+1 = Σn. Clearly consistent and satisfies (ii).
…………….. If Σn ⊬ (¬αn+1), then Σn+1 = Σn ∪ {αn+1}. Clearly satisfies (ii), but is it consistent?
………………….. Suppose not. Then Σn+1 ⊢ (¬αn+1), by explosion.
………………….. So Σn+1 ∪ {αn+1} ⊢ (¬αn+1).
………………….. So Σn+1 ⊢ (αn+1 → (¬αn+1)).
………………….. ⊢ ((α→¬α)→¬α), so Σn+1 ⊢ (¬αn+1). Contradiction!

….. Define Σ’ = ∪ Σn. Σ’ is maximally consistent.
……….. Maximality
…………….. Suppose not. Then for some αn, Σ’ ⊬ αn and Σ’ ∪ {αn} is consistent.
…………….. But Σn ⊆ Σ’, and either Σn ⊢ αn or Σn ⊢ (¬αn).
…………….. If Σn ⊢ αn, by monotonicity Σ ⊢ αn. Contradiction. So Σn ⊢ (¬αn).
…………….. By monotonicity, Σ ⊢ (¬αn), so Σ ∪ {αn} ⊢ (¬αn).
…………….. But Σ ∪ {αn} ⊢ αn. So Σ ∪ {αn} is inconsistent. Contradiction!
……….. Consistency
…………….. Suppose Σ’ is inconsistent. Then for some α, Σ’ ⊢ α and Σ’ ⊢ (¬α).
…………….. So there are proofs of α and (¬α) from Σ’.
…………….. Proofs are finite, so each proof uses only a finite number of assumptions from Σ’.
…………….. So we can choose an n such that Σn contains all the needed assumptions.
…………….. Now both proofs from Σ’ are also proofs from Σn.
…………….. So Σn ⊢ αn and Σ’ ⊢ (¬αn).
…………….. So Σn is inconsistent. Contradiction!

Alright, we’re almost there! So now we have that for any consistent Σ, there’s an extension Σ’ that is maximally consistent. We’ll take it a little further and prove that not only is Σ’ maximally consistent, it’s also complete! (This is the purely syntactic sense of completeness, which is that for every sentence α, either Σ’ proves α or refutes α. This is different from the sense of logical completeness that we’re establishing with the proof.)

Σ’ is complete.
….. Σn ⊆ Σ’, and Σn ⊢ αn or Σn ⊢ (¬αn).
….. So by monotonicity Σ’ ⊢ αn or Σ’ ⊢ (¬αn).

Now we have everything we need to show that Σ’, and thus Σ, is satisfiable.

If Σ is consistent, then Σ is satisfiable.
….. Let Σ’ be a maximally consistent extension of Σ.
….. Define vΣ’(p) over propositional variables p:
….. VΣ’(p) = T if Σ’ ⊢ p and F if Σ’ ⊬ p
….. ṼΣ’(α) = T iff Σ’ ⊢ α
……….. Base case: Let α be a propositional variable. Then ṼΣ’(α) = T iff Σ’ ⊢ α by definition of VΣ’.
……….. Inductive steps:
……….. (¬α)
…………….. If Σ’ ⊢ (¬α), then by consistency Σ’ ⊬ α, so ṼΣ’(α) = F, so ṼΣ’(¬α) = T.
…………….. If Σ’ ⊬ (¬α), then by completeness Σ’ ⊢ α. So ṼΣ’(α) = T, so ṼΣ’(¬α) = F.
……….. (α→β)
…………….. Suppose Σ’ ⊢ (α→β). By completeness Σ’ ⊢ α or Σ’ ⊢ (¬α).
………………….. If Σ’ ⊢ α, then Σ’ ⊢ β, so ṼΣ’(β) = T, so ṼΣ’(α→β) = T.
………………….. If Σ’ ⊢ (¬α), then ṼΣ’(α) = F, so ṼΣ’(α→β) = T.
……………. Suppose Σ’ ⊬ (α→β).
………………….. By completeness Σ’ ⊢ ¬(α→β).
………………….. ⊢ (β→(α→β)), so Σ’ ⊬ β on pain of contradiction. So ṼΣ'(β) = F.
………………….. Suppose ṼΣ’(α→β) = T. Then ṼΣ’(α) = F, so Σ’ ⊢ (¬α).
………………….. ⊢ (¬α→(α→β)). So Σ’ ⊢ (α→β). Contradiction.
………………….. So ṼΣ’(α→β) = F.
….. So vΣ’ satisfies Σ’.
….. Σ ⊆ Σ’, so vΣ’ satisfies Σ.
….. So Σ is satisfiable!

Now our final result becomes a four-line proof.

If Σ ⊨ α, then Σ ⊢ α.
….. Suppose Σ ⊬ α.
….. Then Σ ∪ {¬α} is consistent.
….. So Σ ∪ {¬α} is satisfiable.
….. So Σ ⊭ α.

And we’re done! We’ve shown that if any sentence α is semantically entailed by a set of sentences Σ, then it must also be provable from Σ! If you’ve followed this proof all the way, pat yourself on the back.

With the Completeness Theorem in hand, the proof of the Compactness Theorem goes from several pages to a few lines. It’s so nice and simple that I just have to include it here.

If Σ is finitely satisfiable, then Σ is satisfiable.

….. Suppose Σ is not satisfiable.
….. Then Σ is not consistent.
….. So there is some α for which Σ ⊢ α and Σ ⊢ (¬α).
….. Since proofs are finite, there must be some finite subset Σ* of Σ such that Σ* ⊢ α and Σ* ⊢ (¬α).
….. By soundness, Σ* ⊨ α and Σ* ⊨ (¬α).
….. So Σ* is not satisfiable!

In other words, if Σ is not satisfiable, then there’s some finite subset of Σ that’s also not satisfiable. This is the Compactness Theorem! Previously we proved it entirely based off of the semantics of propositional logic, but now we can see that it is also provable as a consequence of the finite nature of our proof system!

Four Pre-Gödelian Limitations on Mathematics

Even prior to the devastating Incompleteness Theorems there were hints of what was to come. I want to describe and prove four results in mathematical logic that don’t depend on Incompleteness at all, but establish some rather serious limitations on the project of mathematics.

Here are the four. I’ll go through them in order of increasing level of sophistication required to prove them.

  • Indescribable sets of possible worlds
  • Noncompossibility theorem
  • Inevitable nonstandard numbers
  • Mysterious missing subsets

1. Indescribable sets of possible worlds

I already talked about this one here. The basic idea is that even in our safest and least troublesome logic, propositional calculus, it turns out that the language is insufficient to fully capture all the semantic notions. It’s the first hint at something going awry with syntax and semantics, where the semantics can outpace the syntax and leave axiomatic mathematics behind.

So to recap: the result is that in propositional logic there are sets of truth assignments that can not be “described” by any set of propositional sentences (even allowing infinite sets!). A set of sentences is said to “describe” a set of truth assignments if that set of truth assignments is the unique set of truth assignments consistent with all those sentences being true. If we think of truth assignments as possible worlds, and sets of sentences as descriptions of sets of possible worlds, then this result says that there are sets of possible worlds in propositional semantics that cannot be described by any propositional syntax.

The proof of this is astoundingly simple: just look at the cardinality of the set of descriptions and the cardinality of the set of sets of possible worlds. The second is strictly larger than the first, so any mapping from descriptions to sets of possible worlds will of necessity leave some sets of possible worlds out. In fact, it also tells us that virtually all sets of possible worlds are not describable!

2. Noncompossibility Theorem

The noncompossibility theorem is a little-known theorem that establishes a serious limitation on our ability to describe mathematical structures. Here’s what it says. Suppose that you have a description of a countably infinite structure (like, say, the natural numbers, which have a countable infinity of objects) that has the following three properties:

(1) The language has a term for denoting every object in the structure (like 0, 1, 2, 3, 4, and so on)
(2) The axioms in your description are weakly complete: if something is inconsistent with the axioms, it can be proven false.
(3) There is some algorithm for determining whether any given sentence is an axiom.

The noncompossibility theorem tells us that if you have all three of these properties, then your axioms will fail to uniquely pick out your intended structure, and will include models that have extra objects that aren’t in the structure.

Let’s prove this.

We’ll denote the mathematical structure that we’re trying to describe as M and our language as L. We choose L to have sufficient syntactic structure to express the truths of M. From L, we select a decidable set of sentences X with the goal that all these sentences be true of M. We now select a proof system F in L such that for any finite extension L* to L involving only new constant terms, and for any Y ⊆ L*, if X ∪ Y is not satisfiable, then F refutes some finite subset of X ∪ Y.

(As an aside: Why care about this strange weak form of completeness? Well, intuitively all that it’s saying is that our axioms should be able to rule out any set of sentences that are inconsistent with them using some finite proof, as long as those sentences only use finitely many additional constant symbols. This is relatively weak compared to the usual notions of completeness that logicians talk about, which makes it an even better choice for our purposes, as the weaker the axiom the harder to deny.)

Our assumptions can now be written:

(0) |M| is countably infinite.
(1) ∀m ∈ M, ∃tm ∈ terms(X) such that (m = tm)
(2) If we extend L to L* by adding finitely many constant symbols, then for any Y ⊆ L*, if X ∪ Y is not satisfiable, then F refutes some finite subset of X ∪ Y.
(3) X is recursively enumerable.

Our proof starts by adding a new constant term c to our language and constructing an extension of X:

Y = X ∪ {c ≠ tm | m ∈ M}

In other words, Y is X but supplemented with the assertion that there exists an object that isn’t in M. If we can prove that Y is satisfiable, then this entails that X is also satisfiable by the same truth assignment. And this means that there is a model of X in which there are extra objects that aren’t in M.

We proceed with proof by contradiction. Suppose that Y is not satisfiable. Then, by assumption (2), we must be able to refute some finite subset Z of X ∪ Y. But since Z is finite, it involves only finitely many terms tm. And since M is countably infinite, there will always be objects in M that are not equal to any of the chosen terms! So we can’t refute any finite subset of Z! Thus Y is satisfiable.

And if Y is satisfiable, then so must be X, as Y is a superset of X. And since Y is satisfiable, then there’s some truth assignment v that satisfies all of v. But then v also satisfies X, as X is a subset of Y and removing axioms cannot rule out models, only add more! So we’ve proven that X has a model in which there is an object that is not equal to any of the objects in M. That is, X is not categorical: it does not uniquely describe M.

Tennant described this theorem as saying that “in countably infinite realms, you cannot know both where you are and where you are going.” More dully, we cannot have a satisfactory theory of a countably infinite mathematical structure that is both categorical and weakly complete. This isn’t super shocking by today’s standards, but it’s quite cool when you consider how little elaborate theoretical apparatus is required to prove it.

3. Inevitable nonstandard numbers

Suppose we have some first-order theory T that models the natural numbers. Take this theory and append to it a new constant symbol c, as well as an infinite axiom schema saying “c > 0” , “c > 1” , “c > 2”, and so on forever. Call this new theory T*.

Does T* have a model? Well by the compactness theorem, it has a model as long as all its finite subsets have a model. And for every finite subset of T*s axioms, the natural numbers are a model! So T* does have a model. Could this model be the natural numbers? Clearly not, because to satisfy T*, there must be a number greater than all the natural numbers. So whatever the model of T* is, it’s not the standard natural numbers. Let’s call it a nonstandard model, and label it ℕ*.

Here’s the final step of the proof: ℕ* is a model of T*, and T* is a superset of T, so ℕ* must also be a model of T! And thus we find that in any logic with a compactness theorem, a theory of the natural numbers will have models with nonstandard numbers that are greater than all of ℕ.

It’s one of my favorite proofs, because it’s so easy to describe and has such a devastating conclusion. It’s also an example of the compactness theorem using the existence of one type of model (ℕ for each of the finite cases) to prove the existence of something entirely different (ℕ* for the infinite case).

4. Mysterious missing subsets

The Löwenheim-Skolem theorem tells us that if a first-order theory has a model with an infinite cardinality, then it has models with every infinite cardinality. This places a major restriction on our ability to describe an infinite mathematical structure using first order logic. For if we were to try to single out the natural numbers, say, we would inevitably end up failing to rule out models of our axioms that are the cardinality of the real numbers, or worse, or the set of functions from real numbers to real numbers, and so on for all other possible cardinalities.

When applied to set theory, this implies a result that seems on its face to be a straightforward contradiction. Namely, Löwenheim-Skolem tells us that any first order axiomatization of sets will inevitably have a model that contains only a countable infinity of sets. But this seems bizarre, as all we appear to need to rule out countably infinite universes of sets is one axiom that asserts the existence of a countably infinite set, and another that asserts that admits the power-set of any set to the universe of sets. Then we will be forced to admit that there is a set which is the power set of a countably infinite set, and as Cantor’s famous diagonal argument shows, that this set is uncountably large.

So on the one hand, Cantor tells us that there are sets that contain uncountably many objects. And on the other hand, Löwenheim-Skolem tell us that there is a model of set theory with only countably many objects. This dichotomy is known by the name Skolem’s paradox. It appears to be a straightforward contradiction, but it’s not.

What Skolem realized was that the formal notion of a power set, which is something like “the set P(X) such that for all sets Y, if Y is a subset of X, then Y is an element P(X)”, relies on a quantification over all sets, and that in a countable universe of sets, that quantification ranges over only a countable number of objects. In other words, P(X) is only uncountable if our quantifier ranges over all possible sets, but for a countable model, there are sets that are not describable within the model. This means that the notion of a power set is relative to your model of set theory! In fact, there’s no way in first order logic to unambiguously pin down what you mean by “power set” in such a way that all models will agree on what P(X) actually contains. It also means that the notions of cardinality and countability are relative to your model! In Skolem’s words, “even the notions ‘finite’, ‘infinite’, ’simply infinite sequence’ and so forth turn out to be merely relative within axiomatic set theory.”

A challenge to constructivists

Constructive mathematicians do not accept a proof of existence unless it provides a recipe for how to construct the thing whose existence is being asserted. Constructive mathematics is quite interesting, but it also appears to have some big problems. Here’s a challenge for constructivists:

Suppose that I hand you some complicated function f from a set A to another set B. I ask you: “Can every element in B be reached by applying the function to an element in A?” In other words, is f surjective?

Now, it so happens that the cardinality of B is greater than the cardinality of A. That’s sufficient to tell us that f can’t be surjective, as however it maps elements there will always be some left over. So we know that the answer is “no, we can’t reach every element in B.” But we proved this without explicitly constructing the particular element in B that can’t be reached! So a constructivist will be left unsatisfied.

The trick is that I’ve made this function extremely complicated, so that there’s no clever way for them to point to exactly which element is missing. Would they say that even though |B| is strictly larger than |A|, it could still be somehow that every element in B is in the image of f? Imagine asking them to bet on this proposition. I don’t think any sane person would put any money on the proposition that f is onto.

And as a final kicker, our sets don’t even have to be infinite! Let |A| = 20 and |B| = 21. I describe a function from A to B, such that actually computing the “missing element” involves having to calculate the 21st Busy Beaver number or something. And the constructivist gets busy searching for the particular element in B that doesn’t get mapped to, instead of just saying “well of course we can’t map 20 elements to 21 elements!”

Even simpler, let F map {1} to {1,2} as follows: F(1) = 1 if the last digit of the 20th busy beaver number is 0, 1, 2, 3, or 4, and F(1) = 2 otherwise. Now to prove constructively that there is an element in {1, 2} that isn’t in the image of F requires knowing the last digit of the 20th busy beaver number, which humans will most likely never be able to calculate (we’re stuck on the fifth one now). So a constructivist will be remain uncertain on the question of if F is surjective.

But a sane person would just say “look, of course F isn’t surjective; it maps one object to two objects. You can’t do this without leaving something out! It doesn’t matter if we don’t know which element is left out, it has to be one of them!”

And if humanity is about to meet an alien civilization with immense computational power that knows all the digits of the 20th busy beaver number, the standard mathematician could bet their entire life savings on F not being surjective at any odds whatsoever, and the constructive mathematician would bet in favor of F being surjective at some odds. And of course, the constructivist would be wrong and lose money! So this also means that you have a way to make money off of any constructivist mathematicians you encounter, so long as we’re about to make contact with advanced aliens.

Describing the world

Wittgenstein starts his Tractatus Philosophicus with the following two sentences.

1. The world is everything that is the case.

1.1 The world is the totality of facts, not of things.

Let’s take him up on this suggestion and see how far we get. In the process, we’ll discover some deep connections to theorems in mathematical logic, as well as some fascinating limitations on the expressive powers of propositional and first order logic.

We start out with a set of atomic propositions. For a very simple world, we might only need a finite number of these: “Particle 1 out of 3 has property 1 out of 50”, “Particle 2 of 3 has property 17 out of 50”, and so on. More realistically, the set of atomic propositions will be infinite (countable if the universe doesn’t have any continuous properties, and uncountable otherwise).

For simplicity, we’ll imagine labeling our set of atomic propositions P1, P2, P3, and so on (even though this entails that there are at most countably many, nothing important will rest on this assumption.) We combine these atomic propositions with the operators of propositional logic {(, ), ¬, ∧, ∨, →}. This allows us to build up more complicated propositions, like ((P7∧P2)→(¬P13)). This will be the language that we use to describe the world.

Now, the way that the world is is just a consistent assignment of truth values to the set of all grammatical sentences in our language. For example, one simple assignment of truth values is the one that assigns “True” to all atomic propositions. Once we’ve assigned truth values to all the atomic propositions, we get the truth values for the rest of the set of grammatical sentences for free, by the constraint that our truth assignment be consistent. (For instance, if P1 and P2 are both true, then (P1∧P2) must also be true.)

Alright, so the set of ways the world could be corresponds to the set of truth assignments over our atomic propositions. The final ingredient is the notion that we can encode our present knowledge of the world as a set of sentences. Maybe we know by observation that P5 is true, and either P2 or P3 is true but not both. Then to represent this state of knowledge, we can write the following set of sentences:

{P5, (P2∨P3), ¬(P2∧P3)}

Any set of sentences picks out a set of ways the world could be, such that each of these possible worlds is compatible with that knowledge. If you know nothing at all, then the set of sentences representing your knowledge will be the empty set {}, and the set of possible worlds compatible with your knowledge will be the set of all possible worlds (all possible truth assignments). On the other extreme, you might know the truth values of every atomic proposition, in which case your state of knowledge uniquely picks out one possible world.

In general, as you add more sentences to your knowledge-set, you cut out more and more possible worlds. But this is not always true! Ask yourself what the set of possible worlds corresponding to the set {(P1∨¬P1), (P2∨¬P2), (P3∨¬P3)} is. Since each of these sentences is a tautology, no possible worlds are eliminated! So our set of possible worlds is still the set of all worlds.

Now we get to an interesting question: clearly for any knowledge-set of sentences, you can express a set of possible worlds consistent with that knowledge set. But is it the case that for any set of possible worlds, you can find a knowledge-set that uniquely picks it out? If I hand you a set of truth assignment functions and ask you to tell me a set of propositions which are consistent with that set of worlds and ONLY that set, is that always possible? Essentially, what we’re asking is if all sets of possible worlds are describable.

We’ve arrived at the main point of this essay. Take a minute to ponder this and think about whether it’s possible, and why/why not! For clarification, each sentence can only be finitely long. But! You’re allowed to include an infinity of sentences.


(Spoiler-hiding space…)


If there were only a finite number of atomic propositions, then you could pick out any set of possible worlds with just a single sentence in conjunctive normal form. But when we start talking about an infinity of atomic propositions, it turns out that it is not always possible! There are sets of possible worlds that are literally not describable, even though our language includes the capacity to describe each of those words and we’re allowed to include an infinite set of sentences.

There’s a super simple proof of this. Let’s give a name to the cardinality of the set of sentences: call it K. (We’ve been tacitly acting as if the cardinality is countable this whole time, but that doesn’t actually matter.) What’s the cardinality of the set of all truth assignments?

Well, each truth assignment is a function from all sentences to {True, False}. And there are 2K such assignments. 2K is strictly larger than K, so there are more possible worlds than there are sentences. Now, the cardinality of the set of sets of sentences is also 2K. But the set of SETS of truth of assignments is 22^K!

What this means is that we can’t map sets of sentences onto sets of truth assignments without leaving some things out! This proof carries over to predicate logic as well. The language for both propositional and predicate logic is unable to express all sets of possible worlds corresponding to that language!

I love this result. It’s the first hint in mathematical logic that syntax and semantics can come apart.

That result is the climax of this post. What I want to do with the rest of this post is to actually give an explicit example of a set of truth assignments that are “indescribable” by any set of sentences, and to prove it. Warning: If you want to read on, things will get a bit more technical from here.

Alright, so we’ll use a shortcut to denote truth assignments. A truth assignment will be written as a string of “T”s and “F”s, where the nth character corresponds to how the truth assignment evaluates Pn. So the all-true truth assignment will just be written “TTTTTT…” and the all-false truth assignment will be written “FFFFF…”. The truth assignment corresponding to P1 being false and everything else true will be written “FTTTTT…”. And so on.

Now, here’s our un-describable set of truth assignments. {“FFFFFF…”, “TFFFFF…”, “TTFFFF…”, “TTTFFF…”, …}. Formally, define Vn to be the truth assignment that assigns “True” to every atomic proposition up to and including Pn, and “False” to all others. Now our set of truth assignments is just {Vn | n ∈ ℕ}.

Let’s prove that no set of sentences uniquely picks out this set of truth assignments. We prove by contradiction. Suppose that we could find a set of sentences that uniquely pick out these truth assignments and none other. Let’s call this set A. Construct a new set of sentences A’ by appending all atomic propositions A: A’ = A ∪ {P1, P2, P3, …}.

Is there any truth assignment that is consistent with all of A’? Well, we can answer this by using the Compactness Theorem: A’ has a truth assignment if and only if every finite subset of A’ has a truth assignment. But every finite subset of A’ involves sentences from A (which are consistent with Vn for each n by assumption), and a finite number of atomic propositions. Since each finite subset of A’ is only asserting the truth of a finite number of atomic sentences, we can always find a truth assignment Vk in our set that is consistent with it, by choosing one that switches to “False” long after the last atomic proposition that is asserted by our finite subset.

This means that each finite subset of A’ is consistent with at least one of our truth assignments, which means that A’ is consistent with at least one of our truth assignments. But A’ involves the assertion that all atomic propositions are true! The only truth assignment that is consistent with this assertion is the all-true assignment! And is that truth assignment in our set? No! And there we have it, we’ve reached our contradiction!

We cannot actually describe a set of possible worlds in which either all atomic propositions are false, or only the first is true, or only the first two are true, or only the first three are true, and so on forever. But this might prompt the question: didn’t you just describe it? How did you do that, if it’s impossible? Well, technically I didn’t describe it. I just described the first four possibilities and then said “and so on forever”, assuming that you knew what I meant. To have actually fully pinned down this set of possible worlds, I would have had to continue with this sentence forever. And importantly, since this sentence is a disjunction, I could not split this infinite sentence into an infinite set of finite sentences. This fundamental asymmetry between ∨ and ∧ is playing a big role here: while an infinite conjunction can be constructed by simply putting each clause in the conjunction as a separate sentence, an infinite disjunction cannot be. This places a fundamental limit on the ability of a language with only finite sentences to describe the world.

The Surprise-Response Heuristic

Often we judge if somebody else is understanding something that we do not understand by whether the things they say in response to our questions are surprising.

When somebody understands it about as well as you do, the things they say about it will generally be fairly understandable and expected (as they mesh with your current insufficient level of understanding). But if they actually understand it and you don’t, then you should expect to be surprised by the things they say, since you couldn’t have produced those responses yourself or predicted them coming.


Q: “In this step of the proof, are they talking about extending the model or the language?”

A1: “They’re talking about extending the model. Look at the way that they worded the description of the extension in the previous step, it specifically describes adding a character to the model, not the language.”

A2: “No, they couldn’t be extending the model even though the wording suggests that, because then the proof wouldn’t even work; it’s required that we just change the language or else we end up working with a different model and failing to prove that the original model had the desired property. Also, it doesn’t even make sense to talk about adding a character to a model, the characters are a property of the language.”

Even with no context to judge whether the claims are true, I imagine that the second response feels much more convincing than the first, even though it’s probably less likely to be understood. The first is the type of response that is unsurprising and easy to see coming, and indicates only that the person is understanding the grammar of the English sentences they’re reading. It doesn’t strongly discriminate between a person that understands what’s going on and a person that doesn’t. The second is certainly surprising; it suggests that the person objects to the specific wording of the proof because of their understanding of the way it misrepresents the logical structure of the argument. They aren’t just comprehending the grammar, they are comprehending the actual content. Ordinarily, a person wouldn’t be able to off-the-cuff make up a response like that without actually understanding what’s going on.

This is a problem when people are good at saying surprising things without understanding. I’ve met a few people that are very good “contrarians”; they are good at coming up with strange and creative ways to say things that ultimately shed very little insight on the topic at hand. I often found myself in a weird position with such people where I feel like they understand the topic at hand better than me, and yet simultaneously I’m deeply suspicious of every word coming out of their mouth.

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?

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.