Did Descartes make a Diagonal Argument?

Abstract

This paper explores the idea that Descartes’ cogito is a kind of diagonal argument. Using tools from modal logic, it reviews some historical antecedents of this idea from Slezak and Boos and culminates in an orginal result classifying the exact structure of belief frames capable of supporting diagonal arguments and our reconstruction of the cogito.

Appendix

Theorem

(4) Let ${\mathscr{M}}=\langle W,R,v\rangle$ be a model of propositional modal logic. Then the following are equivalent:

• 1. Player II has a well-aligned winning strategy in the diagonal game for ${\mathscr{M}}$; and

• 2. There is a diagonal proposition on ${\mathscr{M}}$: i.e., there is some proposition p such that ${\mathscr{M}}\models p\leftrightarrow \neg \Box p$.

Remark Note also that this is crucially distinct from saying that ${\mathscr{M}}\models \exists p(p\leftrightarrow \neg \Box p)$. This latter formulation merely demands that at every world there is some proposition which is true if and only if it is doubted and nothing about its behaviour at other worlds. We rather want a proposition p which upholds ${\mathscr{M}}\models p\leftrightarrow \neg \Box p$ thus vindicating our intention that p just means that I doubt p. remark

Proof

(2)→(1) Let p be such that for all w ∈ W $w\Vdash p\leftrightarrow \neg \Box p$. We construct a strategy for II as follows.

I’s first move can be anywhere in W, so for all w ∈ W we let:

$$ \sigma(w)=\begin{cases} \text{pass} & \text{if }w\Vdash\Box p\\ \text{some }u\text{ s.t. }u\Vdash\Box p\text{ and }wRu & \text{if }w\Vdash\neg\Box p. \end{cases} $$

Now suppose $\bar {w}=\langle w_{0},...,w_{2n}\rangle$ be a partial play following σ in which I has played last. Then we let

$$ \sigma(\bar{w})=\text{some }u\text{ s.t. }u\Vdash\Box p\text{ and }w_{2n}Ru. $$

The following claim establishes that there is no way of stopping this construction.

Claim

σ is well defined.

Proof

Suppose not. Then let $\bar {w}$ be a play of least length such that I made the last move and $\sigma (\bar {w})$ is not defined. Suppose n = 0. Then $\bar {w}=\langle w_{0}\rangle$. If $w_{0}\Vdash \Box p$, then II can pass. If $w_{0}\nVdash \Box p$, then $w_{0}\Vdash \neg \Box p$ and so there is some u ∈ W with w₀Ru and $u\Vdash \neg p$. Thus II can play such a u and we’ll have $u\Vdash \Box p$.

Suppose n > 0. Then $\bar {w}=\langle w_{0},...,w_{2n}\rangle$ for some n > 0. Since $\sigma (\bar {w})$ is not defined, there is no u such that w_2nRu and $u\Vdash \Box p$. Suppose $w_{2n}\Vdash \neg \Box p$, then for some u with w_2nRu we have $u\Vdash \neg p$ and so $u\Vdash \Box p$: contradiction. Thus $w_{2n}\Vdash \Box p$ and so $w_{2n}\Vdash \neg p$. But the by the minimality of $\bar {w}$ we see that II’s earlier moves are well-defined and so we must have $w_{2n-1}\Vdash \Box p$ and so $w_{2n}\Vdash p$: contradiction. □

The following lemma is useful below:

Lemma

Let $\bar {w}$ be a partial play of the diagonal game played in accordance with σ for II. Then □-assigned (¬□-assigned) worlds w are such that $w\Vdash \Box p$ ($w\Vdash \neg \Box p$).

Proof

Let $\bar {w}$ be a play of minimal length n such that this fails. Suppose n = 0. Then $\bar {w}=\langle w_{0}\rangle$. Suppose II passes using σ. Then w₀ is □-assigned by definition and $w_{0}\Vdash \Box p$ by the definition of σ. Suppose II doesn’t pass using σ. Then w₀ is ¬□-assigned and $w_{0}\Vdash \neg \Box p$ according to σ.

Suppose $\bar {w}=\langle w_{0},...,w_{2n}\rangle$ for some n > 0; i.e., I played last. Then we see that w_2n− 1Rw_2n, $w_{2n-1}\Vdash \Box p$ by minimality, while $w_{2n}\Vdash \Box p$. But then we have $w_{2n}\Vdash p$ and $w_{2n}\Vdash \neg p$: contradiction. Suppose $\bar {w}=\langle w_{0},...,w_{2n+1}\rangle$ for some n > 0; i.e., II played last. Then by definition of σ, $w_{2n+1}\Vdash \Box p$ as required. □

Claim

σ is a winning strategy for II.

Proof

Suppose not. Then there is some play $\bar {w}=\langle w_{0},...,w_{2n}\rangle$ of minimal length such that II has already lost or cannot legally move at this point. Suppose n = 0. Then $\bar {w}=\langle w_{0}\rangle$. Suppose $w_{0}\Vdash \Box p$ so w₀ is □-assigned. Then II legally passes and so has not lost. Suppose $w_{0}\Vdash \neg \Box p$ so w₀ is ¬□-assigned. Then we may fix u with w₀Ru such that $u\Vdash \neg p$. Thus $u\Vdash \Box p$ and w₀ is □-assigned. Since there are no earlier moves and I’s move was appropriated, there is no way for II to lose at this point by playing u.

Suppose n > 0. Then for all all u we have either:

• (1) ¬w_2nRu; or

• (2) one of the following holds:

  • (a) there is a ¬□-assigned w in $\bar {w}$ such that w = u;

  • (b) there is a □-assigned world w in $\bar {w}$ such that wRu; or

  • (c) there is a □-assigned world w in $\bar {w}$ such that w = w_2n.

Since w_2n is ¬□-assigned, we see that $w_{2n}\Vdash \neg \Box p$, and so we may fix we make fix some u such that w_2nRu and $u\Vdash \neg p$. Now suppose that either (a), (b) or (c) are true.

(a) Fix w a ¬□-assigned world in $\bar {w}$ and suppose w = u. Then we have $w\Vdash \neg \Box p$ and $u=w\Vdash p$: contradiction.

(b) Fix w a □-assigned world in $\bar {w}$ and suppose wRu. Then since $w\Vdash \Box p$, we have $u\Vdash p$: contradiction.

(c) Fix w a □-assigned world in $\bar {w}$ and suppose w = w_2n. Then since $w\Vdash \Box p$ and $w_{2n}\Vdash \neg \Box p$: contradiction. □

Finally, it can be seen that σ is such that all and only the worlds w such that $p\Vdash \Box p$ are played (or appropriated) by II; i.e., the □-assigned worlds. Thus σ is well-aligned.

(1)→(2)Suppose II has a well-aligned winning strategy σ for II in ${\mathscr{M}}$. Let p be such that for all w ∈ W

$$ w\Vdash p\ \Leftrightarrow\ w\text{ is }\neg\Box\text{-assigned in every play compatible with }\sigma. $$

Suppose $w_{0}\Vdash p$. So w₀ is ¬□-assigned in all plays compatible with σ. Suppose for a contradiction that $w_{0}\Vdash \Box p$. Then for all w₁ with w₀Rw₁ we have $w_{1}\Vdash p$ and w₁ is ¬□-assigned (by the assignment above). Given that w₀ is ¬□-assigned, it can be seen through the definition that w₀ must have been played by I. Suppose there is some w₁ such that w₀Rw₁. Then we see that no matter how II plays after w₀, we end up with a play containing consecutive worlds which are both ¬□-assigned: contradicting the well-alignment of σ. Suppose there is no w₁ such that w₀Rw₁; i.e., w₀ is a dead end. Then consider the game starting with w₀ played by I. The only way for II to win is to pass and since σ is a winning strategy σ(w₀) = pass. But then we see that w₀ must be □-assigned in a play compatible with σ which is contrary to our initial assumption.

Suppose $w_{0}\Vdash \neg p$. Since σ is well-aligned, we can see that every world is either ¬□-assigned (or □-assigned) in every play compatible with σ; thus, w₀ is □-assigned in all plays compatible with σ. Suppose for a contradiction that $w_{0}\Vdash \neg \Box p$. Then for some w₁ with w₀Rw₁ we have $w_{1}\Vdash \neg p$ and so w₁ is □-assigned. Given that w₀ is □-assigned, either w₀was played by II or II passed and w₀ as the first move by I in some pay of the game. Suppose w₀ was played by II and suppose then that I then plays w₁ after w₀ in some play compatible with σ. But then since w₀ and w₁ are consecutive □-assigned worlds in a partial play, we have a witness to the misalignment of σ. Suppose II passed. Then 〈w₀,w₁〉 is a partial play in which we have consecutive □-assigned worlds, which contradicts the well-alignment of σ. □

Theorem

(9)$\Box \exists p\neg \Box p\dashv \vdash \forall q\Box (\Box q\to \Diamond q)$.

Proof

(⊩)

Then by necessitation get

and so

$$ \Box\exists p\neg\Box p\vdash\forall q\Box(\Box q\to\Diamond q). $$

(⊣)

Then by necessition, we get

and so we have

$$ \forall q\Box(\Box q\to\Diamond q)\vdash\Box\exists p\neg\Box p. $$

□