# Can probability theory be used to model epistemology in mathematics?

Chater and Oaksford (2009) write:

“… logic, and indeed the whole of mathematics, concern determining what beliefs can consistently be held – a mathematical proof shows, given a set of axioms, some theorem must follow; i.e., that the axioms and the negation of the theorem are not consistent. But it seems rather unnatural to see the whole of mathematics as a special case of probability theory!”

First problem: mathematics is bigger than formal logic, as Gödel proved with his incompleteness (really incompletability) theorem. Mathematical proofs as constructed by real mathematicians are not formal. Philosophers debate endlessly what a proof is.

Leaving Gödel aside, yes, classical logic can be used to characterise relations between premises and a conclusion in mathematical proof (a model of epistemology in mathematics which probably isn’t terribly accurate). Even today in the post-Gödel world many researchers are working on formalising segments of mathematics, e.g., in the Isabelle and Coq proof assistants. A big example is the attempt to formalise a proof of Kepler’s conjecture.

So far so good for (non-probabilistic) logic as a model of epistemology.

Consider Fermat’s last conjecture. Before Wiles proved the conjecture, I bet many mathematicians were pretty sure, but not certain, that it held. Or consider the more general case of how mathematicians decide what conjectures to try to prove in the first place. They must use some cues to decide whether the probability of the conjecture is high before spending a lot of time embarking on a proof. Why not use probability to model epistemology in mathematics? I have no idea if such a probability logic of mathematical conjectures (quantifiers, useful structures, etc, plugged into probability) is feasible in practice.

I see no problem in viewing classical logic as a special case of probability theory without a conditional event, and suppositional logic with its “defective” truth-tabled conditional as a special case of probability with the conditional event. A strong reason for this belief is that (as far as I understand) there are proofs of these relationships between logic and probability (or perhaps better phraseology is certain versus uncertain logic).

Complication: the following is a deductive relationship:

P(B1|A1) = p_1,
P(B2|A2) = p_2,

P(Bn-1|An-1) = p_n-1
———————-
P(Bn|An) = p_n

Another complication: the logic used in the mathematical vernacular looks like a bit like informal classical logic, even when used at the meta-level to reason about non-classical logics.

Reference

Chater, N., & Oaksford, M. (2009). Local and global inferential relations: Response to Over (2009). Thinking and Reasoning, 15, 439–446.

Disproof. Formalise as $\forall x : P.~\neg C(x) \rightarrow R(x)$. We want to prove $(\forall x : P.~\neg C(x) \rightarrow R(x)) \rightarrow \bot$, where $P$ is the type of passwords, $C(x)$ is true iff $x$ is the correct password and $R(x)$ means we are redirected to a void by $x$.
Suppose $\forall x : P.~\neg C(x) \rightarrow R(x)$. $\neg C(\mbox{../../..'})$ holds, where $\mbox{../../..'} : P$, thus we must have $R(\mbox{../../..'})$. However by the true constructive meaning of $R$, namely typing the password into the form and observing the result—redirection to http://www.cs.swan.ac.uk/~csetzer/index.html—we can conclude $\neg R(\mbox{../../..'})$, which may be written $R(\mbox{../../..'}) \rightarrow \bot$. From $R(\mbox{../../..'})$ and $R(\mbox{`../../..'}) \rightarrow \bot$ we get $\bot$ by arrow-elimination, thus we can discharge our assumption. QED.