Next, we have the shortest chapter so far, on Quantifier Complexity, which introduces the notions of $latex \Delta_o, \Sigma_1$ and $latex \Pi_1$ wffs.

But there is an intriguing little result in this chapter. If the consistent theory *T* which includes Robinson Arithmetic Q proves a given $latex \Pi_1$ sentence, that sentence must be true. You don’t have to believe the theory *T* is true to accept its $latex \Pi_1$ consequences are true. For example, suppose *T* is the wildly infinitary apparatus that Wiles uses to prove Fermat’s Last Theorem, which is equivalent to a $latex \Pi_1$ sentence. Then you don’t have to believe that infinitary apparatus is actually correct (whatever exactly that means); all you need to believe to accept Fermat’s Last Theorem (assuming Wiles’s corrected derivation from *T *is correctly done) is that *T* is consistent. I still find that rather remarkable.

*Link now removed*]

Sam ButchartA typo in the footnote on p. 49, final sentence. ‘…higher up that the first levels…’ should be ‘… higher up than the first levels…’

On the same page, Theorem 20, ‘… can prove any true E1 sentences…’. Maybe this is just a matter of style but should it be ‘…can prove any true E1 sentence …’?

Rowsety Moidp 47, “Such results show that Q … ‘knows’ …”

I’ve kept changing my mind about whether there’s an issue here or not. Since “such results” appear to be of at least two sorts, it’s not entirely clear what counts as “knowing”. That Theorem 16 is also described in ‘knows’ terms on p 48 seems to bring the number of sorts up to at last 3 (proving an equivalence, like in 1 on p 47; ‘if can prove this, then can prove that’, like in 2 and 3; and captures, as in Theorem 16.)

OTOH, since nothing so far depends on being able to characterise ‘knows’ precisely, such informal use might be ok.

If there is a nice characterisation, though, it might be worth stating it.

p 50, “Which is, in its way, a quite remarkable observation. It means that we don’t have to fully believe a theory T – i.e. don’t have to accept that all its theorems are true on the interpretation built into T ’s language – in order to use it to establish that some ?1 arithmetic generalization is true.”

I’m not sure it’s quite so remarkable when it’s about *belief*. After all, beliefs can be mistaken. So of course what a theory can actually do doesn’t require our belief.

How do things look when belief is taken out of the picture? Since T isn’t required to be sound, it looks T could show a ?1 statement is true even if some T’s theorems are false. (Is that right?) Similarly, some of the heavy-duty infinitary mathematics that Wiles used could be false, and yet he’d still have shown FLT is true.

Peter SmithOn the use of “knows” with (or without) the scare quotes: it’s the sort of vivid lecture-room trope that I

thinkcarries over well enough to written notes so long as it is clear enough from context how to cash it out in more sober terms.On my use of “believe” in the context of discussing the Wiles proof, however, I agree it is better to give that a miss (in fact I’d already revised that passage, to make it clearer).

Rowsety MoidI think Theorem 21 is a great result, though, despite reservations about ‘belief’, and is indeed a remarkable corollary.

I think some of the things that happen around incompleteness (so to speak) are some of the most interesting aspects — Thorems 20 and 21, for example; nonstandard models.

Peter SmithYes, I do agree — which is why IGT rather spreads its wings to fly over some neighbouring territory, even if too rapidly!