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In Peano Arithmetic (PA), let $F(x)$ be the formula that means "$x$ is provable in $PA$". Thus, $F(x)$ is a nontrivial property of closed formulas, as some formulas are provable while others are not.

Now, consider the following formula $Q(x)$ with one free variable $x$:

$$ Q(x): \quad \text{If } x \text{ is not provable, then } 1 = 1; \quad \text{otherwise, if } x \text{ is provable, then } \text{Con}(PA). $$

By the Diagonal Lemma, there exists a fixed point $c$ for $Q(x)$. Specifically, we have $Q(c) \iff c$, which leads to the condition:

$$ Q(c) \iff c. $$

According to the definition of $Q(x)$, this implies:

  • If $c$ is provable, then $Q(c)$ is $\text{Con}(PA)$.
  • If $c$ is not provable, then $Q(c)$ is trivially true because $1 = 1$.

The key observation here is that $c$ is provable if and only if $Q(c)$ is not provable. This seems to suggest that $F(x)$, the provability predicate, is not extensional—that is, it does not preserve logical equivalence because two logically equivalent formulas (i.e., $c$ and $Q(c)$) can behave differently with respect to provability.

However, this reasoning seems problematic because we know that if $a \iff b$ (i.e., $a$ is logically equivalent to $b$), then $a$ is provable if and only if $b$ is provable.

What is the flaw in the reasoning above?

I have taken a look at True vs. Provable but that's not helpful. Because the concern I have is that if two statements imply each other they should be provable or un provable at the same time. But my reasoning contradicts that principle.

Zhiwei Liu
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  • You ought to talk to Craig Feinstein. https://math.stackexchange.com/questions/4179554/puzzle-does-the-following-prove-that-pa-is-inconsistent-part-2 – Gerry Myerson Jan 13 '25 at 16:51
  • Maybe https://math.stackexchange.com/questions/1154271/a-sentence-asserting-about-itself-that-if-it-is-provable-then-it-is-true is relevant. – Gerry Myerson Jan 13 '25 at 16:55
  • The statements Con(PA) and $1=1$ have the same truth value, but the latter is provable in P{A) and the former is not. – Andreas Blass Jan 13 '25 at 17:59
  • @AndreasBlass But they don't imply each other. True statements in general aren't equivalent to each other. So it is not surprising that only one of them is provable. – Zhiwei Liu Jan 13 '25 at 18:02
  • You wrote "if two statements have the same truth value they should be provable or un provable at the same time." My comment was intended as an easy example showing that this "should" isn't what actually happens. – Andreas Blass Jan 13 '25 at 18:07
  • @TankutBeygu The thread you linked to talks about the existence of unprovable true statements. But if two statements imply each other, they should be provable or unprovable at the same time. But my reasoning concludes otherwise. – Zhiwei Liu Jan 13 '25 at 18:09
  • @AndreasBlass Thanks for pointing that out. I just fixed that. Sorry for the confusion. – Zhiwei Liu Jan 13 '25 at 18:17

1 Answers1

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(It would be simpler to just write $Q(x) := \operatorname{Prov}_{\mathsf{PA}}(x)\to \sf Con(PA).$ Or at least that's how I'm interpreting your intention based on what you said in the following... as written, your definition of $Q(x)$ is unclear.)

The observation that $c$ is provable if and only if $Q(c)$ is not provable is incorrect. Just because $c$ is not provable doesn't mean that $\sf PA$ is able to prove that $c$ is not provable (and indeed, how could it, since it can't prove $\sf Con(PA)$), so it won't be able to surmise that all it needs to do is conclude $1=1$ and thereby prove $Q(c).$

spaceisdarkgreen
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  • I mean two statements that imply each other must be provable at the same time. Two statements with the same truth value don't imply each other in general. – Zhiwei Liu Jan 13 '25 at 18:06
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    @ZhiweiLiu If by "imply each other" you mean the bi-implication between the two statements is provable, then yes, the provability of either statement implies the provability of the other. Yes, the diagonalization lemma gives an implication between Q(c) and c in this sense, but as I've said the error is in the assertion that c is provable if and only if Q(c) is not provable. (Here it is just the case that neither is provable.) – spaceisdarkgreen Jan 13 '25 at 18:09
  • No, I'm just saying $1=1$ is provable in $\mathsf{PA}$, so "if $c$ is not provable, then 1=1'' also is. And so $Q(c)$, as in the question, is provable as well. – Léon Probst Jan 13 '25 at 18:19
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    @LéonProbst Yes, I see what you're saying now... I interpreted their intention with the statement as in my "it would be simpler to just write... " If they meant the disjunction "x is not provable implies 1=1 or x is not provable implies Con(PA)" as opposed to the conjunction "x is not provable implies 1=1 and x is provable implies Con(PA)" then the issue with the analysis is different. – spaceisdarkgreen Jan 13 '25 at 18:33