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Smith's "Intro to Godel's Theorems," defines $T$ as a sound theory as everything $T$ proves must be true.

He then mentions a Godel sentence $G_{T}$ which I would characterize as $T\vdash (G_{T}\leftrightarrow\neg\square G_{T})$.

"So $G_{T}$ is true if and only if $T$ can't prove it....Hence if $T$ is sound, $G_{T}$ is unprovable in $T$. Which makes $G_{T}$ true."

I understand the above and that it follows from $T\vdash (G_{T}\leftrightarrow\neg\square G_{T})$.

Now comes my question:

He goes on to say, "Hence $\neg G_{T}$ is false. And so that too can't be proved by $T$, since $T$ only proves truths."

But in this instance, isn't $\neg G_{T}$ being false a truth?

Or, what probably is incorrect: $T\vdash(\neg G_{T}\leftrightarrow \square G_{T})$, and since I started with $G_{T}$ is true, then in this case $\neg G_{T}$ is true.

Thanks

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    $\begingroup$ isn't $¬G_T$ being false a truth? Yes, i.e. $¬¬G_T$ is true. The meta-theory is "classic" and thus double negation holds. $\endgroup$ – Mauro ALLEGRANZA Jun 14 '18 at 18:44
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    $\begingroup$ You've misunderstood "And so that too". Here "that" refers to $\neg G_T$, not to "$\neg G_T$ is false". $\endgroup$ – Andreas Blass Jun 14 '18 at 22:27
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$T$ does indeed prove $\neg G_T\iff\Box G_T$. However, I'm not sure how you get from this to the claim in your last sentence that $\neg G_T$ is true.

I think the key is the following: we're assuming that $T$ only proves true statements, but we're not assuming that $T$ proves all true statements. $G_T$ is true, and if $T$ proved $G_T$ then $G_T$ would be false, but we can't go from "$G_T$ is true" to "$T$ proves $G_T$" (in fact, that's the whole point of $G_T$).

So I think the issue is really linguistic here: that "$T$ only proves truths" is easy to misread as "the things $T$ proves are the truths."

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You understood that $G_T$ is true. Well, then it follows that $\neg G_T$ is false.

Of course, that also means that the statement "$\neg G_T$ is false" is a true statement .. but we're not really interested in that statement here.

Likewise, if we say that $A$ is true, then $\neg A$ is false. And again, it would then also be true that "$\neg A$ is false", but that does not make $\neg A$ true. In fact, if it is true that "$\neg A$ is false", then clearly $\neg A$ is false, not true.

We would only have that $\neg A$ is true if "$\neg A$ is false" is false.

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    $\begingroup$ The second-to-last sentence of this answer seems to be a typo: you've written "$True(False(\neg A))$ implies $True(\neg A)$," that is "$A\implies\neg A$." (The last sentence, though, is correct.) @Andrew The point of this answer is just that "False" and "$\neg$" are the same thing, and double negation cancels; e.g. "(($A$ is false) is false) is false" is just "$\neg\neg\neg A$," which simplifies to "$\neg A$," which is true iff $A$ is false. $\endgroup$ – Noah Schweber Jun 14 '18 at 19:00
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    $\begingroup$ @Andrew Both of the claims mentioned above can be unpacked. "False$(x)$=$\neg(x)$ and True$(x)$=$x$" is the least objectionable (although even it is nontrivial), and of course double negation cancellation is not intuitionistically true. But everything here is taking place in classical logic, which makes things quite simple. $\endgroup$ – Noah Schweber Jun 14 '18 at 19:02
  • $\begingroup$ @NoahSchweber Ha ha, I confused myself :) Too many negations! $\endgroup$ – Bram28 Jun 14 '18 at 19:05

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