# Is provability predicate a truth predicate?

Let T be a consistent, axiomatizable theory containing Q. Because the set of provable sentences of T is semi-recursive, we have a provability predicate of the form $$\exists yPrf(x,y)$$ with $$Prf(x,y)$$ rudimentary. This predicate is correct under the standard interpretation of arithmetic if and only if the sentence represented by $$x$$ is provable. Because this predicate is of the $$\exists$$-rudimentary form, it is correct if and only if it is provable from T, which is to say $$T\vdash A$$ if and only if $$T\vdash\exists yPrf(x,y)$$, or equivalently $$T\vdash A\leftrightarrow\exists yPrf(x,y)$$, with x is the Godel number of the sentence A, for all A. This property of the provability predicate satisfies the condition of being a truth predicate. However, by the diagonal lemma, we can prove that there is no truth predicate. Could you help me explain why provability predicate is not a truth predicate? Thank you a lot!

Your "or equivalently" claim is false: in general "$$(T\vdash\varphi)\leftrightarrow(T\vdash\psi)$$" does not imply "$$T\vdash (\varphi\leftrightarrow\psi)$$."
For example: under the usual assumption that $$\mathsf{PA}$$ is $$\Sigma_1$$-sound (this is stronger than mere consistency) we have $$\mathsf{PA}\not\vdash 0=1$$ and $$\mathsf{PA}\not\vdash \neg Con(\mathsf{PA})$$, so the statement "$$(\mathsf{PA}\vdash 0=1)\leftrightarrow (\mathsf{PA}\vdash \neg Con(\mathsf{PA})$$" is correct, but it is not the case that $$\mathsf{PA}\vdash (0=1\leftrightarrow \neg Con(\mathsf{PA}))$$ since that would contradict the second incompleteness theorem.
First, it is not necessarily the case that $$T\vdash \exists yPrf(\ulcorner A\urcorner, y)$$ if and only if $$T\vdash A.$$ For that we need another assumption, that $$T$$ does not prove any false $$\exists$$-rudimentary sentences. That is a mild correctness assumption in the grand scheme of things, but doesn't follow from $$T$$ being a consistent axiomatizable extension of Q. Consider for example the extension like PA + $$\lnot$$Con(PA). (The other direction, that $$T$$ proves every correct $$\exists$$-rudimentary sentence, does follow, though.)
But the main issue is that it's not true that $$T\vdash \varphi$$ if and only if $$T\vdash \psi$$ means the same thing as that $$T\vdash \varphi\leftrightarrow \psi.$$ (Where $$(\varphi, \psi)$$ ranges over some family of pairs of sentences.)