Prove if theory $T$ is an axiomatizable $\omega$-consistent extension of $\mathbf{Q}$, then if $\vdash_T \text{Prv}(\ulcorner A \urcorner)$ then $\vdash_T A$
This is a textbook proof in Computability and Logic by Boolos et all, page 234.
Typing what the book says:
For if we had $\vdash_T \text{Prv}_T(\ulcorner A \urcorner)$, or in other words $\vdash_T \exists y \text{Prf}_T(\ulcorner A \urcorner, y)$ but did not have $\vdash_T A$, then for each $b$, $\lnot \text{Prf}_T(\ulcorner A \urcorner, \mathbf{b})$ would be correct and hence provable in $\mathbf{Q}$ and hence in $T$, and we would have an $\omega$-inconsistency in $T$.
The piece I'm struggling with is why $\vdash_T \exists y \text{Prf}_T(\ulcorner A \urcorner, y)$ doesn't directly imply $\vdash_T A$, but rather it is required that some natural number $b$ exist such that $\text{Prf}_T(\ulcorner A \urcorner, \mathbf{b})$.
Other than that, I understand the rest
The problem is that the axioms of $T$ may not be true when interpreted in the actual natural numbers. If the sentence $\exists y \text{Prf}_T(\ulcorner A \urcorner, y)$ is true when interpreted in the natural numbers, then that does mean that $\vdash_T A$. But merely having $\vdash_T \exists y \text{Prf}_T(\ulcorner A \urcorner, y)$ does not automatically mean that $\exists y \text{Prf}_T(\ulcorner A \urcorner, y)$ is really true, if you don't assume that the axioms of $T$ are true.