Why we can't use deduction theorem on soundness to contravene second incompleteness with lob's theorem

Question

I'm starting to learn modal logic and there is something that's bothering my mind for a while.
we know from deduction theorem that $((\vdash q) \rightarrow (\vdash p)) \Leftrightarrow(\vdash (q \rightarrow p))$
and also from soundness we know that $\vdash\Box q \rightarrow \vdash q$
so here's my problem, why can't we deduce that $\vdash \Box q\rightarrow q$ if we do that with lob's theorem ($\vdash(\Box q \rightarrow q) \leftrightarrow \vdash q$) we can contravene second incompleteness so it is a wrong thing to do but why ? don't we proof all these sentences in meta so what's the difference ?

Answer 1

We know from deduction theorem that $(\vdash q\rightarrow\vdash p)\iff (\vdash p\rightarrow q)$

This is false. If $\not\vdash q$ then the clause $(\vdash q)\rightarrow(\vdash p)$ (re-parenthesized for clarity) is vacuously true, regardless of what $p$ is. So just take $q$ to be some independent sentence and set $p=\neg q$; we trivially have $\not\vdash p\rightarrow q$ but $(\vdash q)\rightarrow(\vdash p)$, a counterexample to your claim.

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The issue is with your interpretation of the deduction theorem. The deduction theorem says $$\{p\}\vdash q\quad\iff\quad\vdash p\rightarrow q,$$ or more generally $$T\cup\{p\}\vdash q\quad\iff\quad T\vdash p\rightarrow q,$$ but that's very different from what you've written. In particular, the deduction theorem manipulates individual sequents (turning "$\{p\}\vdash q$" into "$\vdash p\rightarrow q$" or similar) rather than hypothetical comparisons of different sequents (such as your "$(\vdash q)\rightarrow(\vdash p)$").