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

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 ?

• Please insert some parentheses, or better still, a lot of them. The formulas are unreadable. Apr 20, 2022 at 13:40
• @AndrejBauer sorry i guess it's fixed now. Apr 20, 2022 at 18:28

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.
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 "$$(\vdash q)\rightarrow(\vdash p)$$").
• i just found out wrote my interpretation wrong at first and i fixed it now but i got your point i guess. my problem was that if $\vdash q$ is correct and we can deduce $\vdash p$ from that then $\{q\} \rightarrow p$ is correct too. but i guess $\vdash q \rightarrow \vdash p$ has nothing to with ${q} \rightarrow p$ because $\vdash p$ might be related to other sentences in our logic and it certainly is in case of soundness. Apr 20, 2022 at 23:01