Q:

deduction for propositonal logic

given problem is,  If $\Sigma\vdash\varphi$ iff $\Sigma\vdash\psi$, then $\Sigma\vdash\varphi\leftrightarrow\psi$.
I can prove this using sound&completeness theorem but I don't know how to do without those theorems.
without them, I proved when $\Sigma\vdash\varphi$ is true, but I couldn't prove when $\Sigma\nvdash\varphi$. without the theorems, I don't know how $\nvdash$  part contributes to prove the problem.
p.s. only member of given set, tautology, and Modus Ponens are allowed for deduction.

A:

No, this is not true. Let $\Sigma$ be empty, and let $\varphi$ and $\psi$ be different propositional variables.
Then $\Sigma\vdash\varphi$ and $\Sigma\vdash\psi$ are both false, and thus $(\Sigma\vdash\varphi)\Leftrightarrow(\Sigma\vdash\psi)$ is true. But $\Sigma
\not\vdash\varphi\leftrightarrow\psi$.

On the other hand, it is true that
$$ (\forall \mathcal M\vDash \Sigma : (\mathcal M\vDash \varphi) \iff (\mathcal M\vDash \psi)) \implies \Sigma\vDash \varphi\leftrightarrow \psi $$

