# deduction for propositonal logic [duplicate]

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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.

## marked as duplicate by Tim Raczkowski, Harish Chandra Rajpoot, Empty, Eclipse Sun, ServaesSep 27 '15 at 3:20

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• This isn't true: if $A$ and $B$ are distinct propositional variables and $\Sigma = \emptyset$, then "$\Sigma \vdash A$ iff $\Sigma \vdash B$" is true, but $\Sigma \vdash A \leftrightarrow B$ is false. – Rob Arthan Sep 24 '15 at 16:55
• Wow! I proved wrong thing! thanks – fbg Sep 24 '15 at 17:06

## 1 Answer

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 $$(\forall \mathcal M\vDash \Sigma : (\mathcal M\vDash \varphi) \iff (\mathcal M\vDash \psi)) \implies \Sigma\vDash \varphi\leftrightarrow \psi$$

• I appriciate it. thanks! – fbg Sep 24 '15 at 17:08