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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, Servaes Sep 27 '15 at 3:20

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  • $\begingroup$ 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. $\endgroup$ – Rob Arthan Sep 24 '15 at 16:55
  • $\begingroup$ Wow! I proved wrong thing! thanks $\endgroup$ – fbg Sep 24 '15 at 17:06

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

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

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