0
$\begingroup$

This question already has an answer here:

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.

$\endgroup$

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

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ 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
2
$\begingroup$

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

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.