Problem is showing or disproving in Sentential logic that:
If $\Sigma$$\vdash$$\varphi$ iff $\Sigma$$\vdash$$\psi$, then $\Sigma$$\vdash$$\varphi\leftrightarrow\psi$
But I wonder how to show that there exists(or there's no) such a deduction.
To show opposite direction, I write as follows:
Suppose that $\Sigma\vdash\varphi\leftrightarrow\psi$ , i.e. there exists a deduction $<a_{1},a_{2},\cdots,a_{n}>$ such that $a_{n}=\varphi\leftrightarrow\psi$ . Then assume $\Sigma\vdash\varphi$ , i.e. there exists a deduction $<b_{1},b_{2},\cdots,b_{m}>$ such that $b_{m}=\varphi$ . Then a sequence $ <a_{1},\cdots,a_{n},b_{1},\cdots,b_{m},\psi>$ is a deduction from $\Sigma$ to $\psi$ . Thus, $\Sigma\vdash\psi$ . Then assume that $\Sigma\vdash\psi$ , i.e. there exists a sequence $<c_{1},c_{2},\cdots,c_{l}>$ such that $c_{l}=\psi$ . Then a sequence $<a_{1},\cdots,a_{n},c_{1},\cdots,c_{l},\varphi>$ is a deduction from $\Sigma$ to $\varphi$ . Thus, $\Sigma\vdash\varphi$ . Thus, If $\Sigma\vdash\varphi\leftrightarrow\psi$ , then $\Sigma\vdash\varphi$ iff $\Sigma\vdash\psi$ .
It seem to me that it is similar to this one. But I don't know how to do it.
This is my new answer:
Comsider $\Sigma=\{p,q\}$ . Then $\Sigma\vdash\varphi$ and $\Sigma\vdash\psi$ . Thus, $\Sigma\vdash\varphi$ iff $\Sigma\vdash\psi$ . But for a truth assignment $\nu$ of $\Sigma$ such that $\bar{\nu}(\varphi)=T$ and $\bar{\nu}(\psi)=F$ , $\bar{\nu}(\varphi\leftrightarrow\psi)=F$ . Thus, $\Sigma\nvDash\varphi\leftrightarrow\psi$ . It means that, by completeness of Sentential Logic, $\Sigma\nvdash\varphi\leftrightarrow\psi$ .
Deduction from $\Sigma$ is a finite sequence of sentences which consists of members of $\Sigma$ , tautology, or sentences obtained by Modus Ponens
And biconditional is not a part of this language.
