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.

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    $\begingroup$ I don't think it is true. Let $\Sigma$ be an incomplete theory and $\varphi, \psi$ two sentences that may or may not be true. For example, the theory of groups, $\varphi$ that the group is abelian, $\psi$ that the group has at least four elements. $\Sigma \vdash \varphi$ iff $\Sigma \vdash \psi$, but $\Sigma \not \vdash \varphi \leftrightarrow \psi$ $\endgroup$ – Ross Millikan Sep 15 '15 at 14:20
  • $\begingroup$ For that matter, suppose $\Sigma=\emptyset$, $\phi=p$, and $\psi=q$. $\endgroup$ – mmw Sep 16 '15 at 18:36
  • $\begingroup$ @Darae-Uri Do you know the completeness theorem yet? Is $\leftrightarrow$ part of the language or is it an abbreviation? If it is part of the language, how exactly do you define deduction? $\endgroup$ – Git Gud Sep 17 '15 at 13:20
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    $\begingroup$ @RossMillikan The problem is in Sentential Logic (A.K.A. Propositional Calculus). $\endgroup$ – Git Gud Sep 17 '15 at 13:21
  • $\begingroup$ @GitGud I know it. Using completeness, I answered it. I was asking new question for my "new answer" to this problem, right befor your comment. I will edit this question to add my new answer but I'm not sure it is correcy $\endgroup$ – Darae-Uri Sep 17 '15 at 13:25

The statement is false. For a counterexample, pick $\Sigma=\emptyset$, $\phi=p$, and $\psi=q$.

The soundness theorem gives that $\emptyset\not\vdash p$ and $\emptyset\not\vdash q$, hence that $\emptyset\vdash p$ iff $\emptyset\vdash q$. But also by soundness, $\emptyset \not\vdash p\leftrightarrow q$.


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