Equivalences are Interderivable

Theorem
If two statement forms are interderivable, they are equivalent:


 * $\left ({p \dashv \vdash q}\right) \iff \left ({p \iff q}\right)$

Proof

 * align="right" | 2 ||
 * align="right" | 1
 * $\left ({p \vdash q}\right) \land \left ({q \vdash p}\right)$
 * SI
 * 1
 * align="right" | 3 ||
 * align="right" | 1
 * $p \vdash q$
 * $\land \mathcal E_1$
 * 2
 * align="right" | 4 ||
 * align="right" | 1
 * $p \implies q$
 * $\implies \mathcal I$
 * 3
 * align="right" | 5 ||
 * align="right" | 1
 * $q \vdash p$
 * $\land \mathcal E_2$
 * 2
 * align="right" | 6 ||
 * align="right" | 1
 * $q \implies p$
 * $\implies \mathcal I$
 * 5
 * $\land \mathcal E_2$
 * 2
 * align="right" | 6 ||
 * align="right" | 1
 * $q \implies p$
 * $\implies \mathcal I$
 * 5
 * $\implies \mathcal I$
 * 5


 * align="right" | 8 ||
 * align="right" | 1
 * $p \iff q$
 * Material Equivalence
 * 7
 * 7

First:


 * align="right" | 2 ||
 * align="right" | 1
 * $\left({p \implies q}\right) \land \left({q \implies p}\right)$
 * Material Equivalence
 * 1
 * align="right" | 3 ||
 * align="right" | 1
 * $p \implies q$
 * $\land \mathcal E_1$
 * 2
 * $p \implies q$
 * $\land \mathcal E_1$
 * 2


 * align="right" | 5 ||
 * align="right" | 1, 4
 * $q$
 * $\implies \mathcal E$
 * 1, 4
 * 1, 4

Similarly:


 * align="right" | 2 ||
 * align="right" | 1
 * $\left({p \implies q}\right) \land \left({q \implies p}\right)$
 * Material Equivalence
 * 1
 * align="right" | 3 ||
 * align="right" | 1
 * $q \implies p$
 * $\land \mathcal E_2$
 * 2
 * $q \implies p$
 * $\land \mathcal E_2$
 * 2


 * align="right" | 5 ||
 * align="right" | 1, 4
 * $p$
 * $\implies \mathcal E$
 * 1, 4
 * 1, 4

Proof by Truth Table
The result follows directly from the truth table for material equivalence:

$\begin{array}{|cc||ccc|} \hline p & q & p & \iff & q \\ \hline F & F & F & T & F \\ F & T & F & F & T \\ T & F & F & F & F \\ T & T & F & T & T \\ \hline \end{array}$

We see that $\mathcal M \left({p \iff q}\right) = T$ precisely when $\mathcal M \left({p}\right) = \mathcal M \left({q}\right)$.