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

Proof by Natural Deduction

By the tableau method:

First, we show that if $p \dashv \vdash q$, then $p \iff q$:

If $p \dashv \vdash q$, then $p \iff q$

Line		Pool	Formula	Rule	Depends upon	Notes
1		1	$p \dashv \vdash q$	P	(None)
2		1	$\left ({p \vdash q}\right) \land \left ({q \vdash p}\right)$	SI	1
3		1	$p \vdash q$	$\land \mathcal E_1$	2
4		1	$p \implies q$	$\implies \mathcal I$	3
5		1	$q \vdash p$	$\land \mathcal E_2$	2
6		1	$q \implies p$	$\implies \mathcal I$	5
7		1	$\left({p \implies q}\right) \land \left({q \implies p}\right)$	$\land \mathcal I$	3, 5
8		1	$p \iff q$	Material Equivalence	7

$\blacksquare$

Next, we show that if $p \iff q$, then $p \dashv \vdash q$:

If $p \iff q$, then $p \vdash q$

Line		Pool	Formula	Rule	Depends upon	Notes
1		1	$p \iff q$	P	(None)
2		1	$\left({p \implies q}\right) \land \left({q \implies p}\right)$	Material Equivalence	1
3		1	$p \implies q$	$\land \mathcal E_1$	2
4		1	$p$	A	(None)
5		1, 4	$q$	$\implies \mathcal E$	1, 4

Similarly:

If $p \iff q$, then $q \vdash p$

Line		Pool	Formula	Rule	Depends upon	Notes
1		1	$p \iff q$	P	(None)
2		1	$\left({p \implies q}\right) \land \left({q \implies p}\right)$	Material Equivalence	1
3		1	$q \implies p$	$\land \mathcal E_2$	2
4		1	$q$	A	(None)
5		1, 4	$p$	$\implies \mathcal E$	1, 4

$\blacksquare$

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 $\mathcal M \left({p}\right) = \mathcal M \left({q}\right)$.

$\blacksquare$