Equivalences are Interderivable

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Theorem

If two propositional formulas are interderivable, they are equivalent:

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


Forward Implication

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

Reverse Implication

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


Proof 1

The result follows directly from the truth table for the biconditional:

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


By inspection, it is seen that $\mathcal M \left({p \iff q}\right) = T$ precisely when $\mathcal M \left({p}\right) = \mathcal M \left({q}\right)$.

$\blacksquare$


Proof 2

Let $v$ be an arbitrary interpretation.

Then by definition of interderivable:

$\map v {p \iff q}$ if and only if $\map v p = \map v q$

Since $v$ is arbitrary, $\map v p = \map v q$ holds in all interpretations.

That is:

$p \dashv \vdash q$

$\blacksquare$


Sources