# 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

- 1964: Donald Kalish and Richard Montague:
*Logic: Techniques of Formal Reasoning*... (previous) ... (next): $\text{II}$: 'AND', 'OR', 'IF AND ONLY IF': $\S 5$: Exercises, Group $\text{III}$