# Equivalences are Interderivable

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## Theorem

If two propositional formulas are interderivable, they are equivalent:

- $\paren {p \dashv \vdash q} \dashv \vdash \paren {p \iff q}$

### 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:

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

By inspection, it is seen that $\map \MM {p \iff q} = \T$ if and only if $\map \MM p = \map \MM q$.

$\blacksquare$

## 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 $\text{III}$