# Exclusive Or as Disjunction of Conjunctions

## Theorem

$p \oplus q \dashv \vdash \paren {\neg p \land q} \lor \paren {p \land \neg q}$

## Proof 1

 $\ds p \oplus q$ $\dashv \vdash$ $\ds \neg \left ({p \iff q}\right)$ Exclusive Or is Negation of Biconditional $\ds$ $\dashv \vdash$ $\ds \left({\neg p \land q}\right) \lor \left({p \land \neg q}\right)$ Non-Equivalence as Disjunction of Conjunctions

$\blacksquare$

## Proof 2

We apply the Method of Truth Tables.

As can be seen by inspection, the truth values under the main connectives match for all boolean interpretations.

$\begin{array}{|ccc||ccccccccc|} \hline p & \oplus & q & (\neg & p & \land & q) & \lor & (p & \land & \neg & q) \\ \hline F & F & F & T & F & F & F & F & F & F & T & F \\ F & T & T & T & F & T & T & T & F & F & F & T \\ T & T & F & F & T & F & F & T & T & T & T & F \\ T & F & T & F & T & F & T & F & T & F & F & T \\ \hline \end{array}$

$\blacksquare$