Non-Equivalence as Conjunction of Disjunction with Negation of Conjunction/Proof by Truth Table
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Theorem
- $\neg \paren {p \iff q} \dashv \vdash \paren {p \lor q} \land \neg \paren {p \land q}$
That is, negation of the biconditional means the same thing as either-or but not both, that is, exclusive or.
Proof
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.
$\begin{array}{|cccc||cccccccc|} \hline
\neg & (p & \iff & q) & (p & \lor & q) & \land & \neg & (p & \land & q) \\
\hline
\F & \F & \T & \F & \F & \F & \F & \F & \T & \F & \F & \F \\
\T & \F & \F & \T & \F & \T & \T & \T & \T & \F & \F & \T \\
\T & \T & \F & \F & \T & \T & \F & \T & \T & \T & \F & \F \\
\F & \T & \T & \T & \T & \T & \T & \F & \F & \T & \T & \T \\
\hline
\end{array}$
$\blacksquare$