Biconditional as Disjunction of Conjunctions/Formulation 2/Proof by Truth Table

Theorem

$\vdash \paren {p \iff q} \iff \paren {\paren {p \land q} \lor \paren {\neg p \land \neg q} }$

Proof

We apply the Method of Truth Tables.

As can be seen by inspection, in all cases the truth values under the main connective is true for all boolean interpretations.

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

$\blacksquare$

Sources

• 1980: D.J. O'Connor and Betty Powell: Elementary Logic ... (previous) ... (next): $\S \text{I}: 13$: Logical Equivalences: Exercise $\text{(A)} \ 12$