Conjunction implies Disjunction/Proof by Truth Table

Theorem

$\vdash \paren {p \land q} \implies \paren {p \lor q}$

Proof

We apply the Method of Truth Tables.

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

$\begin{array}{|ccc|c|ccc|} \hline

(p & \land & q) & \implies & (p & \lor & q) \\ \hline \F & \F & \F & \T & \F & \F & \F \\ \F & \F & \T & \T & \F & \T & \T \\ \T & \F & \F & \T & \T & \T & \F \\ \T & \T & \T & \T & \T & \T & \T \\ \hline \end{array}$

$\blacksquare$

Sources

• 1971: Robert H. Kasriel: Undergraduate Topology ... (previous) ... (next): $\S1.2$: Some Remarks on the Use of the Connectives and, or, implies: Exercise $3$