Factor Principles/Conjunction on Right/Formulation 1/Proof by Truth Table
Theorem
- $p \implies q \vdash \paren {p \land r} \implies \paren {q \land r}$
Proof
Proof by Truth Table:
$\begin{array}{|ccc||ccccccccccc|} \hline p & q & r & (p & \implies & q) & \implies & ((p & \land & r) & \implies & (q & \land & r)) \\ \hline \T & \T & \T & \T & \T & \T & \T & \T & \T & \T & \T & \T & \T & \T \\ \T & \T & \F & \T & \T & \T & \T & \T & \F & \F & \T & \T & \F & \F \\ \T & \F & \T & \T & \F & \F & \T & \T & \T & \T & \F & \F & \F & \T \\ \T & \F & \F & \T & \F & \F & \T & \T & \F & \F & \T & \F & \F & \F \\ \F & \T & \T & \F & \T & \T & \T & \F & \F & \T & \T & \T & \T & \T \\ \F & \T & \F & \F & \T & \T & \T & \F & \F & \F & \T & \T & \F & \F \\ \F & \F & \T & \F & \T & \F & \T & \F & \F & \T & \T & \F & \F & \T \\ \F & \F & \F & \F & \T & \F & \T & \F & \F & \F & \T & \F & \F & \F \\ \hline \end{array}$
$\blacksquare$