Factor Principles/Conjunction on Right/Formulation 1/Proof by Truth Table

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Theorem

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


Proof

Proof by Truth Table:

$\begin{array}{|ccc||ccccccccccc|} \hline p & q & r & (p & \implies & q) & \implies & ((p & \land & r) & \implies & (q & \land & r)) \\ \hline \T & \T & \T & \T & \T & \T & \T & \T & \T & \T & \T & \T & \T & \T \\ \T & \T & \F & \T & \T & \T & \T & \T & \F & \F & \T & \T & \F & \F \\ \T & \F & \T & \T & \F & \F & \T & \T & \T & \T & \F & \F & \F & \T \\ \T & \F & \F & \T & \F & \F & \T & \T & \F & \F & \T & \F & \F & \F \\ \F & \T & \T & \F & \T & \T & \T & \F & \F & \T & \T & \T & \T & \T \\ \F & \T & \F & \F & \T & \T & \T & \F & \F & \F & \T & \T & \F & \F \\ \F & \F & \T & \F & \T & \F & \T & \F & \F & \T & \T & \F & \F & \T \\ \F & \F & \F & \F & \T & \F & \T & \F & \F & \F & \T & \F & \F & \F \\ \hline \end{array}$

$\blacksquare$