Factor Principles/Disjunction on Right/Formulation 1/Proof 4

Theorem

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

Proof

As can be seen for all boolean interpretations by inspection, where the truth value under the main connective on the left hand side is $T$, that under the one on the right hand side is also $T$:

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

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

$\blacksquare$