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

Theorem

 * $p \implies q \vdash \left({p \lor r}\right) \implies \left ({q \lor r}\right)$

Proof
As can be seen for all boolean interpretations by inspection, where the truth value under the main connective on the LHS is $T$, that under the one on the RHS 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}$