Proof by Cases/Formulation 1

Theorem

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

This can be expressed as two separate theorems:

Forward Implication

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

Reverse Implication

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

Proof

We apply the Method of Truth Tables.

As can be seen by inspection, the truth values under the main connectives match for all boolean interpretations.

$\begin{array}{|ccccccc||ccccc|} \hline (p & \implies & r) & \land & (q & \implies & r) & (p & \lor & q) & \implies & r \\ \hline \F & \T & \F & \T & \F & \T & \F & \F & \F & \F & \T & \F \\ \F & \T & \T & \T & \F & \T & \T & \F & \F & \F & \T & \T \\ \F & \T & \F & \F & \T & \F & \F & \F & \T & \T & \F & \F \\ \F & \T & \T & \T & \T & \T & \T & \F & \T & \T & \T & \T \\ \T & \F & \F & \F & \F & \T & \F & \T & \T & \F & \F & \F \\ \T & \T & \T & \T & \F & \T & \T & \T & \T & \F & \T & \T \\ \T & \F & \F & \F & \T & \F & \F & \T & \T & \T & \F & \F \\ \T & \T & \T & \T & \T & \T & \T & \T & \T & \T & \T & \T \\ \hline \end{array}$

$\blacksquare$