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