Proof by Cases/Sequent Form/Proof 2

Theorem

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

Proof
We apply the Method of Truth Tables.

$\begin{array}{|ccc|ccc|ccc||c|} \hline p & \lor & q & p & \implies & r & q & \implies & r & r \\ \hline F & F & F & F & T & F & F & T & F & F \\ F & F & F & F & T & T & F & T & T & T \\ F & T & T & F & T & F & T & F & F & F \\ F & T & T & F & T & T & T & T & T & T \\ T & T & F & T & F & F & F & T & F & F \\ T & T & F & T & T & T & F & T & T & T \\ T & T & T & T & F & F & T & F & F & F \\ T & T & T & T & T & T & T & T & T & T \\ \hline \end{array}$

As can be seen, when $p \lor q$, $p \implies r$ and $q \implies r$ are all true, then so is $r$.