Proof by Cases/Sequent Form/Proof 2
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Theorem
Proof by Cases can be symbolised by the sequent:
- $p \lor q, \paren {p \vdash r}, \paren {q \vdash r} \vdash r$
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 $r$.
$\blacksquare$