# Proof by Cases/Sequent Form

## Theorem

Proof by Cases can be symbolised by the sequent:

$p \lor q, \paren {p \vdash r}, \paren {q \vdash r} \vdash r$

## Proof 1

By the tableau method of natural deduction:

$p \lor q, \paren {p \vdash r}, \paren {q \vdash r} \vdash r$
Line Pool Formula Rule Depends upon Notes
1 1 $p \lor q$ Premise (None)
2 2 $p$ Assumption (None)
3 2 $r$ By hypothesis 2 as $p \vdash r$
4 4 $q$ Assumption (None)
5 4 $r$ By hypothesis 4 as $q \vdash r$
6 1 $r$ Proof by Cases: $\text{PBC}$ 1 , 2 – 3 , 4 – 5 Assumptions 2  and 4 have been discharged

$\blacksquare$

## Proof 2

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 $r$.

$\blacksquare$