Proof by Cases/Sequent Form
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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 1
By the tableau method of natural deduction:
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 $r$.
$\blacksquare$