# Proof by Cases/Sequent Form

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## Theorem

Proof by Cases can be symbolised by the sequent:

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

## 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 $r$.

$\blacksquare$