Conjunction of Disjunctions with Complements implies Disjunction
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Theorem
- $\paren {p \lor r} \land \paren {q \lor \neg r} \vdash p \lor q$
Proof
By the tableau method of natural deduction:
| Line | Pool | Formula | Rule | Depends upon | Notes | |
|---|---|---|---|---|---|---|
| 1 | 1 | $\paren {p \lor r} \land \paren {q \lor \neg r}$ | Premise | (None) | ||
| 2 | 1 | $p \lor r$ | Rule of Simplification: $\land \EE_1$ | 1 | The aim is to use Proof by Cases on this ... | |
| 3 | 3 | $p$ | Assumption | (None) | Assume the first of the disjuncts ... | |
| 4 | 3 | $p \lor q$ | Rule of Addition: $\lor \II_1$ | 3 | ... and demonstrate the conclusion | |
| 5 | 1 | $q \lor \neg r$ | Rule of Simplification: $\land \EE_2$ | 1 | ||
| 6 | 1 | $\neg r \lor q$ | Sequent Introduction | 5 | Disjunction is Commutative | |
| 7 | 1 | $r \implies q$ | Sequent Introduction | 6 | Rule of Material Implication | |
| 8 | 8 | $r$ | Assumption | (None) | then assume the second of the disjuncts ... | |
| 9 | 1, 8 | $q$ | Modus Ponendo Ponens: $\implies \mathcal E$ | 7, 8 | ||
| 10 | 1, 8 | $p \lor q$ | Rule of Addition: $\lor \II_2$ | 9 | ... and demonstrate the conclusion | |
| 11 | 1 | $p \lor q$ | Proof by Cases: $\text{PBC}$ | 2, 3 – 4, 8 – 10 | Assumptions 3 and 8 have been discharged |
$\blacksquare$