Rule of Distribution/Disjunction Distributes over Conjunction/Left Distributive/Formulation 1/Forward Implication
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Theorem
- $p \lor \paren {q \land r} \vdash \paren {p \lor q} \land \paren {p \lor r}$
Proof
By the tableau method of natural deduction:
| Line | Pool | Formula | Rule | Depends upon | Notes | |
|---|---|---|---|---|---|---|
| 1 | 1 | $p \lor \paren {q \land r}$ | Premise | (None) | ||
| 2 | 2 | $p$ | Assumption | (None) | ||
| 3 | 2 | $p \lor q$ | Rule of Addition: $\lor \II_1$ | 2 | ||
| 4 | 2 | $p \lor r$ | Rule of Addition: $\lor \II_1$ | 2 | ||
| 5 | 2 | $\paren {p \lor q} \land \paren {p \lor r}$ | Rule of Conjunction: $\land \II$ | 3, 4 | ||
| 6 | 6 | $q \land r$ | Assumption | (None) | ||
| 7 | 6 | $q$ | Rule of Simplification: $\land \EE_1$ | 6 | ||
| 8 | 6 | $r$ | Rule of Simplification: $\land \EE_2$ | 6 | ||
| 9 | 6 | $p \lor q$ | Rule of Addition: $\lor \II_2$ | 7 | ||
| 10 | 6 | $p \lor r$ | Rule of Addition: $\lor \II_2$ | 8 | ||
| 11 | 6 | $\paren {p \lor q} \land \paren {p \lor r}$ | Rule of Conjunction: $\land \II$ | 7, 8 | ||
| 12 | 1 | $\paren {p \lor q} \land \paren {p \lor r}$ | Proof by Cases: $\text{PBC}$ | 1, 2 – 5, 6 – 11 | Assumptions 2 and 6 have been discharged |
$\blacksquare$