Factor Principles/Disjunction on Right/Formulation 1/Proof 2
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Theorem
- $p \implies q \vdash \paren {p \lor r} \implies \paren {q \lor r}$
Proof
By the tableau method of natural deduction:
| Line | Pool | Formula | Rule | Depends upon | Notes | |
|---|---|---|---|---|---|---|
| 1 | 1 | $p \implies q$ | Premise | (None) | ||
| 2 | 2 | $p \lor r$ | Assumption | (None) | ||
| 3 | 3 | $r$ | Assumption | (None) | ||
| 4 | 3 | $q \lor r$ | Rule of Addition: $\lor \II_2$ | 3 | ||
| 5 | 5 | $p$ | Assumption | (None) | ||
| 6 | 1, 5 | $q$ | Modus Ponendo Ponens: $\implies \mathcal E$ | 1, 5 | ||
| 7 | 1, 5 | $q \lor r$ | Rule of Addition: $\lor \II_1$ | 6 | ||
| 8 | 1, 2 | $q \lor r$ | Proof by Cases: $\text{PBC}$ | 2, 5 – 7, 3 – 4 | Assumptions 5 and 3 have been discharged | |
| 9 | 1 | $\paren {p \lor r} \implies \paren {q \lor r}$ | Rule of Implication: $\implies \II$ | 2 – 8 | Assumption 2 has been discharged |
$\blacksquare$