Conditional is Left Distributive over Disjunction/Formulation 1/Forward Implication
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Theorem
- $p \implies \paren {q \lor r} \vdash \paren {p \implies q} \lor \paren {p \implies r}$
Proof
By the tableau method of natural deduction:
| Line | Pool | Formula | Rule | Depends upon | Notes | |
|---|---|---|---|---|---|---|
| 1 | 1 | $p \implies \paren {q \lor r}$ | Assumption | (None) | ||
| 2 | 2 | $p$ | Assumption | (None) | ||
| 3 | 1, 2 | $q \lor r$ | Modus Ponendo Ponens: $\implies \mathcal E$ | 1, 2 | ||
| 4 | 2 | $p$ | Law of Identity | 2 | ||
| 5 | 5 | $q$ | Assumption | (None) | ||
| 6 | 5 | $p \implies q$ | Rule of Implication: $\implies \II$ | 4 – 5 | Assumption 4 has been discharged | |
| 7 | 5 | $\paren {p \implies q} \lor \paren {p \implies r}$ | Rule of Addition: $\lor \II_1$ | 6 | ||
| 8 | 8 | $r$ | Assumption | (None) | ||
| 9 | 8 | $p \implies r$ | Sequent Introduction | 8 | True Statement is implied by Every Statement | |
| 10 | 8 | $\paren {p \implies q} \lor \paren {p \implies r}$ | Rule of Addition: $\lor \II_2$ | 9 | ||
| 11 | 1 | $\paren {p \implies q} \lor \paren {p \implies r}$ | Proof by Cases: $\text{PBC}$ | 3, 2 – 7, 8 – 10 | Assumptions 2 and 8 have been discharged |
$\blacksquare$