Implication is Left Distributive over Conjunction/Reverse Implication/Formulation 1/Proof
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Theorem
- $\paren {p \implies q} \land \paren {p \implies r} \vdash p \implies \paren {q \land r}$
Proof
By the tableau method of natural deduction:
| Line | Pool | Formula | Rule | Depends upon | Notes | |
|---|---|---|---|---|---|---|
| 1 | 1 | $\paren {p \implies q} \land \paren {p \implies r}$ | Premise | (None) | ||
| 2 | 1 | $\paren {p \land p} \implies \paren {q \land r}$ | Sequent Introduction | 1 | Praeclarum Theorema | |
| 3 | 3 | $p$ | Assumption | (None) | ||
| 4 | 3 | $p \land p$ | Sequent Introduction | 3 | Rule of Idempotence: Conjunction | |
| 5 | 1, 3 | $q \land r$ | Modus Ponendo Ponens: $\implies \mathcal E$ | 2, 4 | ||
| 6 | 1 | $p \implies \paren {q \land r}$ | Rule of Implication: $\implies \II$ | 3 – 5 | Assumption 3 has been discharged |
$\blacksquare$