Implication is Left Distributive over Conjunction/Formulation 1/Proof 1
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Theorem
- $p \implies \paren {q \land r} \dashv \vdash \paren {p \implies q} \land \paren {p \implies r}$
Proof
Proof of Forward Implication
By the tableau method of natural deduction:
Line | Pool | Formula | Rule | Depends upon | Notes | |
---|---|---|---|---|---|---|
1 | 1 | $p \implies \paren {q \land r}$ | Premise | (None) | ||
2 | 2 | $p$ | Assumption | (None) | ||
3 | 1, 2 | $q \land r$ | Modus Ponendo Ponens: $\implies \mathcal E$ | 1, 2 | ||
4 | 1, 2 | $q$ | Rule of Simplification: $\land \EE_1$ | 3 | ||
5 | 1, 2 | $r$ | Rule of Simplification: $\land \EE_2$ | 3 | ||
6 | 1 | $p \implies q$ | Rule of Implication: $\implies \II$ | 2 – 4 | Assumption 2 has been discharged | |
7 | 1 | $p \implies r$ | Rule of Implication: $\implies \II$ | 2 – 5 | Assumption 2 has been discharged | |
8 | 1 | $\paren {p \implies q} \land \paren {p \implies r}$ | Rule of Conjunction: $\land \II$ | 6, 7 |
$\blacksquare$
Proof of Reverse Implication
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 |
$\blacksquare$