Implication is Left Distributive over Conjunction/Reverse Implication/Formulation 1
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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 |
$\blacksquare$
Sources
- 1965: E.J. Lemmon: Beginning Logic ... (previous) ... (next): Chapter $1$: The Propositional Calculus $1$: $3$ Conjunction and Disjunction: Exercise $1 \ \text{(c)}$
- 2000: Michael R.A. Huth and Mark D. Ryan: Logic in Computer Science: Modelling and reasoning about systems ... (previous) ... (next): $\S 1.2.1$: Rules for natural deduction: Exercises $1.4: \ 2 \ \text{(k)}$