Conditional is Left Distributive over Conjunction/Forward Implication/Formulation 1
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Theorem
- $p \implies \paren {q \land r} \vdash \paren {p \implies q} \land \paren {p \implies r}$
Proof
By the tableau method of natural deduction:
| Line | Pool | Formula | Rule | Depends upon | Notes | |
|---|---|---|---|---|---|---|
| 1 | 1 | $p \implies \paren {q \land r}$ | 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 |
$\blacksquare$
Sources
- 1965: E.J. Lemmon: Beginning Logic ... (previous) ... (next): Chapter $1$: The Propositional Calculus $1$: $3$ Conjunction and Disjunction: Exercise $1 \ \text{(i)}$
- 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 $1.4: \ 2 \ \text{(j)}$