Conditional is Left Distributive over Conjunction/Reverse Implication/Formulation 1

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:

$\paren {p \implies q} \land \paren {p \implies r} \vdash p \implies \paren {q \land r} $

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 $1.4: \ 2 \ \text{(k)}$