Conditional is Left Distributive over Conjunction/Formulation 1/Proof 1

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:

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

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:

$\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

$\blacksquare$