Principle of Commutation/Forward Implication/Formulation 1/Proof

Theorem

$p \implies \left({q \implies r}\right) \vdash q \implies \left({p \implies r}\right)$

Proof

By the tableau method of natural deduction:

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

Line		Pool	Formula	Rule	Depends upon	Notes
1		1	$p \implies \paren {q \implies r}$	Premise	(None)
2		2	$q$	Assumption	(None)
3		3	$p$	Assumption	(None)
4		1, 3	$q \implies r$	Modus Ponendo Ponens: $\implies \mathcal E$	1, 3
5		1, 2, 3	$r$	Modus Ponendo Ponens: $\implies \mathcal E$	2, 4
6		1, 2	$p \implies r$	Rule of Implication: $\implies \II$	3 – 5	Assumption 3 has been discharged
7		1	$q \implies \paren {p \implies r}$	Rule of Implication: $\implies \II$	2 – 6	Assumption 2 has been discharged

$\blacksquare$

Sources

• 1965: E.J. Lemmon: Beginning Logic ... (previous) ... (next): Chapter $1$: The Propositional Calculus $1$: $2$ Conditionals and Negation: Theorem $10$

• Metamath: Theorem com12: Commuting antecedents in an implication