Principle of Commutation/Reverse Implication/Formulation 2

Theorem

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

Proof

By the tableau method of natural deduction:

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

Line		Pool	Formula	Rule	Depends upon	Notes
1		1	$q \implies \paren {p \implies r}$	Assumption	(None)
2		1	$p \implies \paren {q \implies r}$	Sequent Introduction	1	Principle of Commutation: Reverse Implication: Formulation 1
3			$\paren {q \implies \paren {p \implies r} } \implies \paren {p \implies \paren {q \implies r} }$	Rule of Implication: $\implies \II$	1 – 2	Assumption 1 has been discharged

$\blacksquare$