Proof by Cases/Formulation 2/Reverse Implication

Theorem

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

Proof

By the tableau method of natural deduction:

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

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

$\blacksquare$