Principle of Dilemma/Formulation 2/Forward Implication/Proof 1

Theorem

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

Proof

By the tableau method of natural deduction:

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

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

$\blacksquare$