Modus Ponendo Ponens/Variant 2/Proof 1

Theorem

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

Proof

By the tableau method of natural deduction:

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

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

$\blacksquare$

Sources

• 1964: Donald Kalish and Richard Montague: Logic: Techniques of Formal Reasoning ... (previous) ... (next): $\text{I}$: 'NOT' and 'IF': $\S 5$: Theorem $\text{T3}$