Negation of Conditional implies Antecedent

Theorem

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

Proof

By the tableau method of natural deduction:

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

Line		Pool	Formula	Rule	Depends upon	Notes
1		1	$\neg \paren {p \implies q}$	Assumption	(None)
2		1	$p \land \neg q$	Sequent Introduction	1	Conjunction with Negative is Equivalent to Negation of Conditional
3		1	$p$	Rule of Simplification: $\land \EE_1$	2
4			$\neg \paren {p \implies q} \implies p$	Rule of Implication: $\implies \II$	1 – 3	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 $\text{T21}$