Proof by Contradiction/Variant 2/Formulation 2

Theorem

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

Proof

By the tableau method of natural deduction:

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

Line		Pool	Formula	Rule	Depends upon	Notes
1		1	$\paren {p \implies q} \land \paren {p \implies \neg q}$	Assumption	(None)
2		1	$p \implies q$	Rule of Simplification: $\land \EE_1$	1
3		1	$p \implies \neg q$	Rule of Simplification: $\land \EE_2$	1
4		1	$\neg p$	Sequent Introduction	2, 3	Proof by Contradiction: Formulation 1
5			$\paren {\paren {p \implies q} \land \paren {p \implies \neg q} } \implies \neg p$	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): $\text{II}$: 'AND', 'OR', 'IF AND ONLY IF': $\S 3$: Theorem $\text{T34}$