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Metamath Proof Explorer

Theorem efald

Description: Deduction based on reductio ad absurdum. (Contributed by Mario Carneiro, 9-Feb-2017)

		Ref	Expression
	Hypothesis	efald.1	$⊢ (φ \land \neg ψ) \to ⊥$
	Assertion	efald	$⊢ φ \to ψ$

Step	Hyp	Ref	Expression
1		efald.1	$⊢ (φ \land \neg ψ) \to ⊥$
2	1	inegd	$⊢ φ \to \neg \neg ψ$
3	2	notnotrd	$⊢ φ \to ψ$