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Metamath Proof Explorer

Theorem bj-abf

Description: Shorter proof of abf (which should be kept as abfALT). (Contributed by BJ, 24-Jul-2019) (Proof modification is discouraged.)

		Ref	Expression
	Hypothesis	bj-abf.1	$⊢ \neg φ$
	Assertion	bj-abf	$⊢ \{x \| φ\} = \emptyset$

Step	Hyp	Ref	Expression
1		bj-abf.1	$⊢ \neg φ$
2		bj-ab0	$⊢ \forall x \neg φ \to \{x \| φ\} = \emptyset$
3	2 1	mpg	$⊢ \{x \| φ\} = \emptyset$