Theorem elnelall 2415

Description: A contradiction concerning membership implies anything. (Contributed by Alexander van der Vekens, 25-Jan-2018.)

Assertion

Ref	Expression
elnelall	|- ( A e. B -> ( A e/ B -> ph ) )

Proof of Theorem elnelall

Step	Hyp	Ref	Expression
1		df-nel 2404	. 2 |- ( A e/ B <-> -. A e. B )
2		pm2.24 610	. 2 |- ( A e. B -> ( -. A e. B -> ph ) )
3	1, 2	syl5bi 151	1 |- ( A e. B -> ( A e/ B -> ph ) )

Syntax hints:   -. wn 3   -> wi 4   e. wcel 1480   e/ wnel 2403

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-in2 604

This theorem depends on definitions:  df-bi 116  df-nel 2404

This theorem is referenced by:  xnn0lenn0nn0  9653