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Theorem elnelall 2416
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
StepHypRef Expression
1 df-nel 2405 . 2  |-  ( A  e/  B  <->  -.  A  e.  B )
2 pm2.24 611 . 2  |-  ( A  e.  B  ->  ( -.  A  e.  B  ->  ph ) )
31, 2syl5bi 151 1  |-  ( A  e.  B  ->  ( A  e/  B  ->  ph )
)
Colors of variables: wff set class
Syntax hints:   -. wn 3    -> wi 4    e. wcel 1481    e/ wnel 2404
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-in2 605
This theorem depends on definitions:  df-bi 116  df-nel 2405
This theorem is referenced by:  xnn0lenn0nn0  9678
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