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Theorem eqneqall 2259
Description: A contradiction concerning equality implies anything. (Contributed by Alexander van der Vekens, 25-Jan-2018.)
Assertion
Ref Expression
eqneqall  |-  ( A  =  B  ->  ( A  =/=  B  ->  ph )
)

Proof of Theorem eqneqall
StepHypRef Expression
1 df-ne 2250 . 2  |-  ( A  =/=  B  <->  -.  A  =  B )
2 pm2.24 584 . 2  |-  ( A  =  B  ->  ( -.  A  =  B  ->  ph ) )
31, 2syl5bi 150 1  |-  ( A  =  B  ->  ( A  =/=  B  ->  ph )
)
Colors of variables: wff set class
Syntax hints:   -. wn 3    -> wi 4    = wceq 1285    =/= wne 2249
This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 104  ax-in2 578
This theorem depends on definitions:  df-bi 115  df-ne 2250
This theorem is referenced by:  modfzo0difsn  9529  nno  10513  prm2orodd  10715
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