Theorem eqneqall 2255

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

Step	Hyp	Ref	Expression
1		df-ne 2246	. 2 |- ( A =/= B <-> -. A = B )
2		pm2.24 583	. 2 |- ( A = B -> ( -. A = B -> ph ) )
3	1, 2	syl5bi 150	1 |- ( A = B -> ( A =/= B -> ph ) )

Syntax hints:   -. wn 3   -> wi 4   = wceq 1284   =/= wne 2245

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 104  ax-in2 577

This theorem depends on definitions:  df-bi 115  df-ne 2246

This theorem is referenced by:  modfzo0difsn  9397  nno  10306  prm2orodd  10508