Theorem eqneqall 2318

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 2309	. 2 |- ( A =/= B <-> -. A = B )
2		pm2.24 610	. 2 |- ( A = B -> ( -. A = B -> ph ) )
3	1, 2	syl5bi 151	1 |- ( A = B -> ( A =/= B -> ph ) )

Syntax hints:   -. wn 3   -> wi 4   = wceq 1331   =/= wne 2308

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-ne 2309

This theorem is referenced by:  eldju2ndl  6957  eldju2ndr  6958  modfzo0difsn  10168  nno  11603  prm2orodd  11807