Theorem eqneqall 2413

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 2404	. 2 |- ( A =/= B <-> -. A = B )
2		pm2.24 626	. 2 |- ( A = B -> ( -. A = B -> ph ) )
3	1, 2	biimtrid 152	1 |- ( A = B -> ( A =/= B -> ph ) )

Syntax hints:   -. wn 3   -> wi 4   = wceq 1398   =/= wne 2403

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-in2 620

This theorem depends on definitions:  df-bi 117  df-ne 2404

This theorem is referenced by:  ssprsseq  3840  eldju2ndl  7331  eldju2ndr  7332  modfzo0difsn  10720  nno  12547  prm2orodd  12778  prm23lt5  12916  dvdsprmpweqnn  12989  logbgcd1irr  15778  gausslemma2dlem0f  15873  gausslemma2dlem0i  15876  2lgs  15923  2lgsoddprm  15932  umgrnloop2  16092  uhgr2edg  16147  umgrclwwlkge2  16343