Theorem eqneqall 2410

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 2401	. 2 |- ( A =/= B <-> -. A = B )
2		pm2.24 624	. 2 |- ( A = B -> ( -. A = B -> ph ) )
3	1, 2	biimtrid 152	1 |- ( A = B -> ( A =/= B -> ph ) )

Syntax hints:   -. wn 3   -> wi 4   = wceq 1395   =/= wne 2400

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-in2 618

This theorem depends on definitions:  df-bi 117  df-ne 2401

This theorem is referenced by:  ssprsseq  3830  eldju2ndl  7239  eldju2ndr  7240  modfzo0difsn  10617  nno  12417  prm2orodd  12648  prm23lt5  12786  dvdsprmpweqnn  12859  logbgcd1irr  15641  gausslemma2dlem0f  15733  gausslemma2dlem0i  15736  2lgs  15783  2lgsoddprm  15792  umgrnloop2  15949  uhgr2edg  16004