Theorem eqneqall 2377

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 2368	. 2 |- ( A =/= B <-> -. A = B )
2		pm2.24 622	. 2 |- ( A = B -> ( -. A = B -> ph ) )
3	1, 2	biimtrid 152	1 |- ( A = B -> ( A =/= B -> ph ) )

Syntax hints:   -. wn 3   -> wi 4   = wceq 1364   =/= wne 2367

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-in2 616

This theorem depends on definitions:  df-bi 117  df-ne 2368

This theorem is referenced by:  eldju2ndl  7147  eldju2ndr  7148  modfzo0difsn  10504  nno  12088  prm2orodd  12319  prm23lt5  12457  dvdsprmpweqnn  12530  logbgcd1irr  15287  gausslemma2dlem0f  15379  gausslemma2dlem0i  15382  2lgs  15429  2lgsoddprm  15438