Theorem eqneqall 2357

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 2348	. 2 |- ( A =/= B <-> -. A = B )
2		pm2.24 621	. 2 |- ( A = B -> ( -. A = B -> ph ) )
3	1, 2	biimtrid 152	1 |- ( A = B -> ( A =/= B -> ph ) )

Syntax hints:   -. wn 3   -> wi 4   = wceq 1353   =/= wne 2347

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-in2 615

This theorem depends on definitions:  df-bi 117  df-ne 2348

This theorem is referenced by:  eldju2ndl  7073  eldju2ndr  7074  modfzo0difsn  10397  nno  11913  prm2orodd  12128  prm23lt5  12265  dvdsprmpweqnn  12337  logbgcd1irr  14470