Theorem eqneqall 2345

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 2336	. 2 |- ( A =/= B <-> -. A = B )
2		pm2.24 611	. 2 |- ( A = B -> ( -. A = B -> ph ) )
3	1, 2	syl5bi 151	1 |- ( A = B -> ( A =/= B -> ph ) )

Syntax hints:   -. wn 3   -> wi 4   = wceq 1343   =/= wne 2335

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-in2 605

This theorem depends on definitions:  df-bi 116  df-ne 2336

This theorem is referenced by:  eldju2ndl  7033  eldju2ndr  7034  modfzo0difsn  10326  nno  11839  prm2orodd  12054  prm23lt5  12191  dvdsprmpweqnn  12263  logbgcd1irr  13485