Theorem neeq1d 2418

Description: Deduction for inequality. (Contributed by NM, 25-Oct-1999.)

Hypothesis

Ref	Expression
neeq1d.1	|- ( ph -> A = B )

Assertion

Ref	Expression
neeq1d	|- ( ph -> ( A =/= C <-> B =/= C ) )

Proof of Theorem neeq1d

Step	Hyp	Ref	Expression
1		neeq1d.1	. 2 |- ( ph -> A = B )
2		neeq1 2413	. 2 |- ( A = B -> ( A =/= C <-> B =/= C ) )
3	1, 2	syl 14	1 |- ( ph -> ( A =/= C <-> B =/= C ) )

Syntax hints:   -> wi 4   <-> wb 105   = wceq 1395   =/= wne 2400

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-in1 617  ax-in2 618  ax-5 1493  ax-gen 1495  ax-4 1556  ax-17 1572  ax-ext 2211

This theorem depends on definitions:  df-bi 117  df-cleq 2222  df-ne 2401

This theorem is referenced by:  neeq12d  2420  eqnetrd  2424  prnzg  3795  pw2f1odclem  7015  hashprg  11062  algcvg  12610  algcvga  12613  eucalgcvga  12620  rpdvds  12661  phibndlem  12778  dfphi2  12782  pcaddlem  12902  ennnfoneleminc  13022  ennnfonelemex  13025  ennnfonelemhom  13026  ennnfonelemnn0  13033  ennnfonelemr  13034  ennnfonelemim  13035  ctinfomlemom  13038  setscomd  13113  lgsne0  15757  umgr2cwwkdifex  16220  dceqnconst  16600  dcapnconst  16601  nconstwlpolem  16605