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  3792  pw2f1odclem  7003  hashprg  11043  algcvg  12585  algcvga  12588  eucalgcvga  12595  rpdvds  12636  phibndlem  12753  dfphi2  12757  pcaddlem  12877  ennnfoneleminc  12997  ennnfonelemex  13000  ennnfonelemhom  13001  ennnfonelemnn0  13008  ennnfonelemr  13009  ennnfonelemim  13010  ctinfomlemom  13013  setscomd  13088  lgsne0  15732  dceqnconst  16488  dcapnconst  16489  nconstwlpolem  16493