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  3791  pw2f1odclem  6991  hashprg  11025  algcvg  12565  algcvga  12568  eucalgcvga  12575  rpdvds  12616  phibndlem  12733  dfphi2  12737  pcaddlem  12857  ennnfoneleminc  12977  ennnfonelemex  12980  ennnfonelemhom  12981  ennnfonelemnn0  12988  ennnfonelemr  12989  ennnfonelemim  12990  ctinfomlemom  12993  setscomd  13068  lgsne0  15711  dceqnconst  16387  dcapnconst  16388  nconstwlpolem  16392