Theorem neeq1d 2420

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 2415	. 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 1397   =/= wne 2402

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 619  ax-in2 620  ax-5 1495  ax-gen 1497  ax-4 1558  ax-17 1574  ax-ext 2213

This theorem depends on definitions:  df-bi 117  df-cleq 2224  df-ne 2403

This theorem is referenced by:  neeq12d  2422  eqnetrd  2426  prnzg  3797  pw2f1odclem  7019  hashprg  11071  algcvg  12619  algcvga  12622  eucalgcvga  12629  rpdvds  12670  phibndlem  12787  dfphi2  12791  pcaddlem  12911  ennnfoneleminc  13031  ennnfonelemex  13034  ennnfonelemhom  13035  ennnfonelemnn0  13042  ennnfonelemr  13043  ennnfonelemim  13044  ctinfomlemom  13047  setscomd  13122  lgsne0  15766  umgr2cwwkdifex  16275  dceqnconst  16664  dcapnconst  16665  nconstwlpolem  16669