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  7020  hashprg  11073  algcvg  12638  algcvga  12641  eucalgcvga  12648  rpdvds  12689  phibndlem  12806  dfphi2  12810  pcaddlem  12930  ennnfoneleminc  13050  ennnfonelemex  13053  ennnfonelemhom  13054  ennnfonelemnn0  13061  ennnfonelemr  13062  ennnfonelemim  13063  ctinfomlemom  13066  setscomd  13141  lgsne0  15786  umgr2cwwkdifex  16295  dceqnconst  16716  dcapnconst  16717  nconstwlpolem  16721