Theorem neeq1d 2438

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 2433	. 2 |- ( A = B -> ( A =/= C <-> B =/= C ) )
3	1, 2	syl 14	1 |- ( ph -> ( A =/= C <-> B =/= C ) )

This proof depends on syntax axioms:   -> wi 4   <-> wb 105   = wceq 1402   =/= wne 2420

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-in1 623  ax-in2 624  ax-5 1500  ax-gen 1502  ax-4 1563  ax-17 1579  ax-ext 2220

This proof depends on definitions:  df-bi 117  df-cleq 2231  df-ne 2421

This theorem is used by:  neeq12d  2440  eqnetrd  2444  prnzg  3838  suppval1  6479  elsuppfng  6482  elsuppfn  6483  suppsnopdc  6490  ressuppss  6494  pw2f1odclem  7134  hashprg  11249  algcvg  12826  algcvga  12829  eucalgcvga  12836  rpdvds  12877  phibndlem  12994  dfphi2  12998  pcaddlem  13118  ennnfoneleminc  13302  ennnfonelemex  13305  ennnfonelemhom  13306  ennnfonelemnn0  13313  ennnfonelemr  13314  ennnfonelemim  13315  ctinfomlemom  13318  setscomd  13393  rrgsupp  14574  pellexlem3  16093  lgsne0  16157  umgr2cwwkdifex  16666  dceqnconst  17110  dcapnconst  17111  nconstwlpolem  17115