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  6483  elsuppfng  6486  elsuppfn  6487  suppsnopdc  6494  ressuppss  6498  pw2f1odclem  7140  hashprg  11270  algcvg  12850  algcvga  12853  eucalgcvga  12860  rpdvds  12901  phibndlem  13022  dfphi2  13026  pcaddlem  13146  ennnfoneleminc  13359  ennnfonelemex  13362  ennnfonelemhom  13363  ennnfonelemnn0  13370  ennnfonelemr  13371  ennnfonelemim  13372  ctinfomlemom  13375  setscomd  13450  rrgsupp  14682  pellexlem3  16248  lgsne0  16379  umgr2cwwkdifex  16888  dceqnconst  17334  dcapnconst  17335  nconstwlpolem  17339