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  11263  algcvg  12842  algcvga  12845  eucalgcvga  12852  rpdvds  12893  phibndlem  13014  dfphi2  13018  pcaddlem  13138  ennnfoneleminc  13351  ennnfonelemex  13354  ennnfonelemhom  13355  ennnfonelemnn0  13362  ennnfonelemr  13363  ennnfonelemim  13364  ctinfomlemom  13367  setscomd  13442  rrgsupp  14623  pellexlem3  16150  lgsne0  16255  umgr2cwwkdifex  16764  dceqnconst  17208  dcapnconst  17209  nconstwlpolem  17213