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  11265  algcvg  12845  algcvga  12848  eucalgcvga  12855  rpdvds  12896  phibndlem  13017  dfphi2  13021  pcaddlem  13141  ennnfoneleminc  13354  ennnfonelemex  13357  ennnfonelemhom  13358  ennnfonelemnn0  13365  ennnfonelemr  13366  ennnfonelemim  13367  ctinfomlemom  13370  setscomd  13445  rrgsupp  14658  pellexlem3  16192  lgsne0  16323  umgr2cwwkdifex  16832  dceqnconst  17277  dcapnconst  17278  nconstwlpolem  17282