Theorem neeq2d 2419

Description: Deduction for inequality. (Contributed by NM, 25-Oct-1999.)

Hypothesis

Ref	Expression
neeq1d.1	|- ( ph -> A = B )

Assertion

Ref	Expression
neeq2d	|- ( ph -> ( C =/= A <-> C =/= B ) )

Proof of Theorem neeq2d

Step	Hyp	Ref	Expression
1		neeq1d.1	. 2 |- ( ph -> A = B )
2		neeq2 2414	. 2 |- ( A = B -> ( C =/= A <-> C =/= B ) )
3	1, 2	syl 14	1 |- ( ph -> ( C =/= A <-> C =/= B ) )

Syntax hints:   -> wi 4   <-> wb 105   = wceq 1395   =/= wne 2400

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 617  ax-in2 618  ax-5 1493  ax-gen 1495  ax-4 1556  ax-17 1572  ax-ext 2211

This theorem depends on definitions:  df-bi 117  df-cleq 2222  df-ne 2401

This theorem is referenced by:  neeq12d  2420  neeqtrd  2428  sqrt2irr  12679  ennnfonelemk  12966  ennnfoneleminc  12977  ennnfonelemex  12980  ennnfonelemnn0  12988  ennnfonelemr  12989  setscomd  13068