Theorem neeq2d 2386

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 2381	. 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 1364   =/= wne 2367

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 615  ax-in2 616  ax-5 1461  ax-gen 1463  ax-4 1524  ax-17 1540  ax-ext 2178

This theorem depends on definitions:  df-bi 117  df-cleq 2189  df-ne 2368

This theorem is referenced by:  neeq12d  2387  neeqtrd  2395  sqrt2irr  12330  ennnfonelemk  12617  ennnfoneleminc  12628  ennnfonelemex  12631  ennnfonelemnn0  12639  ennnfonelemr  12640  setscomd  12719