Theorem neeq1d 1196

Description: Deduction for inequality.

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 1194	. 2 |- (A = B -> (A =/= C <-> B =/= C))
3	1, 2	syl 12	1 |- (ph -> (A =/= C <-> B =/= C))

Syntax hints:   -> wi 2   <-> wb 127   = wceq 1091   =/= wne 1190

This theorem is referenced by:  recneq0z 4232

This theorem was proved from axioms:  ax-1 3  ax-2 4  ax-3 5  ax-mp 6  ax-4 673  ax-5 674  ax-gen 677  ax-17 925  ax-ext 1074

This theorem depends on definitions:  df-bi 128  df-an 198  df-cleq 1097  df-ne 1192

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