Theorem neeq2 1195

Description: Equality theorem for inequality.

Assertion

Ref	Expression
neeq2	⊢ (A = B → (C ≠ A ↔ C ≠ B))

Proof of Theorem neeq2

Step	Hyp	Ref	Expression
1		cleq2 1110	. . 3 ⊢ (A = B → (C = A ↔ C = B))
2	1	negbid 463	. 2 ⊢ (A = B → (¬ C = A ↔ ¬ C = B))
3		df-ne 1192	. 2 ⊢ (C ≠ A ↔ ¬ C = A)
4		df-ne 1192	. 2 ⊢ (C ≠ B ↔ ¬ C = B)
5	2, 3, 4	3bitr4g 428	1 ⊢ (A = B → (C ≠ A ↔ C ≠ B))

Syntax hints:  ¬ wn 1   → wi 2   ↔ wb 127   = wceq 1091   ≠ wne 1190

This theorem is referenced by:  neeq2d 1197  psseq2 1560

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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