Theorem neeq2d 1197

Description: Deduction for inequality.

Hypothesis

Ref	Expression
neeq1d.1	⊢ (φ → A = B)

Assertion

Ref	Expression
neeq2d	⊢ (φ → (C ≠ A ↔ C ≠ B))

Proof of Theorem neeq2d

Step	Hyp	Ref	Expression
1		neeq1d.1	. 2 ⊢ (φ → A = B)
2		neeq2 1195	. 2 ⊢ (A = B → (C ≠ A ↔ C ≠ B))
3	1, 2	syl 12	1 ⊢ (φ → (C ≠ A ↔ C ≠ B))

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

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

metamath.org