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Metamath Proof Explorer

Theorem neeq2d

Description: Deduction for inequality. (Contributed by NM, 25-Oct-1999) (Proof shortened by Wolf Lammen, 19-Nov-2019)

Ref Expression
Hypothesis neeq1d.1 ⊢ φ → A = B
Assertion neeq2d ⊢ φ → C ≠ A ↔ C ≠ B

Proof

Step Hyp Ref Expression
1 neeq1d.1 ⊢ φ → A = B
2 1 eqeq2d ⊢ φ → C = A ↔ C = B
3 2 necon3bid ⊢ φ → C ≠ A ↔ C ≠ B