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Metamath Proof Explorer

Theorem necon3bid

Description: Deduction from equality to inequality. (Contributed by NM, 23-Feb-2005) (Proof shortened by Andrew Salmon, 25-May-2011)

Ref Expression
Hypothesis necon3bid.1 ⊢ φ → A = B ↔ C = D
Assertion necon3bid ⊢ φ → A ≠ B ↔ C ≠ D

Proof

Step Hyp Ref Expression
1 necon3bid.1 ⊢ φ → A = B ↔ C = D
2 df-ne ⊢ A ≠ B ↔ ¬ A = B
3 1 necon3bbid ⊢ φ → ¬ A = B ↔ C ≠ D
4 2 3 bitrid ⊢ φ → A ≠ B ↔ C ≠ D