Metamath Proof Explorer


Theorem necon3abid

Description: Deduction from equality to inequality. (Contributed by NM, 21-Mar-2007)

Ref Expression
Hypothesis necon3abid.1 ⊢ φ → A = B ↔ ψ
Assertion necon3abid ⊢ φ → A ≠ B ↔ ¬ ψ

Proof

Step Hyp Ref Expression
1 necon3abid.1 ⊢ φ → A = B ↔ ψ
2 df-ne ⊢ A ≠ B ↔ ¬ A = B
3 1 notbid ⊢ φ → ¬ A = B ↔ ¬ ψ
4 2 3 bitrid ⊢ φ → A ≠ B ↔ ¬ ψ