Metamath Proof Explorer


Theorem neor

Description: Logical OR with an equality. (Contributed by NM, 29-Apr-2007)

Ref Expression
Assertion neor ⊢ A = B ∨ ψ ↔ A ≠ B → ψ

Proof

Step Hyp Ref Expression
1 df-or ⊢ A = B ∨ ψ ↔ ¬ A = B → ψ
2 df-ne ⊢ A ≠ B ↔ ¬ A = B
3 2 imbi1i ⊢ A ≠ B → ψ ↔ ¬ A = B → ψ
4 1 3 bitr4i ⊢ A = B ∨ ψ ↔ A ≠ B → ψ