Theorem neor 2600

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

Assertion

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

Proof of Theorem neor

Step	Hyp	Ref	Expression
1		df-or 359	. 2 ⊢ ((A = B ∨ ψ) ↔ (¬ A = B → ψ))
2		df-ne 2518	. . 3 ⊢ (A ≠ B ↔ ¬ A = B)
3	2	imbi1i 315	. 2 ⊢ ((A ≠ B → ψ) ↔ (¬ A = B → ψ))
4	1, 3	bitr4i 243	1 ⊢ ((A = B ∨ ψ) ↔ (A ≠ B → ψ))

Syntax hints:  ¬ wn 3   → wi 4   ↔ wb 176   ∨ wo 357   = wceq 1642   ≠ wne 2516

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8

This theorem depends on definitions:  df-bi 177  df-or 359  df-ne 2518

This theorem is referenced by: (None)