Theorem neor 3025

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 848	. 2 ⊢ ((𝐴 = 𝐵 ∨ 𝜓) ↔ (¬ 𝐴 = 𝐵 → 𝜓))
2		df-ne 2934	. . 3 ⊢ (𝐴 ≠ 𝐵 ↔ ¬ 𝐴 = 𝐵)
3	2	imbi1i 349	. 2 ⊢ ((𝐴 ≠ 𝐵 → 𝜓) ↔ (¬ 𝐴 = 𝐵 → 𝜓))
4	1, 3	bitr4i 278	1 ⊢ ((𝐴 = 𝐵 ∨ 𝜓) ↔ (𝐴 ≠ 𝐵 → 𝜓))

Syntax hints:  ¬ wn 3   → wi 4   ↔ wb 206   ∨ wo 847   = wceq 1540   ≠ wne 2933

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 207  df-or 848  df-ne 2934

This theorem is referenced by:  frsn  5747  ord0eln0  6413  fimaxre  12191  fiminre  12194  prime  12679  h1datomi  31567  elat2  32326  bnj563  34779  divrngidl  38057  dmncan1  38105  lkrshp4  39131  cvrcmp  39306  leat2  39317  isat3  39330  2llnmat  39548  2lnat  39808