Theorem neor 3050

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 861	. 2 ⊢ ((𝐴 = 𝐵 ∨ 𝜓) ↔ (¬ 𝐴 = 𝐵 → 𝜓))
2		df-ne 2959	. . 3 ⊢ (𝐴 ≠ 𝐵 ↔ ¬ 𝐴 = 𝐵)
3	2	imbi1i 352	. 2 ⊢ ((𝐴 ≠ 𝐵 → 𝜓) ↔ (¬ 𝐴 = 𝐵 → 𝜓))
4	1, 3	bitr4i 281	1 ⊢ ((𝐴 = 𝐵 ∨ 𝜓) ↔ (𝐴 ≠ 𝐵 → 𝜓))

Syntax hints:  ¬ wn 3   → wi 4   ↔ wb 209   ∨ wo 860   = wceq 1570   ≠ wne 2958

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 210  df-or 861  df-ne 2959

This theorem is referenced by:  frsn  5751  ord0eln0  6419  fimaxre  12160  fiminre  12163  prime  12678  h1datomi  31914  elat2  32673  bnj563  35113  divrngidl  38660  dmncan1  38708  dfdisjALTV5a  39433  dfeldisj5a  39444  lkrshp4  39863  cvrcmp  40038  leat2  40049  isat3  40062  2llnmat  40279  2lnat  40539  idomcanl  49095