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 849	. 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 848   = wceq 1542   ≠ 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 849  df-ne 2934

This theorem is referenced by:  frsn  5720  ord0eln0  6381  fimaxre  12098  fiminre  12101  prime  12585  h1datomi  31668  elat2  32427  bnj563  34919  divrngidl  38276  dmncan1  38324  dfdisjALTV5a  39051  dfeldisj5a  39062  lkrshp4  39481  cvrcmp  39656  leat2  39667  isat3  39680  2llnmat  39897  2lnat  40157