Theorem neor 2600

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

Assertion

Ref	Expression
neor	|- ((A = B \/ ps) <-> (A =/= B -> ps))

Proof of Theorem neor

Step	Hyp	Ref	Expression
1		df-or 359	. 2 |- ((A = B \/ ps) <-> (-. A = B -> ps))
2		df-ne 2518	. . 3 |- (A =/= B <-> -. A = B)
3	2	imbi1i 315	. 2 |- ((A =/= B -> ps) <-> (-. A = B -> ps))
4	1, 3	bitr4i 243	1 |- ((A = B \/ ps) <-> (A =/= B -> ps))

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)