Theorem ud3lem3c 574

Description: Lemma for unified disjunction. (Contributed by NM, 27-Nov-1997.)

Assertion

Ref	Expression
ud3lem3c	((a →3 b)⊥ ∪ (a ∪ b)) = (a ∪ b)

Proof of Theorem ud3lem3c

Step	Hyp	Ref	Expression
1		ud3lem0c 279	. . . 4 (a →3 b)⊥ = (((a ∪ b⊥ ) ∩ (a ∪ b)) ∩ (a⊥ ∪ (a ∩ b⊥ )))
2		an32 83	. . . . 5 (((a ∪ b⊥ ) ∩ (a ∪ b)) ∩ (a⊥ ∪ (a ∩ b⊥ ))) = (((a ∪ b⊥ ) ∩ (a⊥ ∪ (a ∩ b⊥ ))) ∩ (a ∪ b))
3		ancom 74	. . . . 5 (((a ∪ b⊥ ) ∩ (a⊥ ∪ (a ∩ b⊥ ))) ∩ (a ∪ b)) = ((a ∪ b) ∩ ((a ∪ b⊥ ) ∩ (a⊥ ∪ (a ∩ b⊥ ))))
4	2, 3	ax-r2 36	. . . 4 (((a ∪ b⊥ ) ∩ (a ∪ b)) ∩ (a⊥ ∪ (a ∩ b⊥ ))) = ((a ∪ b) ∩ ((a ∪ b⊥ ) ∩ (a⊥ ∪ (a ∩ b⊥ ))))
5	1, 4	ax-r2 36	. . 3 (a →3 b)⊥ = ((a ∪ b) ∩ ((a ∪ b⊥ ) ∩ (a⊥ ∪ (a ∩ b⊥ ))))
6	5	ax-r5 38	. 2 ((a →3 b)⊥ ∪ (a ∪ b)) = (((a ∪ b) ∩ ((a ∪ b⊥ ) ∩ (a⊥ ∪ (a ∩ b⊥ )))) ∪ (a ∪ b))
7		ax-a2 31	. . 3 (((a ∪ b) ∩ ((a ∪ b⊥ ) ∩ (a⊥ ∪ (a ∩ b⊥ )))) ∪ (a ∪ b)) = ((a ∪ b) ∪ ((a ∪ b) ∩ ((a ∪ b⊥ ) ∩ (a⊥ ∪ (a ∩ b⊥ )))))
8		orabs 120	. . 3 ((a ∪ b) ∪ ((a ∪ b) ∩ ((a ∪ b⊥ ) ∩ (a⊥ ∪ (a ∩ b⊥ ))))) = (a ∪ b)
9	7, 8	ax-r2 36	. 2 (((a ∪ b) ∩ ((a ∪ b⊥ ) ∩ (a⊥ ∪ (a ∩ b⊥ )))) ∪ (a ∪ b)) = (a ∪ b)
10	6, 9	ax-r2 36	1 ((a →3 b)⊥ ∪ (a ∪ b)) = (a ∪ b)

This proof depends on syntax axioms:   = wb 1  ^⊥ wn 4   ∪ wo 6   ∩ wa 7   →₃ wi3 14

This proof depends on axioms:  ax-a1 30  ax-a2 31  ax-a3 32  ax-a5 34  ax-r1 35  ax-r2 36  ax-r4 37  ax-r5 38

This proof depends on definitions:  df-a 40  df-i3 46

This theorem is used by:  ud3lem3d  575