Theorem ud3lem3b 573

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

Assertion

Ref	Expression
ud3lem3b	((a →3 b)⊥ ∩ (a ∪ b)⊥ ) = 0

Proof of Theorem ud3lem3b

Step	Hyp	Ref	Expression
1		ud3lem0c 279	. . 3 (a →3 b)⊥ = (((a ∪ b⊥ ) ∩ (a ∪ b)) ∩ (a⊥ ∪ (a ∩ b⊥ )))
2	1	ran 78	. 2 ((a →3 b)⊥ ∩ (a ∪ b)⊥ ) = ((((a ∪ b⊥ ) ∩ (a ∪ b)) ∩ (a⊥ ∪ (a ∩ b⊥ ))) ∩ (a ∪ b)⊥ )
3		an32 83	. . 3 ((((a ∪ b⊥ ) ∩ (a ∪ b)) ∩ (a⊥ ∪ (a ∩ b⊥ ))) ∩ (a ∪ b)⊥ ) = ((((a ∪ b⊥ ) ∩ (a ∪ b)) ∩ (a ∪ b)⊥ ) ∩ (a⊥ ∪ (a ∩ b⊥ )))
4		anass 76	. . . . . 6 (((a ∪ b⊥ ) ∩ (a ∪ b)) ∩ (a ∪ b)⊥ ) = ((a ∪ b⊥ ) ∩ ((a ∪ b) ∩ (a ∪ b)⊥ ))
5		dff 101	. . . . . . . . 9 0 = ((a ∪ b) ∩ (a ∪ b)⊥ )
6	5	ax-r1 35	. . . . . . . 8 ((a ∪ b) ∩ (a ∪ b)⊥ ) = 0
7	6	lan 77	. . . . . . 7 ((a ∪ b⊥ ) ∩ ((a ∪ b) ∩ (a ∪ b)⊥ )) = ((a ∪ b⊥ ) ∩ 0)
8		an0 108	. . . . . . 7 ((a ∪ b⊥ ) ∩ 0) = 0
9	7, 8	ax-r2 36	. . . . . 6 ((a ∪ b⊥ ) ∩ ((a ∪ b) ∩ (a ∪ b)⊥ )) = 0
10	4, 9	ax-r2 36	. . . . 5 (((a ∪ b⊥ ) ∩ (a ∪ b)) ∩ (a ∪ b)⊥ ) = 0
11	10	ran 78	. . . 4 ((((a ∪ b⊥ ) ∩ (a ∪ b)) ∩ (a ∪ b)⊥ ) ∩ (a⊥ ∪ (a ∩ b⊥ ))) = (0 ∩ (a⊥ ∪ (a ∩ b⊥ )))
12		an0r 109	. . . 4 (0 ∩ (a⊥ ∪ (a ∩ b⊥ ))) = 0
13	11, 12	ax-r2 36	. . 3 ((((a ∪ b⊥ ) ∩ (a ∪ b)) ∩ (a ∪ b)⊥ ) ∩ (a⊥ ∪ (a ∩ b⊥ ))) = 0
14	3, 13	ax-r2 36	. 2 ((((a ∪ b⊥ ) ∩ (a ∪ b)) ∩ (a⊥ ∪ (a ∩ b⊥ ))) ∩ (a ∪ b)⊥ ) = 0
15	2, 14	ax-r2 36	1 ((a →3 b)⊥ ∩ (a ∪ b)⊥ ) = 0

This proof depends on syntax axioms:   = wb 1  ^⊥ wn 4   ∪ wo 6   ∩ wa 7  0wf 9   →₃ wi3 14

This proof depends on axioms:  ax-a1 30  ax-a2 31  ax-a3 32  ax-a4 33  ax-r1 35  ax-r2 36  ax-r4 37  ax-r5 38

This proof depends on definitions:  df-a 40  df-t 41  df-f 42  df-i3 46

This theorem is used by:  ud3lem3  576