Theorem wql2lem5 292

Description: Lemma for →₂ WQL axiom. (Contributed by NM, 6-Dec-1998.)

Hypothesis

Ref	Expression
wql2lem5.1	(a →2 b) = 1

Assertion

Ref	Expression
wql2lem5	((b⊥ ∩ (a ∪ b)) →2 a⊥ ) = 1

Proof of Theorem wql2lem5

Step	Hyp	Ref	Expression
1		anor3 90	. . . 4 ((b⊥ ∩ (a ∪ b))⊥ ∩ a⊥ ⊥ ) = ((b⊥ ∩ (a ∪ b)) ∪ a⊥ )⊥
2		oran3 93	. . . . . 6 ((a →2 b)⊥ ∪ a⊥ ) = ((a →2 b) ∩ a)⊥
3		ud2lem0c 278	. . . . . . 7 (a →2 b)⊥ = (b⊥ ∩ (a ∪ b))
4	3	ax-r5 38	. . . . . 6 ((a →2 b)⊥ ∪ a⊥ ) = ((b⊥ ∩ (a ∪ b)) ∪ a⊥ )
5		wql2lem5.1	. . . . . . . . 9 (a →2 b) = 1
6	5	ran 78	. . . . . . . 8 ((a →2 b) ∩ a) = (1 ∩ a)
7		ancom 74	. . . . . . . 8 (1 ∩ a) = (a ∩ 1)
8		an1 106	. . . . . . . 8 (a ∩ 1) = a
9	6, 7, 8	3tr 65	. . . . . . 7 ((a →2 b) ∩ a) = a
10	9	ax-r4 37	. . . . . 6 ((a →2 b) ∩ a)⊥ = a⊥
11	2, 4, 10	3tr2 64	. . . . 5 ((b⊥ ∩ (a ∪ b)) ∪ a⊥ ) = a⊥
12	11	ax-r4 37	. . . 4 ((b⊥ ∩ (a ∪ b)) ∪ a⊥ )⊥ = a⊥ ⊥
13	1, 12	ax-r2 36	. . 3 ((b⊥ ∩ (a ∪ b))⊥ ∩ a⊥ ⊥ ) = a⊥ ⊥
14	13	lor 70	. 2 (a⊥ ∪ ((b⊥ ∩ (a ∪ b))⊥ ∩ a⊥ ⊥ )) = (a⊥ ∪ a⊥ ⊥ )
15		df-i2 45	. 2 ((b⊥ ∩ (a ∪ b)) →2 a⊥ ) = (a⊥ ∪ ((b⊥ ∩ (a ∪ b))⊥ ∩ a⊥ ⊥ ))
16		df-t 41	. 2 1 = (a⊥ ∪ a⊥ ⊥ )
17	14, 15, 16	3tr1 63	1 ((b⊥ ∩ (a ∪ b)) →2 a⊥ ) = 1

This proof depends on syntax axioms:   = wb 1  ^⊥ wn 4   ∪ wo 6   ∩ wa 7  1wt 8   →₂ wi2 13

This proof depends on axioms:  ax-a1 30  ax-a2 31  ax-a5 34  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-i2 45

This theorem is used by: (None)