Theorem oalem2 1006

Description: Lemma. (Contributed by NM, 16-Oct-1998.)

Assertion

Ref	Expression
oalem2	((a →2 b) ∪ ((a →2 c) ∩ ((b ∪ c)⊥ ∪ ((a →2 b) ∩ (a →2 c))))) = (a →2 b)

Proof of Theorem oalem2

Step	Hyp	Ref	Expression
1		ax-a2 31	. . . . . . 7 (b ∪ c) = (c ∪ b)
2	1	ax-r4 37	. . . . . 6 (b ∪ c)⊥ = (c ∪ b)⊥
3		ancom 74	. . . . . 6 ((a →2 b) ∩ (a →2 c)) = ((a →2 c) ∩ (a →2 b))
4	2, 3	2or 72	. . . . 5 ((b ∪ c)⊥ ∪ ((a →2 b) ∩ (a →2 c))) = ((c ∪ b)⊥ ∪ ((a →2 c) ∩ (a →2 b)))
5	4	lan 77	. . . 4 ((a →2 c) ∩ ((b ∪ c)⊥ ∪ ((a →2 b) ∩ (a →2 c)))) = ((a →2 c) ∩ ((c ∪ b)⊥ ∪ ((a →2 c) ∩ (a →2 b))))
6		oath1 1004	. . . 4 ((a →2 c) ∩ ((c ∪ b)⊥ ∪ ((a →2 c) ∩ (a →2 b)))) = ((a →2 c) ∩ (a →2 b))
7	5, 6	ax-r2 36	. . 3 ((a →2 c) ∩ ((b ∪ c)⊥ ∪ ((a →2 b) ∩ (a →2 c)))) = ((a →2 c) ∩ (a →2 b))
8	7	lor 70	. 2 ((a →2 b) ∪ ((a →2 c) ∩ ((b ∪ c)⊥ ∪ ((a →2 b) ∩ (a →2 c))))) = ((a →2 b) ∪ ((a →2 c) ∩ (a →2 b)))
9		ancom 74	. . 3 ((a →2 c) ∩ (a →2 b)) = ((a →2 b) ∩ (a →2 c))
10	9	lor 70	. 2 ((a →2 b) ∪ ((a →2 c) ∩ (a →2 b))) = ((a →2 b) ∪ ((a →2 b) ∩ (a →2 c)))
11		orabs 120	. 2 ((a →2 b) ∪ ((a →2 b) ∩ (a →2 c))) = (a →2 b)
12	8, 10, 11	3tr 65	1 ((a →2 b) ∪ ((a →2 c) ∩ ((b ∪ c)⊥ ∪ ((a →2 b) ∩ (a →2 c))))) = (a →2 b)

Syntax hints:   = wb 1  ^⊥ wn 4   ∪ wo 6   ∩ wa 7   →₂ wi2 13

This theorem was proved from 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  ax-3oa 998

This theorem depends on definitions:  df-a 40  df-t 41  df-f 42  df-i1 44  df-i2 45  df-le1 130  df-le2 131

This theorem is referenced by: (None)