Theorem salem1 837

Description: Lemma for attempt at Sasaki algebra. (Contributed by NM, 4-Jan-1999.)

Assertion

Ref	Expression
salem1	(((a⊥ →1 b) ∪ b) →1 b) = (a →2 b)

Proof of Theorem salem1

Step	Hyp	Ref	Expression
1		u1lemob 630	. . . . . 6 ((a⊥ →1 b) ∪ b) = (a⊥ ⊥ ∪ b)
2	1	ax-r4 37	. . . . 5 ((a⊥ →1 b) ∪ b)⊥ = (a⊥ ⊥ ∪ b)⊥
3		anor1 88	. . . . . 6 (a⊥ ∩ b⊥ ) = (a⊥ ⊥ ∪ b)⊥
4	3	ax-r1 35	. . . . 5 (a⊥ ⊥ ∪ b)⊥ = (a⊥ ∩ b⊥ )
5	2, 4	ax-r2 36	. . . 4 ((a⊥ →1 b) ∪ b)⊥ = (a⊥ ∩ b⊥ )
6	1	ran 78	. . . . 5 (((a⊥ →1 b) ∪ b) ∩ b) = ((a⊥ ⊥ ∪ b) ∩ b)
7		ax-a2 31	. . . . . . 7 (a⊥ ⊥ ∪ b) = (b ∪ a⊥ ⊥ )
8	7	ran 78	. . . . . 6 ((a⊥ ⊥ ∪ b) ∩ b) = ((b ∪ a⊥ ⊥ ) ∩ b)
9		ancom 74	. . . . . 6 ((b ∪ a⊥ ⊥ ) ∩ b) = (b ∩ (b ∪ a⊥ ⊥ ))
10	8, 9	ax-r2 36	. . . . 5 ((a⊥ ⊥ ∪ b) ∩ b) = (b ∩ (b ∪ a⊥ ⊥ ))
11		anabs 121	. . . . 5 (b ∩ (b ∪ a⊥ ⊥ )) = b
12	6, 10, 11	3tr 65	. . . 4 (((a⊥ →1 b) ∪ b) ∩ b) = b
13	5, 12	2or 72	. . 3 (((a⊥ →1 b) ∪ b)⊥ ∪ (((a⊥ →1 b) ∪ b) ∩ b)) = ((a⊥ ∩ b⊥ ) ∪ b)
14		ax-a2 31	. . 3 ((a⊥ ∩ b⊥ ) ∪ b) = (b ∪ (a⊥ ∩ b⊥ ))
15	13, 14	ax-r2 36	. 2 (((a⊥ →1 b) ∪ b)⊥ ∪ (((a⊥ →1 b) ∪ b) ∩ b)) = (b ∪ (a⊥ ∩ b⊥ ))
16		df-i1 44	. 2 (((a⊥ →1 b) ∪ b) →1 b) = (((a⊥ →1 b) ∪ b)⊥ ∪ (((a⊥ →1 b) ∪ b) ∩ b))
17		df-i2 45	. 2 (a →2 b) = (b ∪ (a⊥ ∩ b⊥ ))
18	15, 16, 17	3tr1 63	1 (((a⊥ →1 b) ∪ b) →1 b) = (a →2 b)

This proof depends on syntax axioms:   = wb 1  ^⊥ wn 4   ∪ wo 6   ∩ wa 7   →₁ wi1 12   →₂ wi2 13

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-i1 44  df-i2 45  df-le1 130  df-le2 131

This theorem is used by: (None)