Theorem salem1 819

Description: Lemma for attempt at Sasaki algebra.

Assertion

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

Proof of Theorem salem1

Step	Hyp	Ref	Expression
1		u1lemob 612	. . . . . 6 ((a⊥ →1 b) ∪ b) = (a⊥ ⊥ ∪ b)
2	1	ax-r4 36	. . . . 5 ((a⊥ →1 b) ∪ b)⊥ = (a⊥ ⊥ ∪ b)⊥
3		anor1 80	. . . . . 6 (a⊥ ∩ b⊥ ) = (a⊥ ⊥ ∪ b)⊥
4	3	ax-r1 34	. . . . 5 (a⊥ ⊥ ∪ b)⊥ = (a⊥ ∩ b⊥ )
5	2, 4	ax-r2 35	. . . 4 ((a⊥ →1 b) ∪ b)⊥ = (a⊥ ∩ b⊥ )
6	1	ran 71	. . . . 5 (((a⊥ →1 b) ∪ b) ∩ b) = ((a⊥ ⊥ ∪ b) ∩ b)
7		ax-a2 30	. . . . . . 7 (a⊥ ⊥ ∪ b) = (b ∪ a⊥ ⊥ )
8	7	ran 71	. . . . . 6 ((a⊥ ⊥ ∪ b) ∩ b) = ((b ∪ a⊥ ⊥ ) ∩ b)
9		ancom 68	. . . . . 6 ((b ∪ a⊥ ⊥ ) ∩ b) = (b ∩ (b ∪ a⊥ ⊥ ))
10	8, 9	ax-r2 35	. . . . 5 ((a⊥ ⊥ ∪ b) ∩ b) = (b ∩ (b ∪ a⊥ ⊥ ))
11		a5c 113	. . . . 5 (b ∩ (b ∪ a⊥ ⊥ )) = b
12	6, 10, 11	3tr 62	. . . 4 (((a⊥ →1 b) ∪ b) ∩ b) = b
13	5, 12	2or 67	. . 3 (((a⊥ →1 b) ∪ b)⊥ ∪ (((a⊥ →1 b) ∪ b) ∩ b)) = ((a⊥ ∩ b⊥ ) ∪ b)
14		ax-a2 30	. . 3 ((a⊥ ∩ b⊥ ) ∪ b) = (b ∪ (a⊥ ∩ b⊥ ))
15	13, 14	ax-r2 35	. 2 (((a⊥ →1 b) ∪ b)⊥ ∪ (((a⊥ →1 b) ∪ b) ∩ b)) = (b ∪ (a⊥ ∩ b⊥ ))
16		df-i1 43	. 2 (((a⊥ →1 b) ∪ b) →1 b) = (((a⊥ →1 b) ∪ b)⊥ ∪ (((a⊥ →1 b) ∪ b) ∩ b))
17		df-i2 44	. 2 (a →2 b) = (b ∪ (a⊥ ∩ b⊥ ))
18	15, 16, 17	3tr1 60	1 (((a⊥ →1 b) ∪ b) →1 b) = (a →2 b)

Syntax hints:   = wb 1  ^⊥ wn 4   ∪ wo 6   ∩ wa 7   →₁ wi1 13   →₂ wi2 14

This theorem was proved from axioms:  ax-a1 29  ax-a2 30  ax-a3 31  ax-a5 33  ax-r1 34  ax-r2 35  ax-r4 36  ax-r5 37

This theorem depends on definitions:  df-a 39  df-i1 43  df-i2 44  df-le1 122  df-le2 123

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