Theorem u2lem1 717

Description: Lemma for unified implication study.

Assertion

Ref	Expression
u2lem1	((a →2 b) →2 a) = a

Proof of Theorem u2lem1

Step	Hyp	Ref	Expression
1		df-i2 44	. 2 ((a →2 b) →2 a) = (a ∪ ((a →2 b)⊥ ∩ a⊥ ))
2		ud2lem0c 270	. . . . . 6 (a →2 b)⊥ = (b⊥ ∩ (a ∪ b))
3	2	ran 71	. . . . 5 ((a →2 b)⊥ ∩ a⊥ ) = ((b⊥ ∩ (a ∪ b)) ∩ a⊥ )
4		an32 76	. . . . . 6 ((b⊥ ∩ (a ∪ b)) ∩ a⊥ ) = ((b⊥ ∩ a⊥ ) ∩ (a ∪ b))
5		ax-a2 30	. . . . . . . . 9 (a ∪ b) = (b ∪ a)
6		oran 79	. . . . . . . . 9 (b ∪ a) = (b⊥ ∩ a⊥ )⊥
7	5, 6	ax-r2 35	. . . . . . . 8 (a ∪ b) = (b⊥ ∩ a⊥ )⊥
8	7	lan 70	. . . . . . 7 ((b⊥ ∩ a⊥ ) ∩ (a ∪ b)) = ((b⊥ ∩ a⊥ ) ∩ (b⊥ ∩ a⊥ )⊥ )
9		dff 93	. . . . . . . 8 0 = ((b⊥ ∩ a⊥ ) ∩ (b⊥ ∩ a⊥ )⊥ )
10	9	ax-r1 34	. . . . . . 7 ((b⊥ ∩ a⊥ ) ∩ (b⊥ ∩ a⊥ )⊥ ) = 0
11	8, 10	ax-r2 35	. . . . . 6 ((b⊥ ∩ a⊥ ) ∩ (a ∪ b)) = 0
12	4, 11	ax-r2 35	. . . . 5 ((b⊥ ∩ (a ∪ b)) ∩ a⊥ ) = 0
13	3, 12	ax-r2 35	. . . 4 ((a →2 b)⊥ ∩ a⊥ ) = 0
14	13	lor 66	. . 3 (a ∪ ((a →2 b)⊥ ∩ a⊥ )) = (a ∪ 0)
15		or0 94	. . 3 (a ∪ 0) = a
16	14, 15	ax-r2 35	. 2 (a ∪ ((a →2 b)⊥ ∩ a⊥ )) = a
17	1, 16	ax-r2 35	1 ((a →2 b) →2 a) = a

Syntax hints:   = wb 1  ^⊥ wn 4   ∪ wo 6   ∩ wa 7  0wf 10   →₂ wi2 14

This theorem is referenced by:  u2lem1n 722

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-t 40  df-f 41  df-i2 44

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