Theorem u2lem1 735

Description: Lemma for unified implication study. (Contributed by NM, 14-Dec-1997.)

Assertion

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

Proof of Theorem u2lem1

Step	Hyp	Ref	Expression
1		df-i2 45	. 2 ((a →2 b) →2 a) = (a ∪ ((a →2 b)⊥ ∩ a⊥ ))
2		ud2lem0c 278	. . . . . 6 (a →2 b)⊥ = (b⊥ ∩ (a ∪ b))
3	2	ran 78	. . . . 5 ((a →2 b)⊥ ∩ a⊥ ) = ((b⊥ ∩ (a ∪ b)) ∩ a⊥ )
4		an32 83	. . . . . 6 ((b⊥ ∩ (a ∪ b)) ∩ a⊥ ) = ((b⊥ ∩ a⊥ ) ∩ (a ∪ b))
5		ax-a2 31	. . . . . . . . 9 (a ∪ b) = (b ∪ a)
6		oran 87	. . . . . . . . 9 (b ∪ a) = (b⊥ ∩ a⊥ )⊥
7	5, 6	ax-r2 36	. . . . . . . 8 (a ∪ b) = (b⊥ ∩ a⊥ )⊥
8	7	lan 77	. . . . . . 7 ((b⊥ ∩ a⊥ ) ∩ (a ∪ b)) = ((b⊥ ∩ a⊥ ) ∩ (b⊥ ∩ a⊥ )⊥ )
9		dff 101	. . . . . . . 8 0 = ((b⊥ ∩ a⊥ ) ∩ (b⊥ ∩ a⊥ )⊥ )
10	9	ax-r1 35	. . . . . . 7 ((b⊥ ∩ a⊥ ) ∩ (b⊥ ∩ a⊥ )⊥ ) = 0
11	8, 10	ax-r2 36	. . . . . 6 ((b⊥ ∩ a⊥ ) ∩ (a ∪ b)) = 0
12	4, 11	ax-r2 36	. . . . 5 ((b⊥ ∩ (a ∪ b)) ∩ a⊥ ) = 0
13	3, 12	ax-r2 36	. . . 4 ((a →2 b)⊥ ∩ a⊥ ) = 0
14	13	lor 70	. . 3 (a ∪ ((a →2 b)⊥ ∩ a⊥ )) = (a ∪ 0)
15		or0 102	. . 3 (a ∪ 0) = a
16	14, 15	ax-r2 36	. 2 (a ∪ ((a →2 b)⊥ ∩ a⊥ )) = a
17	1, 16	ax-r2 36	1 ((a →2 b) →2 a) = a

This proof depends on syntax axioms:   = wb 1  ^⊥ wn 4   ∪ wo 6   ∩ wa 7  0wf 9   →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-t 41  df-f 42  df-i2 45

This theorem is used by:  u2lem1n  740