Theorem lem3.3.4 1053

Description: Equation 3.4 of [PavMeg1999] p. 9. (Contributed by Roy F. Longton, 28-Jun-2005.) (Revised by Roy F. Longton, 3-Jul-2005.)

Hypothesis

Ref	Expression
lem3.3.4.1	(b →2 a) = 1

Assertion

Ref	Expression
lem3.3.4	(a →2 (a ≡5 b)) = (a ≡5 b)

Proof of Theorem lem3.3.4

Step	Hyp	Ref	Expression
1		df-i2 45	. 2 (a →2 (a ≡5 b)) = ((a ≡5 b) ∪ (a⊥ ∩ (a ≡5 b)⊥ ))
2		df-id5 1047	. . . . . . 7 (a ≡5 b) = ((a ∩ b) ∪ (a⊥ ∩ b⊥ ))
3	2	ax-r4 37	. . . . . 6 (a ≡5 b)⊥ = ((a ∩ b) ∪ (a⊥ ∩ b⊥ ))⊥
4	3	lan 77	. . . . 5 (a⊥ ∩ (a ≡5 b)⊥ ) = (a⊥ ∩ ((a ∩ b) ∪ (a⊥ ∩ b⊥ ))⊥ )
5		anor3 90	. . . . . . . 8 ((a ∩ b)⊥ ∩ (a⊥ ∩ b⊥ )⊥ ) = ((a ∩ b) ∪ (a⊥ ∩ b⊥ ))⊥
6	5	ax-r1 35	. . . . . . 7 ((a ∩ b) ∪ (a⊥ ∩ b⊥ ))⊥ = ((a ∩ b)⊥ ∩ (a⊥ ∩ b⊥ )⊥ )
7		oran3 93	. . . . . . . . 9 (a⊥ ∪ b⊥ ) = (a ∩ b)⊥
8	7	ax-r1 35	. . . . . . . 8 (a ∩ b)⊥ = (a⊥ ∪ b⊥ )
9		oran 87	. . . . . . . . 9 (a ∪ b) = (a⊥ ∩ b⊥ )⊥
10	9	ax-r1 35	. . . . . . . 8 (a⊥ ∩ b⊥ )⊥ = (a ∪ b)
11	8, 10	2an 79	. . . . . . 7 ((a ∩ b)⊥ ∩ (a⊥ ∩ b⊥ )⊥ ) = ((a⊥ ∪ b⊥ ) ∩ (a ∪ b))
12	6, 11	ax-r2 36	. . . . . 6 ((a ∩ b) ∪ (a⊥ ∩ b⊥ ))⊥ = ((a⊥ ∪ b⊥ ) ∩ (a ∪ b))
13	12	lan 77	. . . . 5 (a⊥ ∩ ((a ∩ b) ∪ (a⊥ ∩ b⊥ ))⊥ ) = (a⊥ ∩ ((a⊥ ∪ b⊥ ) ∩ (a ∪ b)))
14		anabs 121	. . . . . . 7 (a⊥ ∩ (a⊥ ∪ b⊥ )) = a⊥
15	14	ran 78	. . . . . 6 ((a⊥ ∩ (a⊥ ∪ b⊥ )) ∩ (a ∪ b)) = (a⊥ ∩ (a ∪ b))
16		anass 76	. . . . . 6 ((a⊥ ∩ (a⊥ ∪ b⊥ )) ∩ (a ∪ b)) = (a⊥ ∩ ((a⊥ ∪ b⊥ ) ∩ (a ∪ b)))
17		lem3.3.4.1	. . . . . . . 8 (b →2 a) = 1
18	17	ax-r4 37	. . . . . . 7 (b →2 a)⊥ = 1⊥
19	9	con2 67	. . . . . . . . . . 11 (a ∪ b)⊥ = (a⊥ ∩ b⊥ )
20		ancom 74	. . . . . . . . . . 11 (a⊥ ∩ b⊥ ) = (b⊥ ∩ a⊥ )
21	19, 20	ax-r2 36	. . . . . . . . . 10 (a ∪ b)⊥ = (b⊥ ∩ a⊥ )
22	21	lor 70	. . . . . . . . 9 (a ∪ (a ∪ b)⊥ ) = (a ∪ (b⊥ ∩ a⊥ ))
23		oran1 91	. . . . . . . . . 10 (a ∪ (a ∪ b)⊥ ) = (a⊥ ∩ (a ∪ b))⊥
24	23	ax-r1 35	. . . . . . . . 9 (a⊥ ∩ (a ∪ b))⊥ = (a ∪ (a ∪ b)⊥ )
25		df-i2 45	. . . . . . . . 9 (b →2 a) = (a ∪ (b⊥ ∩ a⊥ ))
26	22, 24, 25	3tr1 63	. . . . . . . 8 (a⊥ ∩ (a ∪ b))⊥ = (b →2 a)
27	26	con3 68	. . . . . . 7 (a⊥ ∩ (a ∪ b)) = (b →2 a)⊥
28		df-f 42	. . . . . . 7 0 = 1⊥
29	18, 27, 28	3tr1 63	. . . . . 6 (a⊥ ∩ (a ∪ b)) = 0
30	15, 16, 29	3tr2 64	. . . . 5 (a⊥ ∩ ((a⊥ ∪ b⊥ ) ∩ (a ∪ b))) = 0
31	4, 13, 30	3tr 65	. . . 4 (a⊥ ∩ (a ≡5 b)⊥ ) = 0
32	31	lor 70	. . 3 ((a ≡5 b) ∪ (a⊥ ∩ (a ≡5 b)⊥ )) = ((a ≡5 b) ∪ 0)
33		ax-a2 31	. . 3 ((a ≡5 b) ∪ 0) = (0 ∪ (a ≡5 b))
34	32, 33	ax-r2 36	. 2 ((a ≡5 b) ∪ (a⊥ ∩ (a ≡5 b)⊥ )) = (0 ∪ (a ≡5 b))
35		or0r 103	. 2 (0 ∪ (a ≡5 b)) = (a ≡5 b)
36	1, 34, 35	3tr 65	1 (a →2 (a ≡5 b)) = (a ≡5 b)

Syntax hints:   = wb 1  ^⊥ wn 4   ∪ wo 6   ∩ wa 7  1wt 8  0wf 9   →₂ wi2 13   ≡₅ wid5 22

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

This theorem depends on definitions:  df-a 40  df-t 41  df-f 42  df-i2 45  df-id5 1047

This theorem is referenced by:  lem3.4.4  1077