Theorem u3lem10 767

Description: Lemma for unified implication study.

Assertion

Ref	Expression
u3lem10	(a →3 (a⊥ ∩ (a ∪ b))) = a⊥

Proof of Theorem u3lem10

Step	Hyp	Ref	Expression
1		df-i3 45	. 2 (a →3 (a⊥ ∩ (a ∪ b))) = (((a⊥ ∩ (a⊥ ∩ (a ∪ b))) ∪ (a⊥ ∩ (a⊥ ∩ (a ∪ b))⊥ )) ∪ (a ∩ (a⊥ ∪ (a⊥ ∩ (a ∪ b)))))
2		anass 69	. . . . . . . 8 ((a⊥ ∩ a⊥ ) ∩ (a ∪ b)) = (a⊥ ∩ (a⊥ ∩ (a ∪ b)))
3	2	ax-r1 34	. . . . . . 7 (a⊥ ∩ (a⊥ ∩ (a ∪ b))) = ((a⊥ ∩ a⊥ ) ∩ (a ∪ b))
4		anidm 103	. . . . . . . 8 (a⊥ ∩ a⊥ ) = a⊥
5	4	ran 71	. . . . . . 7 ((a⊥ ∩ a⊥ ) ∩ (a ∪ b)) = (a⊥ ∩ (a ∪ b))
6	3, 5	ax-r2 35	. . . . . 6 (a⊥ ∩ (a⊥ ∩ (a ∪ b))) = (a⊥ ∩ (a ∪ b))
7		anor3 82	. . . . . . . . . . 11 (a⊥ ∩ b⊥ ) = (a ∪ b)⊥
8	7	lor 66	. . . . . . . . . 10 (a ∪ (a⊥ ∩ b⊥ )) = (a ∪ (a ∪ b)⊥ )
9		oran1 83	. . . . . . . . . 10 (a ∪ (a ∪ b)⊥ ) = (a⊥ ∩ (a ∪ b))⊥
10	8, 9	ax-r2 35	. . . . . . . . 9 (a ∪ (a⊥ ∩ b⊥ )) = (a⊥ ∩ (a ∪ b))⊥
11	10	ax-r1 34	. . . . . . . 8 (a⊥ ∩ (a ∪ b))⊥ = (a ∪ (a⊥ ∩ b⊥ ))
12	11	lan 70	. . . . . . 7 (a⊥ ∩ (a⊥ ∩ (a ∪ b))⊥ ) = (a⊥ ∩ (a ∪ (a⊥ ∩ b⊥ )))
13		omlan 430	. . . . . . 7 (a⊥ ∩ (a ∪ (a⊥ ∩ b⊥ ))) = (a⊥ ∩ b⊥ )
14	12, 13	ax-r2 35	. . . . . 6 (a⊥ ∩ (a⊥ ∩ (a ∪ b))⊥ ) = (a⊥ ∩ b⊥ )
15	6, 14	2or 67	. . . . 5 ((a⊥ ∩ (a⊥ ∩ (a ∪ b))) ∪ (a⊥ ∩ (a⊥ ∩ (a ∪ b))⊥ )) = ((a⊥ ∩ (a ∪ b)) ∪ (a⊥ ∩ b⊥ ))
16		comanr1 446	. . . . . . 7 a⊥ C (a⊥ ∩ b⊥ )
17		comorr 176	. . . . . . . 8 a C (a ∪ b)
18	17	comcom3 436	. . . . . . 7 a⊥ C (a ∪ b)
19	16, 18	fh4r 458	. . . . . 6 ((a⊥ ∩ (a ∪ b)) ∪ (a⊥ ∩ b⊥ )) = ((a⊥ ∪ (a⊥ ∩ b⊥ )) ∩ ((a ∪ b) ∪ (a⊥ ∩ b⊥ )))
20		a5b 112	. . . . . . . 8 (a⊥ ∪ (a⊥ ∩ b⊥ )) = a⊥
21	7	lor 66	. . . . . . . . 9 ((a ∪ b) ∪ (a⊥ ∩ b⊥ )) = ((a ∪ b) ∪ (a ∪ b)⊥ )
22		df-t 40	. . . . . . . . . 10 1 = ((a ∪ b) ∪ (a ∪ b)⊥ )
23	22	ax-r1 34	. . . . . . . . 9 ((a ∪ b) ∪ (a ∪ b)⊥ ) = 1
24	21, 23	ax-r2 35	. . . . . . . 8 ((a ∪ b) ∪ (a⊥ ∩ b⊥ )) = 1
25	20, 24	2an 72	. . . . . . 7 ((a⊥ ∪ (a⊥ ∩ b⊥ )) ∩ ((a ∪ b) ∪ (a⊥ ∩ b⊥ ))) = (a⊥ ∩ 1)
26		an1 98	. . . . . . 7 (a⊥ ∩ 1) = a⊥
27	25, 26	ax-r2 35	. . . . . 6 ((a⊥ ∪ (a⊥ ∩ b⊥ )) ∩ ((a ∪ b) ∪ (a⊥ ∩ b⊥ ))) = a⊥
28	19, 27	ax-r2 35	. . . . 5 ((a⊥ ∩ (a ∪ b)) ∪ (a⊥ ∩ b⊥ )) = a⊥
29	15, 28	ax-r2 35	. . . 4 ((a⊥ ∩ (a⊥ ∩ (a ∪ b))) ∪ (a⊥ ∩ (a⊥ ∩ (a ∪ b))⊥ )) = a⊥
30		a5b 112	. . . . . 6 (a⊥ ∪ (a⊥ ∩ (a ∪ b))) = a⊥
31	30	lan 70	. . . . 5 (a ∩ (a⊥ ∪ (a⊥ ∩ (a ∪ b)))) = (a ∩ a⊥ )
32		ancom 68	. . . . 5 (a ∩ a⊥ ) = (a⊥ ∩ a)
33	31, 32	ax-r2 35	. . . 4 (a ∩ (a⊥ ∪ (a⊥ ∩ (a ∪ b)))) = (a⊥ ∩ a)
34	29, 33	2or 67	. . 3 (((a⊥ ∩ (a⊥ ∩ (a ∪ b))) ∪ (a⊥ ∩ (a⊥ ∩ (a ∪ b))⊥ )) ∪ (a ∩ (a⊥ ∪ (a⊥ ∩ (a ∪ b))))) = (a⊥ ∪ (a⊥ ∩ a))
35		a5b 112	. . 3 (a⊥ ∪ (a⊥ ∩ a)) = a⊥
36	34, 35	ax-r2 35	. 2 (((a⊥ ∩ (a⊥ ∩ (a ∪ b))) ∪ (a⊥ ∩ (a⊥ ∩ (a ∪ b))⊥ )) ∪ (a ∩ (a⊥ ∪ (a⊥ ∩ (a ∪ b))))) = a⊥
37	1, 36	ax-r2 35	1 (a →3 (a⊥ ∩ (a ∪ b))) = a⊥

Syntax hints:   = wb 1  ^⊥ wn 4   ∪ wo 6   ∩ wa 7  1wt 9   →₃ wi3 15

This theorem was proved from axioms:  ax-a1 29  ax-a2 30  ax-a3 31  ax-a4 32  ax-a5 33  ax-r1 34  ax-r2 35  ax-r4 36  ax-r5 37  ax-r3 421

This theorem depends on definitions:  df-b 38  df-a 39  df-t 40  df-f 41  df-i3 45  df-le1 122  df-le2 123  df-c1 124  df-c2 125

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