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Theorem oa3-1lem 962

Description: Lemma for 3-OA(1). Equivalence with substitution into 6-OA dual.

**Assertion**
Ref	Expression
oa3-1lem	(1 ∩ (0 ∪ (a ∩ (((0 ∩ a) ∪ (1 ∩ (a →₁c))) ∪ (((0 ∩ b) ∪ (1 ∩ (b →₁c))) ∩ ((a ∩ b) ∪ ((a →₁c) ∩ (b →₁c)))))))) = (a ∩ ((a →₁c) ∪ ((b →₁c) ∩ ((a ∩ b) ∪ ((a →₁c) ∩ (b →₁c))))))

**Proof of Theorem oa3-1lem**
Step	Hyp	Ref	Expression
1		ancom 68	. 2 (1 ∩ (0 ∪ (a ∩ (((0 ∩ a) ∪ (1 ∩ (a →₁c))) ∪ (((0 ∩ b) ∪ (1 ∩ (b →₁c))) ∩ ((a ∩ b) ∪ ((a →₁c) ∩ (b →₁c)))))))) = ((0 ∪ (a ∩ (((0 ∩ a) ∪ (1 ∩ (a →₁c))) ∪ (((0 ∩ b) ∪ (1 ∩ (b →₁c))) ∩ ((a ∩ b) ∪ ((a →₁c) ∩ (b →₁c))))))) ∩ 1)
2		an1 98	. 2 ((0 ∪ (a ∩ (((0 ∩ a) ∪ (1 ∩ (a →₁c))) ∪ (((0 ∩ b) ∪ (1 ∩ (b →₁c))) ∩ ((a ∩ b) ∪ ((a →₁c) ∩ (b →₁c))))))) ∩ 1) = (0 ∪ (a ∩ (((0 ∩ a) ∪ (1 ∩ (a →₁c))) ∪ (((0 ∩ b) ∪ (1 ∩ (b →₁c))) ∩ ((a ∩ b) ∪ ((a →₁c) ∩ (b →₁c)))))))
3		ax-a2 30	. . 3 (0 ∪ (a ∩ (((0 ∩ a) ∪ (1 ∩ (a →₁c))) ∪ (((0 ∩ b) ∪ (1 ∩ (b →₁c))) ∩ ((a ∩ b) ∪ ((a →₁c) ∩ (b →₁c))))))) = ((a ∩ (((0 ∩ a) ∪ (1 ∩ (a →₁c))) ∪ (((0 ∩ b) ∪ (1 ∩ (b →₁c))) ∩ ((a ∩ b) ∪ ((a →₁c) ∩ (b →₁c)))))) ∪ 0)
4		or0 94	. . 3 ((a ∩ (((0 ∩ a) ∪ (1 ∩ (a →₁c))) ∪ (((0 ∩ b) ∪ (1 ∩ (b →₁c))) ∩ ((a ∩ b) ∪ ((a →₁c) ∩ (b →₁c)))))) ∪ 0) = (a ∩ (((0 ∩ a) ∪ (1 ∩ (a →₁c))) ∪ (((0 ∩ b) ∪ (1 ∩ (b →₁c))) ∩ ((a ∩ b) ∪ ((a →₁c) ∩ (b →₁c))))))
5		ancom 68	. . . . . . . . 9 (0 ∩ a) = (a ∩ 0)
6		an0 100	. . . . . . . . 9 (a ∩ 0) = 0
7	5, 6	ax-r2 35	. . . . . . . 8 (0 ∩ a) = 0
8		ancom 68	. . . . . . . . 9 (1 ∩ (a →₁c)) = ((a →₁c) ∩ 1)
9		an1 98	. . . . . . . . 9 ((a →₁c) ∩ 1) = (a →₁c)
10	8, 9	ax-r2 35	. . . . . . . 8 (1 ∩ (a →₁c)) = (a →₁c)
11	7, 10	2or 67	. . . . . . 7 ((0 ∩ a) ∪ (1 ∩ (a →₁c))) = (0 ∪ (a →₁c))
12		ax-a2 30	. . . . . . 7 (0 ∪ (a →₁c)) = ((a →₁c) ∪ 0)
13		or0 94	. . . . . . 7 ((a →₁c) ∪ 0) = (a →₁c)
14	11, 12, 13	3tr 62	. . . . . 6 ((0 ∩ a) ∪ (1 ∩ (a →₁c))) = (a →₁c)
15	14	ax-r5 37	. . . . 5 (((0 ∩ a) ∪ (1 ∩ (a →₁c))) ∪ (((0 ∩ b) ∪ (1 ∩ (b →₁c))) ∩ ((a ∩ b) ∪ ((a →₁c) ∩ (b →₁c))))) = ((a →₁c) ∪ (((0 ∩ b) ∪ (1 ∩ (b →₁c))) ∩ ((a ∩ b) ∪ ((a →₁c) ∩ (b →₁c)))))
16		ax-a2 30	. . . . . . . 8 ((0 ∩ b) ∪ (1 ∩ (b →₁c))) = ((1 ∩ (b →₁c)) ∪ (0 ∩ b))
17		ancom 68	. . . . . . . . . 10 (1 ∩ (b →₁c)) = ((b →₁c) ∩ 1)
18		an1 98	. . . . . . . . . 10 ((b →₁c) ∩ 1) = (b →₁c)
19	17, 18	ax-r2 35	. . . . . . . . 9 (1 ∩ (b →₁c)) = (b →₁c)
20		ancom 68	. . . . . . . . . 10 (0 ∩ b) = (b ∩ 0)
21		an0 100	. . . . . . . . . 10 (b ∩ 0) = 0
22	20, 21	ax-r2 35	. . . . . . . . 9 (0 ∩ b) = 0
23	19, 22	2or 67	. . . . . . . 8 ((1 ∩ (b →₁c)) ∪ (0 ∩ b)) = ((b →₁c) ∪ 0)
24		or0 94	. . . . . . . 8 ((b →₁c) ∪ 0) = (b →₁c)
25	16, 23, 24	3tr 62	. . . . . . 7 ((0 ∩ b) ∪ (1 ∩ (b →₁c))) = (b →₁c)
26	25	ran 71	. . . . . 6 (((0 ∩ b) ∪ (1 ∩ (b →₁c))) ∩ ((a ∩ b) ∪ ((a →₁c) ∩ (b →₁c)))) = ((b →₁c) ∩ ((a ∩ b) ∪ ((a →₁c) ∩ (b →₁c))))
27	26	lor 66	. . . . 5 ((a →₁c) ∪ (((0 ∩ b) ∪ (1 ∩ (b →₁c))) ∩ ((a ∩ b) ∪ ((a →₁c) ∩ (b →₁c))))) = ((a →₁c) ∪ ((b →₁c) ∩ ((a ∩ b) ∪ ((a →₁c) ∩ (b →₁c)))))
28	15, 27	ax-r2 35	. . . 4 (((0 ∩ a) ∪ (1 ∩ (a →₁c))) ∪ (((0 ∩ b) ∪ (1 ∩ (b →₁c))) ∩ ((a ∩ b) ∪ ((a →₁c) ∩ (b →₁c))))) = ((a →₁c) ∪ ((b →₁c) ∩ ((a ∩ b) ∪ ((a →₁c) ∩ (b →₁c)))))
29	28	lan 70	. . 3 (a ∩ (((0 ∩ a) ∪ (1 ∩ (a →₁c))) ∪ (((0 ∩ b) ∪ (1 ∩ (b →₁c))) ∩ ((a ∩ b) ∪ ((a →₁c) ∩ (b →₁c)))))) = (a ∩ ((a →₁c) ∪ ((b →₁c) ∩ ((a ∩ b) ∪ ((a →₁c) ∩ (b →₁c))))))
30	3, 4, 29	3tr 62	. 2 (0 ∪ (a ∩ (((0 ∩ a) ∪ (1 ∩ (a →₁c))) ∪ (((0 ∩ b) ∪ (1 ∩ (b →₁c))) ∩ ((a ∩ b) ∪ ((a →₁c) ∩ (b →₁c))))))) = (a ∩ ((a →₁c) ∪ ((b →₁c) ∩ ((a ∩ b) ∪ ((a →₁c) ∩ (b →₁c))))))
31	1, 2, 30	3tr 62	1 (1 ∩ (0 ∪ (a ∩ (((0 ∩ a) ∪ (1 ∩ (a →₁c))) ∪ (((0 ∩ b) ∪ (1 ∩ (b →₁c))) ∩ ((a ∩ b) ∪ ((a →₁c) ∩ (b →₁c)))))))) = (a ∩ ((a →₁c) ∪ ((b →₁c) ∩ ((a ∩ b) ∪ ((a →₁c) ∩ (b →₁c))))))

Colors of variables: term

Syntax hints: = wb 1 ∪ wo 6 ∩ wa 7 1wt 9 0wf 10 →₁wi1 13

This theorem was proved from axioms: ax-a1 29 ax-a2 30 ax-a4 32 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

metamath.org