oa3-5lem - Quantum Logic Explorer

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Theorem oa3-5lem 984

Description: Lemma for 3-OA(5). Equivalence with substitution into 6-OA dual.

**Assertion**
Ref	Expression
oa3-5lem	((a →₁c) ∩ (a ∪ (c ∩ (((a ∩ c) ∪ ((a →₁c) ∩ 1)) ∪ (((a ∩ b) ∪ ((a →₁c) ∩ (b →₁c))) ∩ ((c ∩ b) ∪ (1 ∩ (b →₁c)))))))) = ((a →₁c) ∩ (a ∪ (c ∩ ((a →₁c) ∪ ((b →₁c) ∩ ((a ∩ b) ∪ ((a →₁c) ∩ (b →₁c))))))))

**Proof of Theorem oa3-5lem**
Step	Hyp	Ref	Expression
1		or12 80	. . . . . . 7 ((a ∩ c) ∪ (a^⊥ ∪ (a ∩ c))) = (a^⊥ ∪ ((a ∩ c) ∪ (a ∩ c)))
2		oridm 110	. . . . . . . 8 ((a ∩ c) ∪ (a ∩ c)) = (a ∩ c)
3	2	lor 70	. . . . . . 7 (a^⊥ ∪ ((a ∩ c) ∪ (a ∩ c))) = (a^⊥ ∪ (a ∩ c))
4	1, 3	ax-r2 36	. . . . . 6 ((a ∩ c) ∪ (a^⊥ ∪ (a ∩ c))) = (a^⊥ ∪ (a ∩ c))
5		an1 106	. . . . . . . 8 ((a →₁c) ∩ 1) = (a →₁c)
6		df-i1 44	. . . . . . . 8 (a →₁c) = (a^⊥ ∪ (a ∩ c))
7	5, 6	ax-r2 36	. . . . . . 7 ((a →₁c) ∩ 1) = (a^⊥ ∪ (a ∩ c))
8	7	lor 70	. . . . . 6 ((a ∩ c) ∪ ((a →₁c) ∩ 1)) = ((a ∩ c) ∪ (a^⊥ ∪ (a ∩ c)))
9	4, 8, 6	3tr1 63	. . . . 5 ((a ∩ c) ∪ ((a →₁c) ∩ 1)) = (a →₁c)
10		or12 80	. . . . . . . . 9 ((c ∩ b) ∪ (b^⊥ ∪ (b ∩ c))) = (b^⊥ ∪ ((c ∩ b) ∪ (b ∩ c)))
11		ancom 74	. . . . . . . . . . . 12 (c ∩ b) = (b ∩ c)
12	11	ax-r5 38	. . . . . . . . . . 11 ((c ∩ b) ∪ (b ∩ c)) = ((b ∩ c) ∪ (b ∩ c))
13		oridm 110	. . . . . . . . . . 11 ((b ∩ c) ∪ (b ∩ c)) = (b ∩ c)
14	12, 13	ax-r2 36	. . . . . . . . . 10 ((c ∩ b) ∪ (b ∩ c)) = (b ∩ c)
15	14	lor 70	. . . . . . . . 9 (b^⊥ ∪ ((c ∩ b) ∪ (b ∩ c))) = (b^⊥ ∪ (b ∩ c))
16	10, 15	ax-r2 36	. . . . . . . 8 ((c ∩ b) ∪ (b^⊥ ∪ (b ∩ c))) = (b^⊥ ∪ (b ∩ c))
17		ancom 74	. . . . . . . . . 10 (1 ∩ (b →₁c)) = ((b →₁c) ∩ 1)
18		an1 106	. . . . . . . . . 10 ((b →₁c) ∩ 1) = (b →₁c)
19		df-i1 44	. . . . . . . . . 10 (b →₁c) = (b^⊥ ∪ (b ∩ c))
20	17, 18, 19	3tr 65	. . . . . . . . 9 (1 ∩ (b →₁c)) = (b^⊥ ∪ (b ∩ c))
21	20	lor 70	. . . . . . . 8 ((c ∩ b) ∪ (1 ∩ (b →₁c))) = ((c ∩ b) ∪ (b^⊥ ∪ (b ∩ c)))
22	16, 21, 19	3tr1 63	. . . . . . 7 ((c ∩ b) ∪ (1 ∩ (b →₁c))) = (b →₁c)
23	22	lan 77	. . . . . 6 (((a ∩ b) ∪ ((a →₁c) ∩ (b →₁c))) ∩ ((c ∩ b) ∪ (1 ∩ (b →₁c)))) = (((a ∩ b) ∪ ((a →₁c) ∩ (b →₁c))) ∩ (b →₁c))
24		ancom 74	. . . . . 6 (((a ∩ b) ∪ ((a →₁c) ∩ (b →₁c))) ∩ (b →₁c)) = ((b →₁c) ∩ ((a ∩ b) ∪ ((a →₁c) ∩ (b →₁c))))
25	23, 24	ax-r2 36	. . . . 5 (((a ∩ b) ∪ ((a →₁c) ∩ (b →₁c))) ∩ ((c ∩ b) ∪ (1 ∩ (b →₁c)))) = ((b →₁c) ∩ ((a ∩ b) ∪ ((a →₁c) ∩ (b →₁c))))
26	9, 25	2or 72	. . . 4 (((a ∩ c) ∪ ((a →₁c) ∩ 1)) ∪ (((a ∩ b) ∪ ((a →₁c) ∩ (b →₁c))) ∩ ((c ∩ b) ∪ (1 ∩ (b →₁c))))) = ((a →₁c) ∪ ((b →₁c) ∩ ((a ∩ b) ∪ ((a →₁c) ∩ (b →₁c)))))
27	26	lan 77	. . 3 (c ∩ (((a ∩ c) ∪ ((a →₁c) ∩ 1)) ∪ (((a ∩ b) ∪ ((a →₁c) ∩ (b →₁c))) ∩ ((c ∩ b) ∪ (1 ∩ (b →₁c)))))) = (c ∩ ((a →₁c) ∪ ((b →₁c) ∩ ((a ∩ b) ∪ ((a →₁c) ∩ (b →₁c))))))
28	27	lor 70	. 2 (a ∪ (c ∩ (((a ∩ c) ∪ ((a →₁c) ∩ 1)) ∪ (((a ∩ b) ∪ ((a →₁c) ∩ (b →₁c))) ∩ ((c ∩ b) ∪ (1 ∩ (b →₁c))))))) = (a ∪ (c ∩ ((a →₁c) ∪ ((b →₁c) ∩ ((a ∩ b) ∪ ((a →₁c) ∩ (b →₁c)))))))
29	28	lan 77	1 ((a →₁c) ∩ (a ∪ (c ∩ (((a ∩ c) ∪ ((a →₁c) ∩ 1)) ∪ (((a ∩ b) ∪ ((a →₁c) ∩ (b →₁c))) ∩ ((c ∩ b) ∪ (1 ∩ (b →₁c)))))))) = ((a →₁c) ∩ (a ∪ (c ∩ ((a →₁c) ∪ ((b →₁c) ∩ ((a ∩ b) ∪ ((a →₁c) ∩ (b →₁c))))))))

Colors of variables: term

Syntax hints: = wb 1 ^⊥ wn 4 ∪ wo 6 ∩ wa 7 1wt 8 →₁wi1 12

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-i1 44

This theorem is referenced by: (None)

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