Theorem oa3-5lem 984

Description: Lemma for 3-OA(5). Equivalence with substitution into 6-OA dual. (Contributed by NM, 25-Dec-1998.)

Assertion

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

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 →1 c) ∩ 1) = (a →1 c)
6		df-i1 44	. . . . . . . 8 (a →1 c) = (a⊥ ∪ (a ∩ c))
7	5, 6	ax-r2 36	. . . . . . 7 ((a →1 c) ∩ 1) = (a⊥ ∪ (a ∩ c))
8	7	lor 70	. . . . . 6 ((a ∩ c) ∪ ((a →1 c) ∩ 1)) = ((a ∩ c) ∪ (a⊥ ∪ (a ∩ c)))
9	4, 8, 6	3tr1 63	. . . . 5 ((a ∩ c) ∪ ((a →1 c) ∩ 1)) = (a →1 c)
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 →1 c)) = ((b →1 c) ∩ 1)
18		an1 106	. . . . . . . . . 10 ((b →1 c) ∩ 1) = (b →1 c)
19		df-i1 44	. . . . . . . . . 10 (b →1 c) = (b⊥ ∪ (b ∩ c))
20	17, 18, 19	3tr 65	. . . . . . . . 9 (1 ∩ (b →1 c)) = (b⊥ ∪ (b ∩ c))
21	20	lor 70	. . . . . . . 8 ((c ∩ b) ∪ (1 ∩ (b →1 c))) = ((c ∩ b) ∪ (b⊥ ∪ (b ∩ c)))
22	16, 21, 19	3tr1 63	. . . . . . 7 ((c ∩ b) ∪ (1 ∩ (b →1 c))) = (b →1 c)
23	22	lan 77	. . . . . 6 (((a ∩ b) ∪ ((a →1 c) ∩ (b →1 c))) ∩ ((c ∩ b) ∪ (1 ∩ (b →1 c)))) = (((a ∩ b) ∪ ((a →1 c) ∩ (b →1 c))) ∩ (b →1 c))
24		ancom 74	. . . . . 6 (((a ∩ b) ∪ ((a →1 c) ∩ (b →1 c))) ∩ (b →1 c)) = ((b →1 c) ∩ ((a ∩ b) ∪ ((a →1 c) ∩ (b →1 c))))
25	23, 24	ax-r2 36	. . . . 5 (((a ∩ b) ∪ ((a →1 c) ∩ (b →1 c))) ∩ ((c ∩ b) ∪ (1 ∩ (b →1 c)))) = ((b →1 c) ∩ ((a ∩ b) ∪ ((a →1 c) ∩ (b →1 c))))
26	9, 25	2or 72	. . . 4 (((a ∩ c) ∪ ((a →1 c) ∩ 1)) ∪ (((a ∩ b) ∪ ((a →1 c) ∩ (b →1 c))) ∩ ((c ∩ b) ∪ (1 ∩ (b →1 c))))) = ((a →1 c) ∪ ((b →1 c) ∩ ((a ∩ b) ∪ ((a →1 c) ∩ (b →1 c)))))
27	26	lan 77	. . 3 (c ∩ (((a ∩ c) ∪ ((a →1 c) ∩ 1)) ∪ (((a ∩ b) ∪ ((a →1 c) ∩ (b →1 c))) ∩ ((c ∩ b) ∪ (1 ∩ (b →1 c)))))) = (c ∩ ((a →1 c) ∪ ((b →1 c) ∩ ((a ∩ b) ∪ ((a →1 c) ∩ (b →1 c))))))
28	27	lor 70	. 2 (a ∪ (c ∩ (((a ∩ c) ∪ ((a →1 c) ∩ 1)) ∪ (((a ∩ b) ∪ ((a →1 c) ∩ (b →1 c))) ∩ ((c ∩ b) ∪ (1 ∩ (b →1 c))))))) = (a ∪ (c ∩ ((a →1 c) ∪ ((b →1 c) ∩ ((a ∩ b) ∪ ((a →1 c) ∩ (b →1 c)))))))
29	28	lan 77	1 ((a →1 c) ∩ (a ∪ (c ∩ (((a ∩ c) ∪ ((a →1 c) ∩ 1)) ∪ (((a ∩ b) ∪ ((a →1 c) ∩ (b →1 c))) ∩ ((c ∩ b) ∪ (1 ∩ (b →1 c)))))))) = ((a →1 c) ∩ (a ∪ (c ∩ ((a →1 c) ∪ ((b →1 c) ∩ ((a ∩ b) ∪ ((a →1 c) ∩ (b →1 c))))))))

This proof depends on syntax axioms:   = wb 1  ^⊥ wn 4   ∪ wo 6   ∩ wa 7  1wt 8   →₁ wi1 12

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-i1 44

This theorem is used by: (None)