Theorem oa4to6lem2 961

Description: Lemma for orthoarguesian law (4-variable to 6-variable proof). (Contributed by NM, 18-Dec-1998.)

Hypotheses

Ref	Expression
oa4to6lem.1	a⊥ ≤ b
oa4to6lem.2	c⊥ ≤ d
oa4to6lem.3	e⊥ ≤ f
oa4to6lem.4	g = (((a ∩ b) ∪ (c ∩ d)) ∪ (e ∩ f))

Assertion

Ref	Expression
oa4to6lem2	d ≤ (c →1 g)

Proof of Theorem oa4to6lem2

Step	Hyp	Ref	Expression
1		leor 159	. . . 4 d ≤ (c⊥ ∪ d)
2		comid 187	. . . . . . . . 9 c C c
3	2	comcom3 454	. . . . . . . 8 c⊥ C c
4		oa4to6lem.2	. . . . . . . . 9 c⊥ ≤ d
5	4	lecom 180	. . . . . . . 8 c⊥ C d
6	3, 5	fh3 471	. . . . . . 7 (c⊥ ∪ (c ∩ d)) = ((c⊥ ∪ c) ∩ (c⊥ ∪ d))
7		ancom 74	. . . . . . . 8 (1 ∩ (c⊥ ∪ d)) = ((c⊥ ∪ d) ∩ 1)
8		df-t 41	. . . . . . . . . 10 1 = (c ∪ c⊥ )
9		ax-a2 31	. . . . . . . . . 10 (c ∪ c⊥ ) = (c⊥ ∪ c)
10	8, 9	ax-r2 36	. . . . . . . . 9 1 = (c⊥ ∪ c)
11	10	ran 78	. . . . . . . 8 (1 ∩ (c⊥ ∪ d)) = ((c⊥ ∪ c) ∩ (c⊥ ∪ d))
12		an1 106	. . . . . . . 8 ((c⊥ ∪ d) ∩ 1) = (c⊥ ∪ d)
13	7, 11, 12	3tr2 64	. . . . . . 7 ((c⊥ ∪ c) ∩ (c⊥ ∪ d)) = (c⊥ ∪ d)
14	6, 13	ax-r2 36	. . . . . 6 (c⊥ ∪ (c ∩ d)) = (c⊥ ∪ d)
15	14	ax-r1 35	. . . . 5 (c⊥ ∪ d) = (c⊥ ∪ (c ∩ d))
16		anidm 111	. . . . . . . . 9 (c ∩ c) = c
17	16	ran 78	. . . . . . . 8 ((c ∩ c) ∩ d) = (c ∩ d)
18	17	ax-r1 35	. . . . . . 7 (c ∩ d) = ((c ∩ c) ∩ d)
19		anass 76	. . . . . . 7 ((c ∩ c) ∩ d) = (c ∩ (c ∩ d))
20	18, 19	ax-r2 36	. . . . . 6 (c ∩ d) = (c ∩ (c ∩ d))
21	20	lor 70	. . . . 5 (c⊥ ∪ (c ∩ d)) = (c⊥ ∪ (c ∩ (c ∩ d)))
22	15, 21	ax-r2 36	. . . 4 (c⊥ ∪ d) = (c⊥ ∪ (c ∩ (c ∩ d)))
23	1, 22	lbtr 139	. . 3 d ≤ (c⊥ ∪ (c ∩ (c ∩ d)))
24		leor 159	. . . . . 6 (c ∩ d) ≤ (((a ∩ b) ∪ (e ∩ f)) ∪ (c ∩ d))
25		or32 82	. . . . . 6 (((a ∩ b) ∪ (e ∩ f)) ∪ (c ∩ d)) = (((a ∩ b) ∪ (c ∩ d)) ∪ (e ∩ f))
26	24, 25	lbtr 139	. . . . 5 (c ∩ d) ≤ (((a ∩ b) ∪ (c ∩ d)) ∪ (e ∩ f))
27	26	lelan 167	. . . 4 (c ∩ (c ∩ d)) ≤ (c ∩ (((a ∩ b) ∪ (c ∩ d)) ∪ (e ∩ f)))
28	27	lelor 166	. . 3 (c⊥ ∪ (c ∩ (c ∩ d))) ≤ (c⊥ ∪ (c ∩ (((a ∩ b) ∪ (c ∩ d)) ∪ (e ∩ f))))
29	23, 28	letr 137	. 2 d ≤ (c⊥ ∪ (c ∩ (((a ∩ b) ∪ (c ∩ d)) ∪ (e ∩ f))))
30		oa4to6lem.4	. . . . 5 g = (((a ∩ b) ∪ (c ∩ d)) ∪ (e ∩ f))
31	30	ud1lem0a 255	. . . 4 (c →1 g) = (c →1 (((a ∩ b) ∪ (c ∩ d)) ∪ (e ∩ f)))
32		df-i1 44	. . . 4 (c →1 (((a ∩ b) ∪ (c ∩ d)) ∪ (e ∩ f))) = (c⊥ ∪ (c ∩ (((a ∩ b) ∪ (c ∩ d)) ∪ (e ∩ f))))
33	31, 32	ax-r2 36	. . 3 (c →1 g) = (c⊥ ∪ (c ∩ (((a ∩ b) ∪ (c ∩ d)) ∪ (e ∩ f))))
34	33	ax-r1 35	. 2 (c⊥ ∪ (c ∩ (((a ∩ b) ∪ (c ∩ d)) ∪ (e ∩ f)))) = (c →1 g)
35	29, 34	lbtr 139	1 d ≤ (c →1 g)

This proof depends on syntax axioms:   = wb 1   ≤ wle 2  ^⊥ 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-a4 33  ax-a5 34  ax-r1 35  ax-r2 36  ax-r4 37  ax-r5 38  ax-r3 439

This proof depends on definitions:  df-b 39  df-a 40  df-t 41  df-f 42  df-i1 44  df-le1 130  df-le2 131  df-c1 132  df-c2 133

This theorem is used by:  oa4to6lem4  963