Theorem gomaex3lem8 921

Description: Lemma for Godowski 6-var -> Mayet Example 3. (Contributed by NM, 29-Nov-1999.)

Hypotheses

Ref	Expression
gomaex3lem5.1	a ≤ b⊥
gomaex3lem5.2	b ≤ c⊥
gomaex3lem5.3	c ≤ d⊥
gomaex3lem5.5	e ≤ f⊥
gomaex3lem5.6	f ≤ a⊥
gomaex3lem5.8	(((i →2 g) ∩ (g →2 y)) ∩ (((y →2 w) ∩ (w →2 n)) ∩ ((n →2 k) ∩ (k →2 i)))) ≤ (g →2 i)
gomaex3lem5.9	p = ((a ∪ b) →1 (d ∪ e)⊥ )⊥
gomaex3lem5.10	q = ((e ∪ f) →1 (b ∪ c)⊥ )⊥
gomaex3lem5.11	r = ((p⊥ →1 q)⊥ ∩ (c ∪ d))
gomaex3lem5.12	g = a
gomaex3lem5.13	h = b
gomaex3lem5.14	i = c
gomaex3lem5.15	j = (c ∪ d)⊥
gomaex3lem5.16	k = r
gomaex3lem5.17	m = (p⊥ →1 q)
gomaex3lem5.18	n = (p⊥ →1 q)⊥
gomaex3lem5.19	u = (p⊥ ∩ q)
gomaex3lem5.20	w = q⊥
gomaex3lem5.21	x = q
gomaex3lem5.22	y = (e ∪ f)⊥
gomaex3lem5.23	z = f

Assertion

Ref	Expression
gomaex3lem8	(((a ∪ b) ∩ (d ∪ e)⊥ ) ∩ ((r ∪ (p⊥ →1 q)) ∩ p⊥ )) ≤ (b ∪ c)

Proof of Theorem gomaex3lem8

Step	Hyp	Ref	Expression
1		an32 83	. . 3 (((a ∪ b) ∩ (d⊥ ∩ e⊥ )) ∩ ((r ∪ (p⊥ →1 q)) ∩ p⊥ )) = (((a ∪ b) ∩ ((r ∪ (p⊥ →1 q)) ∩ p⊥ )) ∩ (d⊥ ∩ e⊥ ))
2		anor3 90	. . . . 5 (d⊥ ∩ e⊥ ) = (d ∪ e)⊥
3	2	lan 77	. . . 4 ((a ∪ b) ∩ (d⊥ ∩ e⊥ )) = ((a ∪ b) ∩ (d ∪ e)⊥ )
4	3	ran 78	. . 3 (((a ∪ b) ∩ (d⊥ ∩ e⊥ )) ∩ ((r ∪ (p⊥ →1 q)) ∩ p⊥ )) = (((a ∪ b) ∩ (d ∪ e)⊥ ) ∩ ((r ∪ (p⊥ →1 q)) ∩ p⊥ ))
5		an4 86	. . 3 (((a ∪ b) ∩ ((r ∪ (p⊥ →1 q)) ∩ p⊥ )) ∩ (d⊥ ∩ e⊥ )) = (((a ∪ b) ∩ d⊥ ) ∩ (((r ∪ (p⊥ →1 q)) ∩ p⊥ ) ∩ e⊥ ))
6	1, 4, 5	3tr2 64	. 2 (((a ∪ b) ∩ (d ∪ e)⊥ ) ∩ ((r ∪ (p⊥ →1 q)) ∩ p⊥ )) = (((a ∪ b) ∩ d⊥ ) ∩ (((r ∪ (p⊥ →1 q)) ∩ p⊥ ) ∩ e⊥ ))
7		gomaex3lem5.1	. . 3 a ≤ b⊥
8		gomaex3lem5.2	. . 3 b ≤ c⊥
9		gomaex3lem5.3	. . 3 c ≤ d⊥
10		gomaex3lem5.5	. . 3 e ≤ f⊥
11		gomaex3lem5.6	. . 3 f ≤ a⊥
12		gomaex3lem5.8	. . 3 (((i →2 g) ∩ (g →2 y)) ∩ (((y →2 w) ∩ (w →2 n)) ∩ ((n →2 k) ∩ (k →2 i)))) ≤ (g →2 i)
13		gomaex3lem5.9	. . 3 p = ((a ∪ b) →1 (d ∪ e)⊥ )⊥
14		gomaex3lem5.10	. . 3 q = ((e ∪ f) →1 (b ∪ c)⊥ )⊥
15		gomaex3lem5.11	. . 3 r = ((p⊥ →1 q)⊥ ∩ (c ∪ d))
16		gomaex3lem5.12	. . 3 g = a
17		gomaex3lem5.13	. . 3 h = b
18		gomaex3lem5.14	. . 3 i = c
19		gomaex3lem5.15	. . 3 j = (c ∪ d)⊥
20		gomaex3lem5.16	. . 3 k = r
21		gomaex3lem5.17	. . 3 m = (p⊥ →1 q)
22		gomaex3lem5.18	. . 3 n = (p⊥ →1 q)⊥
23		gomaex3lem5.19	. . 3 u = (p⊥ ∩ q)
24		gomaex3lem5.20	. . 3 w = q⊥
25		gomaex3lem5.21	. . 3 x = q
26		gomaex3lem5.22	. . 3 y = (e ∪ f)⊥
27		gomaex3lem5.23	. . 3 z = f
28	7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27	gomaex3lem7 920	. 2 (((a ∪ b) ∩ d⊥ ) ∩ (((r ∪ (p⊥ →1 q)) ∩ p⊥ ) ∩ e⊥ )) ≤ (b ∪ c)
29	6, 28	bltr 138	1 (((a ∪ b) ∩ (d ∪ e)⊥ ) ∩ ((r ∪ (p⊥ →1 q)) ∩ p⊥ )) ≤ (b ∪ c)

This proof depends on syntax axioms:   = wb 1   ≤ wle 2  ^⊥ wn 4   ∪ wo 6   ∩ wa 7   →₁ wi1 12   →₂ wi2 13

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-i2 45  df-le1 130  df-le2 131  df-c1 132  df-c2 133

This theorem is used by:  gomaex3lem9  922