Theorem 4oagen1 1042

Description: "Generalized" 4-OA. (Contributed by NM, 29-Dec-1998.)

Hypotheses

Ref	Expression
4oa.1	e = (((a ∩ c) ∪ ((a →1 d) ∩ (c →1 d))) ∩ ((b ∩ c) ∪ ((b →1 d) ∩ (c →1 d))))
4oa.2	f = (((a ∩ b) ∪ ((a →1 d) ∩ (b →1 d))) ∪ e)
4oagen1.1	g ≤ f

Assertion

Ref	Expression
4oagen1	((a →1 d) ∩ (g ∪ ((a →1 d) ∩ (b →1 d)))) = ((a →1 d) ∩ (b →1 d))

Proof of Theorem 4oagen1

Step	Hyp	Ref	Expression
1		4oagen1.1	. . . . . . 7 g ≤ f
2		4oa.2	. . . . . . . 8 f = (((a ∩ b) ∪ ((a →1 d) ∩ (b →1 d))) ∪ e)
3		or32 82	. . . . . . . 8 (((a ∩ b) ∪ ((a →1 d) ∩ (b →1 d))) ∪ e) = (((a ∩ b) ∪ e) ∪ ((a →1 d) ∩ (b →1 d)))
4	2, 3	ax-r2 36	. . . . . . 7 f = (((a ∩ b) ∪ e) ∪ ((a →1 d) ∩ (b →1 d)))
5	1, 4	lbtr 139	. . . . . 6 g ≤ (((a ∩ b) ∪ e) ∪ ((a →1 d) ∩ (b →1 d)))
6	5	leror 152	. . . . 5 (g ∪ ((a →1 d) ∩ (b →1 d))) ≤ ((((a ∩ b) ∪ e) ∪ ((a →1 d) ∩ (b →1 d))) ∪ ((a →1 d) ∩ (b →1 d)))
7		ax-a3 32	. . . . . 6 ((((a ∩ b) ∪ e) ∪ ((a →1 d) ∩ (b →1 d))) ∪ ((a →1 d) ∩ (b →1 d))) = (((a ∩ b) ∪ e) ∪ (((a →1 d) ∩ (b →1 d)) ∪ ((a →1 d) ∩ (b →1 d))))
8		oridm 110	. . . . . . . 8 (((a →1 d) ∩ (b →1 d)) ∪ ((a →1 d) ∩ (b →1 d))) = ((a →1 d) ∩ (b →1 d))
9	8	lor 70	. . . . . . 7 (((a ∩ b) ∪ e) ∪ (((a →1 d) ∩ (b →1 d)) ∪ ((a →1 d) ∩ (b →1 d)))) = (((a ∩ b) ∪ e) ∪ ((a →1 d) ∩ (b →1 d)))
10	4	ax-r1 35	. . . . . . 7 (((a ∩ b) ∪ e) ∪ ((a →1 d) ∩ (b →1 d))) = f
11	9, 10	ax-r2 36	. . . . . 6 (((a ∩ b) ∪ e) ∪ (((a →1 d) ∩ (b →1 d)) ∪ ((a →1 d) ∩ (b →1 d)))) = f
12	7, 11	ax-r2 36	. . . . 5 ((((a ∩ b) ∪ e) ∪ ((a →1 d) ∩ (b →1 d))) ∪ ((a →1 d) ∩ (b →1 d))) = f
13	6, 12	lbtr 139	. . . 4 (g ∪ ((a →1 d) ∩ (b →1 d))) ≤ f
14	13	lelan 167	. . 3 ((a →1 d) ∩ (g ∪ ((a →1 d) ∩ (b →1 d)))) ≤ ((a →1 d) ∩ f)
15		4oa.1	. . . 4 e = (((a ∩ c) ∪ ((a →1 d) ∩ (c →1 d))) ∩ ((b ∩ c) ∪ ((b →1 d) ∩ (c →1 d))))
16	15, 2	4oath1 1041	. . 3 ((a →1 d) ∩ f) = ((a →1 d) ∩ (b →1 d))
17	14, 16	lbtr 139	. 2 ((a →1 d) ∩ (g ∪ ((a →1 d) ∩ (b →1 d)))) ≤ ((a →1 d) ∩ (b →1 d))
18		lea 160	. . 3 ((a →1 d) ∩ (b →1 d)) ≤ (a →1 d)
19		leor 159	. . 3 ((a →1 d) ∩ (b →1 d)) ≤ (g ∪ ((a →1 d) ∩ (b →1 d)))
20	18, 19	ler2an 173	. 2 ((a →1 d) ∩ (b →1 d)) ≤ ((a →1 d) ∩ (g ∪ ((a →1 d) ∩ (b →1 d))))
21	17, 20	lebi 145	1 ((a →1 d) ∩ (g ∪ ((a →1 d) ∩ (b →1 d)))) = ((a →1 d) ∩ (b →1 d))

This proof depends on syntax axioms:   = wb 1   ≤ wle 2   ∪ wo 6   ∩ wa 7   →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  ax-4oa 1033

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:  4oagen1b  1043