Theorem oadistc 1002

Description: Distributive law.

Hypotheses

Ref	Expression
oadistc.1	d ≤ ((a →2 b) ∩ (a →2 c))
oadistc.2	((a →2 b) ∩ ((b ∪ c)⊥ ∪ d)) ≤ (((a →2 b) ∩ (b ∪ c)⊥ ) ∪ d)

Assertion

Ref	Expression
oadistc	((a →2 b) ∩ ((b ∪ c)⊥ ∪ d)) = (((a →2 b) ∩ (b ∪ c)⊥ ) ∪ ((a →2 b) ∩ d))

Proof of Theorem oadistc

Step	Hyp	Ref	Expression
1		oadistc.2	. . 3 ((a →2 b) ∩ ((b ∪ c)⊥ ∪ d)) ≤ (((a →2 b) ∩ (b ∪ c)⊥ ) ∪ d)
2		oadistc.1	. . . . . . . 8 d ≤ ((a →2 b) ∩ (a →2 c))
3		lea 152	. . . . . . . 8 ((a →2 b) ∩ (a →2 c)) ≤ (a →2 b)
4	2, 3	letr 129	. . . . . . 7 d ≤ (a →2 b)
5	4	df2le2 128	. . . . . 6 (d ∩ (a →2 b)) = d
6	5	ax-r1 34	. . . . 5 d = (d ∩ (a →2 b))
7		ancom 68	. . . . 5 (d ∩ (a →2 b)) = ((a →2 b) ∩ d)
8	6, 7	ax-r2 35	. . . 4 d = ((a →2 b) ∩ d)
9	8	lor 66	. . 3 (((a →2 b) ∩ (b ∪ c)⊥ ) ∪ d) = (((a →2 b) ∩ (b ∪ c)⊥ ) ∪ ((a →2 b) ∩ d))
10	1, 9	lbtr 131	. 2 ((a →2 b) ∩ ((b ∪ c)⊥ ∪ d)) ≤ (((a →2 b) ∩ (b ∪ c)⊥ ) ∪ ((a →2 b) ∩ d))
11		ledi 166	. 2 (((a →2 b) ∩ (b ∪ c)⊥ ) ∪ ((a →2 b) ∩ d)) ≤ ((a →2 b) ∩ ((b ∪ c)⊥ ∪ d))
12	10, 11	lebi 137	1 ((a →2 b) ∩ ((b ∪ c)⊥ ∪ d)) = (((a →2 b) ∩ (b ∪ c)⊥ ) ∪ ((a →2 b) ∩ d))

Syntax hints:   = wb 1   ≤ wle 2  ^⊥ wn 4   ∪ wo 6   ∩ wa 7   →₂ wi2 14

This theorem was proved from axioms:  ax-a1 29  ax-a2 30  ax-a3 31  ax-a5 33  ax-r1 34  ax-r2 35  ax-r4 36  ax-r5 37

This theorem depends on definitions:  df-a 39  df-t 40  df-f 41  df-le1 122  df-le2 123

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