Theorem oacom 1011

Description: Commutation law requiring OA. (Contributed by NM, 19-Nov-1998.)

Hypotheses

Ref	Expression
oacom.1	d C ((b ∪ c) →0 ((a →2 b) ∩ (a →2 c)))
oacom.2	(d ∩ ((b ∪ c) →0 ((a →2 b) ∩ (a →2 c)))) C (a →2 b)

Assertion

Ref	Expression
oacom	d C ((a →2 b) ∩ (a →2 c))

Proof of Theorem oacom

Step	Hyp	Ref	Expression
1		oacom.1	. . . . 5 d C ((b ∪ c) →0 ((a →2 b) ∩ (a →2 c)))
2	1	comcom 453	. . . 4 ((b ∪ c) →0 ((a →2 b) ∩ (a →2 c))) C d
3		ancom 74	. . . . . 6 (((b ∪ c) →0 ((a →2 b) ∩ (a →2 c))) ∩ d) = (d ∩ ((b ∪ c) →0 ((a →2 b) ∩ (a →2 c))))
4		oacom.2	. . . . . 6 (d ∩ ((b ∪ c) →0 ((a →2 b) ∩ (a →2 c)))) C (a →2 b)
5	3, 4	bctr 181	. . . . 5 (((b ∪ c) →0 ((a →2 b) ∩ (a →2 c))) ∩ d) C (a →2 b)
6	5	comcom 453	. . . 4 (a →2 b) C (((b ∪ c) →0 ((a →2 b) ∩ (a →2 c))) ∩ d)
7	2, 6	gsth2 490	. . 3 ((a →2 b) ∩ ((b ∪ c) →0 ((a →2 b) ∩ (a →2 c)))) C d
8	7	comcom 453	. 2 d C ((a →2 b) ∩ ((b ∪ c) →0 ((a →2 b) ∩ (a →2 c))))
9		df-i0 43	. . . 4 ((b ∪ c) →0 ((a →2 b) ∩ (a →2 c))) = ((b ∪ c)⊥ ∪ ((a →2 b) ∩ (a →2 c)))
10	9	lan 77	. . 3 ((a →2 b) ∩ ((b ∪ c) →0 ((a →2 b) ∩ (a →2 c)))) = ((a →2 b) ∩ ((b ∪ c)⊥ ∪ ((a →2 b) ∩ (a →2 c))))
11		oath1 1004	. . 3 ((a →2 b) ∩ ((b ∪ c)⊥ ∪ ((a →2 b) ∩ (a →2 c)))) = ((a →2 b) ∩ (a →2 c))
12	10, 11	ax-r2 36	. 2 ((a →2 b) ∩ ((b ∪ c) →0 ((a →2 b) ∩ (a →2 c)))) = ((a →2 b) ∩ (a →2 c))
13	8, 12	cbtr 182	1 d C ((a →2 b) ∩ (a →2 c))

Syntax hints:   C wc 3  ^⊥ wn 4   ∪ wo 6   ∩ wa 7   →₀ wi0 11   →₂ wi2 13

This theorem was proved from 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-3oa 998

This theorem depends on definitions:  df-b 39  df-a 40  df-t 41  df-f 42  df-i0 43  df-i1 44  df-i2 45  df-le1 130  df-le2 131  df-c1 132  df-c2 133

This theorem is referenced by:  oacom2  1012