Theorem oacom3 1013

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

Hypotheses

Ref	Expression
oacom3.1	(d ∩ (a →2 b)) C ((b ∪ c) →0 ((a →2 b) ∩ (a →2 c)))
oacom3.2	d C (a →2 b)

Assertion

Ref	Expression
oacom3	d C ((a →2 b) ∩ (a →2 c))

Proof of Theorem oacom3

Step	Hyp	Ref	Expression
1		oacom3.2	. . . . 5 d C (a →2 b)
2	1	comcom 453	. . . 4 (a →2 b) C d
3		ancom 74	. . . . . 6 ((a →2 b) ∩ d) = (d ∩ (a →2 b))
4		oacom3.1	. . . . . 6 (d ∩ (a →2 b)) C ((b ∪ c) →0 ((a →2 b) ∩ (a →2 c)))
5	3, 4	bctr 181	. . . . 5 ((a →2 b) ∩ d) C ((b ∪ c) →0 ((a →2 b) ∩ (a →2 c)))
6	5	comcom 453	. . . 4 ((b ∪ c) →0 ((a →2 b) ∩ (a →2 c))) C ((a →2 b) ∩ d)
7	2, 6	gsth2 490	. . 3 (((b ∪ c) →0 ((a →2 b) ∩ (a →2 c))) ∩ (a →2 b)) C d
8	7	comcom 453	. 2 d C (((b ∪ c) →0 ((a →2 b) ∩ (a →2 c))) ∩ (a →2 b))
9		df-i0 43	. . . 4 ((b ∪ c) →0 ((a →2 b) ∩ (a →2 c))) = ((b ∪ c)⊥ ∪ ((a →2 b) ∩ (a →2 c)))
10	9	ran 78	. . 3 (((b ∪ c) →0 ((a →2 b) ∩ (a →2 c))) ∩ (a →2 b)) = (((b ∪ c)⊥ ∪ ((a →2 b) ∩ (a →2 c))) ∩ (a →2 b))
11		ancom 74	. . 3 (((b ∪ c)⊥ ∪ ((a →2 b) ∩ (a →2 c))) ∩ (a →2 b)) = ((a →2 b) ∩ ((b ∪ c)⊥ ∪ ((a →2 b) ∩ (a →2 c))))
12		oath1 1004	. . 3 ((a →2 b) ∩ ((b ∪ c)⊥ ∪ ((a →2 b) ∩ (a →2 c)))) = ((a →2 b) ∩ (a →2 c))
13	10, 11, 12	3tr 65	. 2 (((b ∪ c) →0 ((a →2 b) ∩ (a →2 c))) ∩ (a →2 b)) = ((a →2 b) ∩ (a →2 c))
14	8, 13	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: (None)