Theorem u1lemab 592

Description: Lemma for Sasaki implication study.

Assertion

Ref	Expression
u1lemab	((a →1 b) ∩ b) = ((a ∩ b) ∪ (a⊥ ∩ b))

Proof of Theorem u1lemab

Step	Hyp	Ref	Expression
1		df-i1 43	. . 3 (a →1 b) = (a⊥ ∪ (a ∩ b))
2	1	ran 71	. 2 ((a →1 b) ∩ b) = ((a⊥ ∪ (a ∩ b)) ∩ b)
3		ax-a2 30	. . . . 5 (a⊥ ∪ (a ∩ b)) = ((a ∩ b) ∪ a⊥ )
4	3	ran 71	. . . 4 ((a⊥ ∪ (a ∩ b)) ∩ b) = (((a ∩ b) ∪ a⊥ ) ∩ b)
5		coman2 178	. . . . 5 (a ∩ b) C b
6		coman1 177	. . . . . 6 (a ∩ b) C a
7	6	comcom2 175	. . . . 5 (a ∩ b) C a⊥
8	5, 7	fh2r 456	. . . 4 (((a ∩ b) ∪ a⊥ ) ∩ b) = (((a ∩ b) ∩ b) ∪ (a⊥ ∩ b))
9	4, 8	ax-r2 35	. . 3 ((a⊥ ∪ (a ∩ b)) ∩ b) = (((a ∩ b) ∩ b) ∪ (a⊥ ∩ b))
10		anass 69	. . . . 5 ((a ∩ b) ∩ b) = (a ∩ (b ∩ b))
11		anidm 103	. . . . . 6 (b ∩ b) = b
12	11	lan 70	. . . . 5 (a ∩ (b ∩ b)) = (a ∩ b)
13	10, 12	ax-r2 35	. . . 4 ((a ∩ b) ∩ b) = (a ∩ b)
14	13	ax-r5 37	. . 3 (((a ∩ b) ∩ b) ∪ (a⊥ ∩ b)) = ((a ∩ b) ∪ (a⊥ ∩ b))
15	9, 14	ax-r2 35	. 2 ((a⊥ ∪ (a ∩ b)) ∩ b) = ((a ∩ b) ∪ (a⊥ ∩ b))
16	2, 15	ax-r2 35	1 ((a →1 b) ∩ b) = ((a ∩ b) ∪ (a⊥ ∩ b))

Syntax hints:   = wb 1  ^⊥ wn 4   ∪ wo 6   ∩ wa 7   →₁ wi1 13

This theorem is referenced by:  u1lemnonb 657  i1com 690  u1lem3 731  u1lem11 762  sadm3 820  negantlem2 831  negantlem3 832  negantlem10 843  neg3antlem1 846  oa4to4u 953  oa3-6lem 960  oa3-u1lem 965  oa3-u2lem 966  oa3-1to5 973

This theorem was proved from axioms:  ax-a1 29  ax-a2 30  ax-a3 31  ax-a4 32  ax-a5 33  ax-r1 34  ax-r2 35  ax-r4 36  ax-r5 37  ax-r3 421

This theorem depends on definitions:  df-b 38  df-a 39  df-t 40  df-f 41  df-i1 43  df-le1 122  df-le2 123  df-c1 124  df-c2 125

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