Theorem u1lemaa 582

Description: Lemma for Sasaki implication study.

Assertion

Ref	Expression
u1lemaa	((a →1 b) ∩ a) = (a ∩ b)

Proof of Theorem u1lemaa

Step	Hyp	Ref	Expression
1		df-i1 43	. . 3 (a →1 b) = (a⊥ ∪ (a ∩ b))
2	1	ran 71	. 2 ((a →1 b) ∩ a) = ((a⊥ ∪ (a ∩ b)) ∩ a)
3		comid 179	. . . . 5 a C a
4	3	comcom2 175	. . . 4 a C a⊥
5		comanr1 446	. . . 4 a C (a ∩ b)
6	4, 5	fh1r 455	. . 3 ((a⊥ ∪ (a ∩ b)) ∩ a) = ((a⊥ ∩ a) ∪ ((a ∩ b) ∩ a))
7		ax-a2 30	. . . . 5 ((a⊥ ∩ a) ∪ ((a ∩ b) ∩ a)) = (((a ∩ b) ∩ a) ∪ (a⊥ ∩ a))
8		an32 76	. . . . . . 7 ((a ∩ b) ∩ a) = ((a ∩ a) ∩ b)
9		anidm 103	. . . . . . . 8 (a ∩ a) = a
10	9	ran 71	. . . . . . 7 ((a ∩ a) ∩ b) = (a ∩ b)
11	8, 10	ax-r2 35	. . . . . 6 ((a ∩ b) ∩ a) = (a ∩ b)
12		ancom 68	. . . . . . 7 (a⊥ ∩ a) = (a ∩ a⊥ )
13		dff 93	. . . . . . . 8 0 = (a ∩ a⊥ )
14	13	ax-r1 34	. . . . . . 7 (a ∩ a⊥ ) = 0
15	12, 14	ax-r2 35	. . . . . 6 (a⊥ ∩ a) = 0
16	11, 15	2or 67	. . . . 5 (((a ∩ b) ∩ a) ∪ (a⊥ ∩ a)) = ((a ∩ b) ∪ 0)
17	7, 16	ax-r2 35	. . . 4 ((a⊥ ∩ a) ∪ ((a ∩ b) ∩ a)) = ((a ∩ b) ∪ 0)
18		or0 94	. . . 4 ((a ∩ b) ∪ 0) = (a ∩ b)
19	17, 18	ax-r2 35	. . 3 ((a⊥ ∩ a) ∪ ((a ∩ b) ∩ a)) = (a ∩ b)
20	6, 19	ax-r2 35	. 2 ((a⊥ ∪ (a ∩ b)) ∩ a) = (a ∩ b)
21	2, 20	ax-r2 35	1 ((a →1 b) ∩ a) = (a ∩ b)

Syntax hints:   = wb 1  ^⊥ wn 4   ∪ wo 6   ∩ wa 7  0wf 10   →₁ wi1 13

This theorem is referenced by:  u1lemnona 647  u12lembi 708  u1lem5 743  negantlem2 831  kb10iii 875  oas 905  oau 909  oaur 910  oa6to4 938  oa8to5 952

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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