Theorem u1lemc4 683

Description: Lemma for Sasaki implication study.

Hypothesis

Ref	Expression
ulemc3.1	a C b

Assertion

Ref	Expression
u1lemc4	(a →1 b) = (a⊥ ∪ b)

Proof of Theorem u1lemc4

Step	Hyp	Ref	Expression
1		df-i1 43	. 2 (a →1 b) = (a⊥ ∪ (a ∩ b))
2		comid 179	. . . . 5 a C a
3	2	comcom2 175	. . . 4 a C a⊥
4		ulemc3.1	. . . 4 a C b
5	3, 4	fh4 454	. . 3 (a⊥ ∪ (a ∩ b)) = ((a⊥ ∪ a) ∩ (a⊥ ∪ b))
6		ancom 68	. . . 4 ((a⊥ ∪ a) ∩ (a⊥ ∪ b)) = ((a⊥ ∪ b) ∩ (a⊥ ∪ a))
7		ax-a2 30	. . . . . . 7 (a⊥ ∪ a) = (a ∪ a⊥ )
8		df-t 40	. . . . . . . 8 1 = (a ∪ a⊥ )
9	8	ax-r1 34	. . . . . . 7 (a ∪ a⊥ ) = 1
10	7, 9	ax-r2 35	. . . . . 6 (a⊥ ∪ a) = 1
11	10	lan 70	. . . . 5 ((a⊥ ∪ b) ∩ (a⊥ ∪ a)) = ((a⊥ ∪ b) ∩ 1)
12		an1 98	. . . . 5 ((a⊥ ∪ b) ∩ 1) = (a⊥ ∪ b)
13	11, 12	ax-r2 35	. . . 4 ((a⊥ ∪ b) ∩ (a⊥ ∪ a)) = (a⊥ ∪ b)
14	6, 13	ax-r2 35	. . 3 ((a⊥ ∪ a) ∩ (a⊥ ∪ b)) = (a⊥ ∪ b)
15	5, 14	ax-r2 35	. 2 (a⊥ ∪ (a ∩ b)) = (a⊥ ∪ b)
16	1, 15	ax-r2 35	1 (a →1 b) = (a⊥ ∪ b)

Syntax hints:   = wb 1   C wc 3  ^⊥ wn 4   ∪ wo 6   ∩ wa 7  1wt 9   →₁ wi1 13

This theorem is referenced by:  u1lemle1 692  u1lem1 716

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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