Theorem u1lemanb 615

Description: Lemma for Sasaki implication study. (Contributed by NM, 14-Dec-1997.)

Assertion

Ref	Expression
u1lemanb	((a →1 b) ∩ b⊥ ) = (a⊥ ∩ b⊥ )

Proof of Theorem u1lemanb

Step	Hyp	Ref	Expression
1		df-i1 44	. . 3 (a →1 b) = (a⊥ ∪ (a ∩ b))
2	1	ran 78	. 2 ((a →1 b) ∩ b⊥ ) = ((a⊥ ∪ (a ∩ b)) ∩ b⊥ )
3		ax-a2 31	. . . 4 (a⊥ ∪ (a ∩ b)) = ((a ∩ b) ∪ a⊥ )
4	3	ran 78	. . 3 ((a⊥ ∪ (a ∩ b)) ∩ b⊥ ) = (((a ∩ b) ∪ a⊥ ) ∩ b⊥ )
5		coman2 186	. . . . . 6 (a ∩ b) C b
6	5	comcom2 183	. . . . 5 (a ∩ b) C b⊥
7		coman1 185	. . . . . 6 (a ∩ b) C a
8	7	comcom2 183	. . . . 5 (a ∩ b) C a⊥
9	6, 8	fh2r 474	. . . 4 (((a ∩ b) ∪ a⊥ ) ∩ b⊥ ) = (((a ∩ b) ∩ b⊥ ) ∪ (a⊥ ∩ b⊥ ))
10		ax-a2 31	. . . . 5 (((a ∩ b) ∩ b⊥ ) ∪ (a⊥ ∩ b⊥ )) = ((a⊥ ∩ b⊥ ) ∪ ((a ∩ b) ∩ b⊥ ))
11		anass 76	. . . . . . . 8 ((a ∩ b) ∩ b⊥ ) = (a ∩ (b ∩ b⊥ ))
12		dff 101	. . . . . . . . . . 11 0 = (b ∩ b⊥ )
13	12	lan 77	. . . . . . . . . 10 (a ∩ 0) = (a ∩ (b ∩ b⊥ ))
14	13	ax-r1 35	. . . . . . . . 9 (a ∩ (b ∩ b⊥ )) = (a ∩ 0)
15		an0 108	. . . . . . . . 9 (a ∩ 0) = 0
16	14, 15	ax-r2 36	. . . . . . . 8 (a ∩ (b ∩ b⊥ )) = 0
17	11, 16	ax-r2 36	. . . . . . 7 ((a ∩ b) ∩ b⊥ ) = 0
18	17	lor 70	. . . . . 6 ((a⊥ ∩ b⊥ ) ∪ ((a ∩ b) ∩ b⊥ )) = ((a⊥ ∩ b⊥ ) ∪ 0)
19		or0 102	. . . . . 6 ((a⊥ ∩ b⊥ ) ∪ 0) = (a⊥ ∩ b⊥ )
20	18, 19	ax-r2 36	. . . . 5 ((a⊥ ∩ b⊥ ) ∪ ((a ∩ b) ∩ b⊥ )) = (a⊥ ∩ b⊥ )
21	10, 20	ax-r2 36	. . . 4 (((a ∩ b) ∩ b⊥ ) ∪ (a⊥ ∩ b⊥ )) = (a⊥ ∩ b⊥ )
22	9, 21	ax-r2 36	. . 3 (((a ∩ b) ∪ a⊥ ) ∩ b⊥ ) = (a⊥ ∩ b⊥ )
23	4, 22	ax-r2 36	. 2 ((a⊥ ∪ (a ∩ b)) ∩ b⊥ ) = (a⊥ ∩ b⊥ )
24	2, 23	ax-r2 36	1 ((a →1 b) ∩ b⊥ ) = (a⊥ ∩ b⊥ )

This proof depends on syntax axioms:   = wb 1  ^⊥ wn 4   ∪ wo 6   ∩ wa 7  0wf 9   →₁ wi1 12

This proof depends on 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

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

This theorem is used by:  u1lemnob  670  u3lem14a  791  negantlem5  853