Theorem u2lemanb 598

Description: Lemma for Dishkant implication study.

Assertion

Ref	Expression
u2lemanb	((a →2 b) ∩ b⊥ ) = (a⊥ ∩ b⊥ )

Proof of Theorem u2lemanb

Step	Hyp	Ref	Expression
1		df-i2 44	. . 3 (a →2 b) = (b ∪ (a⊥ ∩ b⊥ ))
2	1	ran 71	. 2 ((a →2 b) ∩ b⊥ ) = ((b ∪ (a⊥ ∩ b⊥ )) ∩ b⊥ )
3		comid 179	. . . . 5 b C b
4	3	comcom3 436	. . . 4 b⊥ C b
5		comanr2 447	. . . 4 b⊥ C (a⊥ ∩ b⊥ )
6	4, 5	fh1r 455	. . 3 ((b ∪ (a⊥ ∩ b⊥ )) ∩ b⊥ ) = ((b ∩ b⊥ ) ∪ ((a⊥ ∩ b⊥ ) ∩ b⊥ ))
7		ax-a2 30	. . . 4 ((b ∩ b⊥ ) ∪ ((a⊥ ∩ b⊥ ) ∩ b⊥ )) = (((a⊥ ∩ b⊥ ) ∩ b⊥ ) ∪ (b ∩ b⊥ ))
8		anass 69	. . . . . . 7 ((a⊥ ∩ b⊥ ) ∩ b⊥ ) = (a⊥ ∩ (b⊥ ∩ b⊥ ))
9		anidm 103	. . . . . . . 8 (b⊥ ∩ b⊥ ) = b⊥
10	9	lan 70	. . . . . . 7 (a⊥ ∩ (b⊥ ∩ b⊥ )) = (a⊥ ∩ b⊥ )
11	8, 10	ax-r2 35	. . . . . 6 ((a⊥ ∩ b⊥ ) ∩ b⊥ ) = (a⊥ ∩ b⊥ )
12		dff 93	. . . . . . 7 0 = (b ∩ b⊥ )
13	12	ax-r1 34	. . . . . 6 (b ∩ b⊥ ) = 0
14	11, 13	2or 67	. . . . 5 (((a⊥ ∩ b⊥ ) ∩ b⊥ ) ∪ (b ∩ b⊥ )) = ((a⊥ ∩ b⊥ ) ∪ 0)
15		or0 94	. . . . 5 ((a⊥ ∩ b⊥ ) ∪ 0) = (a⊥ ∩ b⊥ )
16	14, 15	ax-r2 35	. . . 4 (((a⊥ ∩ b⊥ ) ∩ b⊥ ) ∪ (b ∩ b⊥ )) = (a⊥ ∩ b⊥ )
17	7, 16	ax-r2 35	. . 3 ((b ∩ b⊥ ) ∪ ((a⊥ ∩ b⊥ ) ∩ b⊥ )) = (a⊥ ∩ b⊥ )
18	6, 17	ax-r2 35	. 2 ((b ∪ (a⊥ ∩ b⊥ )) ∩ b⊥ ) = (a⊥ ∩ b⊥ )
19	2, 18	ax-r2 35	1 ((a →2 b) ∩ b⊥ ) = (a⊥ ∩ b⊥ )

Syntax hints:   = wb 1  ^⊥ wn 4   ∪ wo 6   ∩ wa 7  0wf 10   →₂ wi2 14

This theorem is referenced by:  u2lemnob 653  u21lembi 709  bi3 821  bi4 822  imp3 823  oal42 915  oa23 916

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-i2 44  df-le1 122  df-le2 123  df-c1 124  df-c2 125

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