Theorem u2lemanb 616

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

Assertion

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

Proof of Theorem u2lemanb

Step	Hyp	Ref	Expression
1		df-i2 45	. . 3 (a →2 b) = (b ∪ (a⊥ ∩ b⊥ ))
2	1	ran 78	. 2 ((a →2 b) ∩ b⊥ ) = ((b ∪ (a⊥ ∩ b⊥ )) ∩ b⊥ )
3		comid 187	. . . . 5 b C b
4	3	comcom3 454	. . . 4 b⊥ C b
5		comanr2 465	. . . 4 b⊥ C (a⊥ ∩ b⊥ )
6	4, 5	fh1r 473	. . 3 ((b ∪ (a⊥ ∩ b⊥ )) ∩ b⊥ ) = ((b ∩ b⊥ ) ∪ ((a⊥ ∩ b⊥ ) ∩ b⊥ ))
7		ax-a2 31	. . . 4 ((b ∩ b⊥ ) ∪ ((a⊥ ∩ b⊥ ) ∩ b⊥ )) = (((a⊥ ∩ b⊥ ) ∩ b⊥ ) ∪ (b ∩ b⊥ ))
8		anass 76	. . . . . . 7 ((a⊥ ∩ b⊥ ) ∩ b⊥ ) = (a⊥ ∩ (b⊥ ∩ b⊥ ))
9		anidm 111	. . . . . . . 8 (b⊥ ∩ b⊥ ) = b⊥
10	9	lan 77	. . . . . . 7 (a⊥ ∩ (b⊥ ∩ b⊥ )) = (a⊥ ∩ b⊥ )
11	8, 10	ax-r2 36	. . . . . 6 ((a⊥ ∩ b⊥ ) ∩ b⊥ ) = (a⊥ ∩ b⊥ )
12		dff 101	. . . . . . 7 0 = (b ∩ b⊥ )
13	12	ax-r1 35	. . . . . 6 (b ∩ b⊥ ) = 0
14	11, 13	2or 72	. . . . 5 (((a⊥ ∩ b⊥ ) ∩ b⊥ ) ∪ (b ∩ b⊥ )) = ((a⊥ ∩ b⊥ ) ∪ 0)
15		or0 102	. . . . 5 ((a⊥ ∩ b⊥ ) ∪ 0) = (a⊥ ∩ b⊥ )
16	14, 15	ax-r2 36	. . . 4 (((a⊥ ∩ b⊥ ) ∩ b⊥ ) ∪ (b ∩ b⊥ )) = (a⊥ ∩ b⊥ )
17	7, 16	ax-r2 36	. . 3 ((b ∩ b⊥ ) ∪ ((a⊥ ∩ b⊥ ) ∩ b⊥ )) = (a⊥ ∩ b⊥ )
18	6, 17	ax-r2 36	. 2 ((b ∪ (a⊥ ∩ b⊥ )) ∩ b⊥ ) = (a⊥ ∩ b⊥ )
19	2, 18	ax-r2 36	1 ((a →2 b) ∩ b⊥ ) = (a⊥ ∩ b⊥ )

This proof depends on syntax axioms:   = wb 1  ^⊥ wn 4   ∪ wo 6   ∩ wa 7  0wf 9   →₂ wi2 13

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-i2 45  df-le1 130  df-le2 131  df-c1 132  df-c2 133

This theorem is used by:  u2lemnob  671  u21lembi  727  bi3  839  bi4  840  imp3  841  oal42  935  oa23  936