Theorem u21lembi 727

Description: Dishkant/Sasaki implication and biconditional. (Contributed by NM, 3-Mar-2000.)

Assertion

Ref	Expression
u21lembi	((a →2 b) ∩ (b →1 a)) = (a ≡ b)

Proof of Theorem u21lembi

Step	Hyp	Ref	Expression
1		u2lemc1 681	. . . . 5 b C (a →2 b)
2	1	comcom3 454	. . . 4 b⊥ C (a →2 b)
3		comanr1 464	. . . . 5 b C (b ∩ a)
4	3	comcom3 454	. . . 4 b⊥ C (b ∩ a)
5	2, 4	fh2 470	. . 3 ((a →2 b) ∩ (b⊥ ∪ (b ∩ a))) = (((a →2 b) ∩ b⊥ ) ∪ ((a →2 b) ∩ (b ∩ a)))
6		u2lemanb 616	. . . 4 ((a →2 b) ∩ b⊥ ) = (a⊥ ∩ b⊥ )
7		u2lemab 611	. . . . . 6 ((a →2 b) ∩ b) = b
8	7	ran 78	. . . . 5 (((a →2 b) ∩ b) ∩ a) = (b ∩ a)
9		anass 76	. . . . 5 (((a →2 b) ∩ b) ∩ a) = ((a →2 b) ∩ (b ∩ a))
10		ancom 74	. . . . 5 (b ∩ a) = (a ∩ b)
11	8, 9, 10	3tr2 64	. . . 4 ((a →2 b) ∩ (b ∩ a)) = (a ∩ b)
12	6, 11	2or 72	. . 3 (((a →2 b) ∩ b⊥ ) ∪ ((a →2 b) ∩ (b ∩ a))) = ((a⊥ ∩ b⊥ ) ∪ (a ∩ b))
13		ax-a2 31	. . 3 ((a⊥ ∩ b⊥ ) ∪ (a ∩ b)) = ((a ∩ b) ∪ (a⊥ ∩ b⊥ ))
14	5, 12, 13	3tr 65	. 2 ((a →2 b) ∩ (b⊥ ∪ (b ∩ a))) = ((a ∩ b) ∪ (a⊥ ∩ b⊥ ))
15		df-i1 44	. . 3 (b →1 a) = (b⊥ ∪ (b ∩ a))
16	15	lan 77	. 2 ((a →2 b) ∩ (b →1 a)) = ((a →2 b) ∩ (b⊥ ∪ (b ∩ a)))
17		dfb 94	. 2 (a ≡ b) = ((a ∩ b) ∪ (a⊥ ∩ b⊥ ))
18	14, 16, 17	3tr1 63	1 ((a →2 b) ∩ (b →1 a)) = (a ≡ b)

This proof depends on syntax axioms:   = wb 1  ^⊥ wn 4   ≡ tb 5   ∪ wo 6   ∩ wa 7   →₁ wi1 12   →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-i1 44  df-i2 45  df-le1 130  df-le2 131  df-c1 132  df-c2 133

This theorem is used by: (None)