Theorem u21lembi 709

Description: Dishkant/Sasaki implication and biconditional.

Assertion

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

Proof of Theorem u21lembi

Step	Hyp	Ref	Expression
1		u2lemc1 663	. . . . 5 b C (a →2 b)
2	1	comcom3 436	. . . 4 b⊥ C (a →2 b)
3		comanr1 446	. . . . 5 b C (b ∩ a)
4	3	comcom3 436	. . . 4 b⊥ C (b ∩ a)
5	2, 4	fh2 452	. . 3 ((a →2 b) ∩ (b⊥ ∪ (b ∩ a))) = (((a →2 b) ∩ b⊥ ) ∪ ((a →2 b) ∩ (b ∩ a)))
6		u2lemanb 598	. . . 4 ((a →2 b) ∩ b⊥ ) = (a⊥ ∩ b⊥ )
7		u2lemab 593	. . . . . 6 ((a →2 b) ∩ b) = b
8	7	ran 71	. . . . 5 (((a →2 b) ∩ b) ∩ a) = (b ∩ a)
9		anass 69	. . . . 5 (((a →2 b) ∩ b) ∩ a) = ((a →2 b) ∩ (b ∩ a))
10		ancom 68	. . . . 5 (b ∩ a) = (a ∩ b)
11	8, 9, 10	3tr2 61	. . . 4 ((a →2 b) ∩ (b ∩ a)) = (a ∩ b)
12	6, 11	2or 67	. . 3 (((a →2 b) ∩ b⊥ ) ∪ ((a →2 b) ∩ (b ∩ a))) = ((a⊥ ∩ b⊥ ) ∪ (a ∩ b))
13		ax-a2 30	. . 3 ((a⊥ ∩ b⊥ ) ∪ (a ∩ b)) = ((a ∩ b) ∪ (a⊥ ∩ b⊥ ))
14	5, 12, 13	3tr 62	. 2 ((a →2 b) ∩ (b⊥ ∪ (b ∩ a))) = ((a ∩ b) ∪ (a⊥ ∩ b⊥ ))
15		df-i1 43	. . 3 (b →1 a) = (b⊥ ∪ (b ∩ a))
16	15	lan 70	. 2 ((a →2 b) ∩ (b →1 a)) = ((a →2 b) ∩ (b⊥ ∪ (b ∩ a)))
17		dfb 86	. 2 (a ≡ b) = ((a ∩ b) ∪ (a⊥ ∩ b⊥ ))
18	14, 16, 17	3tr1 60	1 ((a →2 b) ∩ (b →1 a)) = (a ≡ b)

Syntax hints:   = wb 1  ^⊥ wn 4   ≡ tb 5   ∪ wo 6   ∩ wa 7   →₁ wi1 13   →₂ wi2 14

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

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