Theorem u1lembi 702

Description: Sasaki implication and biconditional.

Assertion

Ref	Expression
u1lembi	((a →1 b) ∩ (b →1 a)) = (a ≡ b)

Proof of Theorem u1lembi

Step	Hyp	Ref	Expression
1		ax-a2 30	. . . 4 (a⊥ ∪ (a ∩ b)) = ((a ∩ b) ∪ a⊥ )
2		ax-a2 30	. . . 4 (b⊥ ∪ (a ∩ b)) = ((a ∩ b) ∪ b⊥ )
3	1, 2	2an 72	. . 3 ((a⊥ ∪ (a ∩ b)) ∩ (b⊥ ∪ (a ∩ b))) = (((a ∩ b) ∪ a⊥ ) ∩ ((a ∩ b) ∪ b⊥ ))
4		coman1 177	. . . . . 6 (a ∩ b) C a
5	4	comcom2 175	. . . . 5 (a ∩ b) C a⊥
6		coman2 178	. . . . . 6 (a ∩ b) C b
7	6	comcom2 175	. . . . 5 (a ∩ b) C b⊥
8	5, 7	fh3 453	. . . 4 ((a ∩ b) ∪ (a⊥ ∩ b⊥ )) = (((a ∩ b) ∪ a⊥ ) ∩ ((a ∩ b) ∪ b⊥ ))
9	8	ax-r1 34	. . 3 (((a ∩ b) ∪ a⊥ ) ∩ ((a ∩ b) ∪ b⊥ )) = ((a ∩ b) ∪ (a⊥ ∩ b⊥ ))
10	3, 9	ax-r2 35	. 2 ((a⊥ ∪ (a ∩ b)) ∩ (b⊥ ∪ (a ∩ b))) = ((a ∩ b) ∪ (a⊥ ∩ b⊥ ))
11		df-i1 43	. . 3 (a →1 b) = (a⊥ ∪ (a ∩ b))
12		df-i1 43	. . . 4 (b →1 a) = (b⊥ ∪ (b ∩ a))
13		ancom 68	. . . . 5 (b ∩ a) = (a ∩ b)
14	13	lor 66	. . . 4 (b⊥ ∪ (b ∩ a)) = (b⊥ ∪ (a ∩ b))
15	12, 14	ax-r2 35	. . 3 (b →1 a) = (b⊥ ∪ (a ∩ b))
16	11, 15	2an 72	. 2 ((a →1 b) ∩ (b →1 a)) = ((a⊥ ∪ (a ∩ b)) ∩ (b⊥ ∪ (a ∩ b)))
17		dfb 86	. 2 (a ≡ b) = ((a ∩ b) ∪ (a⊥ ∩ b⊥ ))
18	10, 16, 17	3tr1 60	1 ((a →1 b) ∩ (b →1 a)) = (a ≡ b)

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

This theorem is referenced by:  mlaoml 815  comanblem1 852

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

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