Theorem u1lembi 720

Description: Sasaki implication and biconditional. (Contributed by NM, 17-Jan-1998.)

Assertion

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

Proof of Theorem u1lembi

Step	Hyp	Ref	Expression
1		ax-a2 31	. . . 4 (a⊥ ∪ (a ∩ b)) = ((a ∩ b) ∪ a⊥ )
2		ax-a2 31	. . . 4 (b⊥ ∪ (a ∩ b)) = ((a ∩ b) ∪ b⊥ )
3	1, 2	2an 79	. . 3 ((a⊥ ∪ (a ∩ b)) ∩ (b⊥ ∪ (a ∩ b))) = (((a ∩ b) ∪ a⊥ ) ∩ ((a ∩ b) ∪ b⊥ ))
4		coman1 185	. . . . . 6 (a ∩ b) C a
5	4	comcom2 183	. . . . 5 (a ∩ b) C a⊥
6		coman2 186	. . . . . 6 (a ∩ b) C b
7	6	comcom2 183	. . . . 5 (a ∩ b) C b⊥
8	5, 7	fh3 471	. . . 4 ((a ∩ b) ∪ (a⊥ ∩ b⊥ )) = (((a ∩ b) ∪ a⊥ ) ∩ ((a ∩ b) ∪ b⊥ ))
9	8	ax-r1 35	. . 3 (((a ∩ b) ∪ a⊥ ) ∩ ((a ∩ b) ∪ b⊥ )) = ((a ∩ b) ∪ (a⊥ ∩ b⊥ ))
10	3, 9	ax-r2 36	. 2 ((a⊥ ∪ (a ∩ b)) ∩ (b⊥ ∪ (a ∩ b))) = ((a ∩ b) ∪ (a⊥ ∩ b⊥ ))
11		df-i1 44	. . 3 (a →1 b) = (a⊥ ∪ (a ∩ b))
12		df-i1 44	. . . 4 (b →1 a) = (b⊥ ∪ (b ∩ a))
13		ancom 74	. . . . 5 (b ∩ a) = (a ∩ b)
14	13	lor 70	. . . 4 (b⊥ ∪ (b ∩ a)) = (b⊥ ∪ (a ∩ b))
15	12, 14	ax-r2 36	. . 3 (b →1 a) = (b⊥ ∪ (a ∩ b))
16	11, 15	2an 79	. 2 ((a →1 b) ∩ (b →1 a)) = ((a⊥ ∪ (a ∩ b)) ∩ (b⊥ ∪ (a ∩ b)))
17		dfb 94	. 2 (a ≡ b) = ((a ∩ b) ∪ (a⊥ ∩ b⊥ ))
18	10, 16, 17	3tr1 63	1 ((a →1 b) ∩ (b →1 a)) = (a ≡ b)

Syntax hints:   = wb 1  ^⊥ wn 4   ≡ tb 5   ∪ wo 6   ∩ wa 7   →₁ wi1 12

This theorem was proved from 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 theorem depends on definitions:  df-b 39  df-a 40  df-t 41  df-f 42  df-i1 44  df-le1 130  df-le2 131  df-c1 132  df-c2 133

This theorem is referenced by:  mlaoml  833  comanblem1  870