Theorem u2lembi 721

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

Assertion

Ref	Expression
u2lembi	((a →2 b) ∩ (b →2 a)) = (a ≡ b)

Proof of Theorem u2lembi

Step	Hyp	Ref	Expression
1		ancom 74	. . 3 ((b ∪ (a⊥ ∩ b⊥ )) ∩ (a ∪ (a⊥ ∩ b⊥ ))) = ((a ∪ (a⊥ ∩ b⊥ )) ∩ (b ∪ (a⊥ ∩ b⊥ )))
2		coman1 185	. . . . . 6 (a⊥ ∩ b⊥ ) C a⊥
3	2	comcom7 460	. . . . 5 (a⊥ ∩ b⊥ ) C a
4		coman2 186	. . . . . 6 (a⊥ ∩ b⊥ ) C b⊥
5	4	comcom7 460	. . . . 5 (a⊥ ∩ b⊥ ) C b
6	3, 5	fh3r 475	. . . 4 ((a ∩ b) ∪ (a⊥ ∩ b⊥ )) = ((a ∪ (a⊥ ∩ b⊥ )) ∩ (b ∪ (a⊥ ∩ b⊥ )))
7	6	ax-r1 35	. . 3 ((a ∪ (a⊥ ∩ b⊥ )) ∩ (b ∪ (a⊥ ∩ b⊥ ))) = ((a ∩ b) ∪ (a⊥ ∩ b⊥ ))
8	1, 7	ax-r2 36	. 2 ((b ∪ (a⊥ ∩ b⊥ )) ∩ (a ∪ (a⊥ ∩ b⊥ ))) = ((a ∩ b) ∪ (a⊥ ∩ b⊥ ))
9		df-i2 45	. . 3 (a →2 b) = (b ∪ (a⊥ ∩ b⊥ ))
10		df-i2 45	. . . 4 (b →2 a) = (a ∪ (b⊥ ∩ a⊥ ))
11		ancom 74	. . . . 5 (b⊥ ∩ a⊥ ) = (a⊥ ∩ b⊥ )
12	11	lor 70	. . . 4 (a ∪ (b⊥ ∩ a⊥ )) = (a ∪ (a⊥ ∩ b⊥ ))
13	10, 12	ax-r2 36	. . 3 (b →2 a) = (a ∪ (a⊥ ∩ b⊥ ))
14	9, 13	2an 79	. 2 ((a →2 b) ∩ (b →2 a)) = ((b ∪ (a⊥ ∩ b⊥ )) ∩ (a ∪ (a⊥ ∩ b⊥ )))
15		dfb 94	. 2 (a ≡ b) = ((a ∩ b) ∪ (a⊥ ∩ b⊥ ))
16	8, 14, 15	3tr1 63	1 ((a →2 b) ∩ (b →2 a)) = (a ≡ b)

This proof depends on syntax axioms:   = wb 1  ^⊥ wn 4   ≡ tb 5   ∪ wo 6   ∩ wa 7   →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:  i2bi  722  mloa  1018