Theorem i5con 272

Description: Converse of →₅ . (Contributed by NM, 7-Feb-1999.)

Assertion

Ref	Expression
i5con	(a →5 b) = (b⊥ →5 a⊥ )

Proof of Theorem i5con

Step	Hyp	Ref	Expression
1		ancom 74	. . . 4 (a⊥ ∩ b⊥ ) = (b⊥ ∩ a⊥ )
2		ax-a2 31	. . . . 5 ((a ∩ b) ∪ (a⊥ ∩ b)) = ((a⊥ ∩ b) ∪ (a ∩ b))
3		ancom 74	. . . . . . 7 (a⊥ ∩ b) = (b ∩ a⊥ )
4		ax-a1 30	. . . . . . . 8 b = b⊥ ⊥
5	4	ran 78	. . . . . . 7 (b ∩ a⊥ ) = (b⊥ ⊥ ∩ a⊥ )
6	3, 5	ax-r2 36	. . . . . 6 (a⊥ ∩ b) = (b⊥ ⊥ ∩ a⊥ )
7		ancom 74	. . . . . . 7 (a ∩ b) = (b ∩ a)
8		ax-a1 30	. . . . . . . 8 a = a⊥ ⊥
9	4, 8	2an 79	. . . . . . 7 (b ∩ a) = (b⊥ ⊥ ∩ a⊥ ⊥ )
10	7, 9	ax-r2 36	. . . . . 6 (a ∩ b) = (b⊥ ⊥ ∩ a⊥ ⊥ )
11	6, 10	2or 72	. . . . 5 ((a⊥ ∩ b) ∪ (a ∩ b)) = ((b⊥ ⊥ ∩ a⊥ ) ∪ (b⊥ ⊥ ∩ a⊥ ⊥ ))
12	2, 11	ax-r2 36	. . . 4 ((a ∩ b) ∪ (a⊥ ∩ b)) = ((b⊥ ⊥ ∩ a⊥ ) ∪ (b⊥ ⊥ ∩ a⊥ ⊥ ))
13	1, 12	2or 72	. . 3 ((a⊥ ∩ b⊥ ) ∪ ((a ∩ b) ∪ (a⊥ ∩ b))) = ((b⊥ ∩ a⊥ ) ∪ ((b⊥ ⊥ ∩ a⊥ ) ∪ (b⊥ ⊥ ∩ a⊥ ⊥ )))
14		ax-a2 31	. . 3 (((a ∩ b) ∪ (a⊥ ∩ b)) ∪ (a⊥ ∩ b⊥ )) = ((a⊥ ∩ b⊥ ) ∪ ((a ∩ b) ∪ (a⊥ ∩ b)))
15		ax-a3 32	. . 3 (((b⊥ ∩ a⊥ ) ∪ (b⊥ ⊥ ∩ a⊥ )) ∪ (b⊥ ⊥ ∩ a⊥ ⊥ )) = ((b⊥ ∩ a⊥ ) ∪ ((b⊥ ⊥ ∩ a⊥ ) ∪ (b⊥ ⊥ ∩ a⊥ ⊥ )))
16	13, 14, 15	3tr1 63	. 2 (((a ∩ b) ∪ (a⊥ ∩ b)) ∪ (a⊥ ∩ b⊥ )) = (((b⊥ ∩ a⊥ ) ∪ (b⊥ ⊥ ∩ a⊥ )) ∪ (b⊥ ⊥ ∩ a⊥ ⊥ ))
17		df-i5 48	. 2 (a →5 b) = (((a ∩ b) ∪ (a⊥ ∩ b)) ∪ (a⊥ ∩ b⊥ ))
18		df-i5 48	. 2 (b⊥ →5 a⊥ ) = (((b⊥ ∩ a⊥ ) ∪ (b⊥ ⊥ ∩ a⊥ )) ∪ (b⊥ ⊥ ∩ a⊥ ⊥ ))
19	16, 17, 18	3tr1 63	1 (a →5 b) = (b⊥ →5 a⊥ )

This proof depends on syntax axioms:   = wb 1  ^⊥ wn 4   ∪ wo 6   ∩ wa 7   →₅ wi5 16

This proof depends on axioms:  ax-a1 30  ax-a2 31  ax-a3 32  ax-r1 35  ax-r2 36  ax-r4 37  ax-r5 38

This proof depends on definitions:  df-a 40  df-i5 48

This theorem is used by:  nom45  330