Theorem i5con 264

Description: Converse of →₅ .

Assertion

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

Proof of Theorem i5con

Step	Hyp	Ref	Expression
1		ancom 68	. . . 4 (a⊥ ∩ b⊥ ) = (b⊥ ∩ a⊥ )
2		ax-a2 30	. . . . 5 ((a ∩ b) ∪ (a⊥ ∩ b)) = ((a⊥ ∩ b) ∪ (a ∩ b))
3		ancom 68	. . . . . . 7 (a⊥ ∩ b) = (b ∩ a⊥ )
4		ax-a1 29	. . . . . . . 8 b = b⊥ ⊥
5	4	ran 71	. . . . . . 7 (b ∩ a⊥ ) = (b⊥ ⊥ ∩ a⊥ )
6	3, 5	ax-r2 35	. . . . . 6 (a⊥ ∩ b) = (b⊥ ⊥ ∩ a⊥ )
7		ancom 68	. . . . . . 7 (a ∩ b) = (b ∩ a)
8		ax-a1 29	. . . . . . . 8 a = a⊥ ⊥
9	4, 8	2an 72	. . . . . . 7 (b ∩ a) = (b⊥ ⊥ ∩ a⊥ ⊥ )
10	7, 9	ax-r2 35	. . . . . 6 (a ∩ b) = (b⊥ ⊥ ∩ a⊥ ⊥ )
11	6, 10	2or 67	. . . . 5 ((a⊥ ∩ b) ∪ (a ∩ b)) = ((b⊥ ⊥ ∩ a⊥ ) ∪ (b⊥ ⊥ ∩ a⊥ ⊥ ))
12	2, 11	ax-r2 35	. . . 4 ((a ∩ b) ∪ (a⊥ ∩ b)) = ((b⊥ ⊥ ∩ a⊥ ) ∪ (b⊥ ⊥ ∩ a⊥ ⊥ ))
13	1, 12	2or 67	. . 3 ((a⊥ ∩ b⊥ ) ∪ ((a ∩ b) ∪ (a⊥ ∩ b))) = ((b⊥ ∩ a⊥ ) ∪ ((b⊥ ⊥ ∩ a⊥ ) ∪ (b⊥ ⊥ ∩ a⊥ ⊥ )))
14		ax-a2 30	. . 3 (((a ∩ b) ∪ (a⊥ ∩ b)) ∪ (a⊥ ∩ b⊥ )) = ((a⊥ ∩ b⊥ ) ∪ ((a ∩ b) ∪ (a⊥ ∩ b)))
15		ax-a3 31	. . 3 (((b⊥ ∩ a⊥ ) ∪ (b⊥ ⊥ ∩ a⊥ )) ∪ (b⊥ ⊥ ∩ a⊥ ⊥ )) = ((b⊥ ∩ a⊥ ) ∪ ((b⊥ ⊥ ∩ a⊥ ) ∪ (b⊥ ⊥ ∩ a⊥ ⊥ )))
16	13, 14, 15	3tr1 60	. 2 (((a ∩ b) ∪ (a⊥ ∩ b)) ∪ (a⊥ ∩ b⊥ )) = (((b⊥ ∩ a⊥ ) ∪ (b⊥ ⊥ ∩ a⊥ )) ∪ (b⊥ ⊥ ∩ a⊥ ⊥ ))
17		df-i5 47	. 2 (a →5 b) = (((a ∩ b) ∪ (a⊥ ∩ b)) ∪ (a⊥ ∩ b⊥ ))
18		df-i5 47	. 2 (b⊥ →5 a⊥ ) = (((b⊥ ∩ a⊥ ) ∪ (b⊥ ⊥ ∩ a⊥ )) ∪ (b⊥ ⊥ ∩ a⊥ ⊥ ))
19	16, 17, 18	3tr1 60	1 (a →5 b) = (b⊥ →5 a⊥ )

Syntax hints:   = wb 1  ^⊥ wn 4   ∪ wo 6   ∩ wa 7   →₅ wi5 17

This theorem is referenced by:  nom45 322

This theorem was proved from axioms:  ax-a1 29  ax-a2 30  ax-a3 31  ax-r1 34  ax-r2 35  ax-r4 36  ax-r5 37

This theorem depends on definitions:  df-a 39  df-i5 47

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