Theorem i3i4 270

Description: Correspondence between Kalmbach and non-tollens conditionals. (Contributed by NM, 7-Feb-1999.)

Assertion

Ref	Expression
i3i4	(a →3 b) = (b⊥ →4 a⊥ )

Proof of Theorem i3i4

Step	Hyp	Ref	Expression
1		ax-a2 31	. . . 4 ((a⊥ ∩ b) ∪ (a⊥ ∩ b⊥ )) = ((a⊥ ∩ b⊥ ) ∪ (a⊥ ∩ b))
2		ancom 74	. . . . 5 (a⊥ ∩ b⊥ ) = (b⊥ ∩ a⊥ )
3		ancom 74	. . . . . 6 (a⊥ ∩ b) = (b ∩ a⊥ )
4		ax-a1 30	. . . . . . 7 b = b⊥ ⊥
5	4	ran 78	. . . . . 6 (b ∩ a⊥ ) = (b⊥ ⊥ ∩ a⊥ )
6	3, 5	ax-r2 36	. . . . 5 (a⊥ ∩ b) = (b⊥ ⊥ ∩ a⊥ )
7	2, 6	2or 72	. . . 4 ((a⊥ ∩ b⊥ ) ∪ (a⊥ ∩ b)) = ((b⊥ ∩ a⊥ ) ∪ (b⊥ ⊥ ∩ a⊥ ))
8	1, 7	ax-r2 36	. . 3 ((a⊥ ∩ b) ∪ (a⊥ ∩ b⊥ )) = ((b⊥ ∩ a⊥ ) ∪ (b⊥ ⊥ ∩ a⊥ ))
9		ancom 74	. . . 4 (a ∩ (a⊥ ∪ b)) = ((a⊥ ∪ b) ∩ a)
10		ax-a2 31	. . . . . 6 (a⊥ ∪ b) = (b ∪ a⊥ )
11	4	ax-r5 38	. . . . . 6 (b ∪ a⊥ ) = (b⊥ ⊥ ∪ a⊥ )
12	10, 11	ax-r2 36	. . . . 5 (a⊥ ∪ b) = (b⊥ ⊥ ∪ a⊥ )
13		ax-a1 30	. . . . 5 a = a⊥ ⊥
14	12, 13	2an 79	. . . 4 ((a⊥ ∪ b) ∩ a) = ((b⊥ ⊥ ∪ a⊥ ) ∩ a⊥ ⊥ )
15	9, 14	ax-r2 36	. . 3 (a ∩ (a⊥ ∪ b)) = ((b⊥ ⊥ ∪ a⊥ ) ∩ a⊥ ⊥ )
16	8, 15	2or 72	. 2 (((a⊥ ∩ b) ∪ (a⊥ ∩ b⊥ )) ∪ (a ∩ (a⊥ ∪ b))) = (((b⊥ ∩ a⊥ ) ∪ (b⊥ ⊥ ∩ a⊥ )) ∪ ((b⊥ ⊥ ∪ a⊥ ) ∩ a⊥ ⊥ ))
17		df-i3 46	. 2 (a →3 b) = (((a⊥ ∩ b) ∪ (a⊥ ∩ b⊥ )) ∪ (a ∩ (a⊥ ∪ b)))
18		df-i4 47	. 2 (b⊥ →4 a⊥ ) = (((b⊥ ∩ a⊥ ) ∪ (b⊥ ⊥ ∩ a⊥ )) ∪ ((b⊥ ⊥ ∪ a⊥ ) ∩ a⊥ ⊥ ))
19	16, 17, 18	3tr1 63	1 (a →3 b) = (b⊥ →4 a⊥ )

This proof depends on syntax axioms:   = wb 1  ^⊥ wn 4   ∪ wo 6   ∩ wa 7   →₃ wi3 14   →₄ wi4 15

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

This proof depends on definitions:  df-a 40  df-i3 46  df-i4 47

This theorem is used by:  i4i3  271  nom43  328