Theorem i1i2 266

Description: Correspondence between Sasaki and Dishkant conditionals. (Contributed by NM, 25-Nov-1998.)

Assertion

Ref	Expression
i1i2	(a →1 b) = (b⊥ →2 a⊥ )

Proof of Theorem i1i2

Step	Hyp	Ref	Expression
1		ax-a1 30	. . . . 5 a = a⊥ ⊥
2		ax-a1 30	. . . . 5 b = b⊥ ⊥
3	1, 2	2an 79	. . . 4 (a ∩ b) = (a⊥ ⊥ ∩ b⊥ ⊥ )
4		ancom 74	. . . 4 (a⊥ ⊥ ∩ b⊥ ⊥ ) = (b⊥ ⊥ ∩ a⊥ ⊥ )
5	3, 4	ax-r2 36	. . 3 (a ∩ b) = (b⊥ ⊥ ∩ a⊥ ⊥ )
6	5	lor 70	. 2 (a⊥ ∪ (a ∩ b)) = (a⊥ ∪ (b⊥ ⊥ ∩ a⊥ ⊥ ))
7		df-i1 44	. 2 (a →1 b) = (a⊥ ∪ (a ∩ b))
8		df-i2 45	. 2 (b⊥ →2 a⊥ ) = (a⊥ ∪ (b⊥ ⊥ ∩ a⊥ ⊥ ))
9	6, 7, 8	3tr1 63	1 (a →1 b) = (b⊥ →2 a⊥ )

This proof depends on syntax axioms:   = wb 1  ^⊥ wn 4   ∪ wo 6   ∩ wa 7   →₁ wi1 12   →₂ wi2 13

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-i1 44  df-i2 45

This theorem is used by:  i2i1  267  i1i2con1  268  i1i2con2  269  nom41  326  1oai1  821  2oath1i1  827  oal1  1000