Theorem i1i2 258

Description: Correspondence between Sasaki and Dishkant conditionals.

Assertion

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

Proof of Theorem i1i2

Step	Hyp	Ref	Expression
1		ax-a1 29	. . . . 5 a = a⊥ ⊥
2		ax-a1 29	. . . . 5 b = b⊥ ⊥
3	1, 2	2an 72	. . . 4 (a ∩ b) = (a⊥ ⊥ ∩ b⊥ ⊥ )
4		ancom 68	. . . 4 (a⊥ ⊥ ∩ b⊥ ⊥ ) = (b⊥ ⊥ ∩ a⊥ ⊥ )
5	3, 4	ax-r2 35	. . 3 (a ∩ b) = (b⊥ ⊥ ∩ a⊥ ⊥ )
6	5	lor 66	. 2 (a⊥ ∪ (a ∩ b)) = (a⊥ ∪ (b⊥ ⊥ ∩ a⊥ ⊥ ))
7		df-i1 43	. 2 (a →1 b) = (a⊥ ∪ (a ∩ b))
8		df-i2 44	. 2 (b⊥ →2 a⊥ ) = (a⊥ ∪ (b⊥ ⊥ ∩ a⊥ ⊥ ))
9	6, 7, 8	3tr1 60	1 (a →1 b) = (b⊥ →2 a⊥ )

Syntax hints:   = wb 1  ^⊥ wn 4   ∪ wo 6   ∩ wa 7   →₁ wi1 13   →₂ wi2 14

This theorem is referenced by:  i2i1 259  i1i2con1 260  i1i2con2 261  nom41 318  1oai1 803  2oath1i1 809  oal1 980

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

This theorem depends on definitions:  df-a 39  df-i1 43  df-i2 44

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