Theorem i1i2con2 261

Description: Correspondence between Sasaki and Dishkant conditionals.

Assertion

Ref	Expression
i1i2con2	(a⊥ →1 b) = (b⊥ →2 a)

Proof of Theorem i1i2con2

Step	Hyp	Ref	Expression
1		i1i2 258	. 2 (a⊥ →1 b) = (b⊥ →2 a⊥ ⊥ )
2		ax-a1 29	. . . 4 a = a⊥ ⊥
3	2	ax-r1 34	. . 3 a⊥ ⊥ = a
4	3	ud2lem0a 250	. 2 (b⊥ →2 a⊥ ⊥ ) = (b⊥ →2 a)
5	1, 4	ax-r2 35	1 (a⊥ →1 b) = (b⊥ →2 a)

Syntax hints:   = wb 1  ^⊥ wn 4   →₁ wi1 13   →₂ wi2 14

This theorem is referenced by:  2oai1u 804  1oath1i1u 810

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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