Theorem i1i2con1 260

Description: Correspondence between Sasaki and Dishkant conditionals.

Assertion

Ref	Expression
i1i2con1	(a ->1 b_|_) = (b ->2 a_|_)

Proof of Theorem i1i2con1

Step	Hyp	Ref	Expression
1		i1i2 258	. 2 (a ->1 b_|_) = (b_|__|_ ->2 a_|_)
2		ax-a1 29	. . . 4 b = b_|__|_
3	2	ax-r1 34	. . 3 b_|__|_ = b
4	3	ud2lem0b 251	. 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 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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