Theorem i1i2con2 269

Description: Correspondence between Sasaki and Dishkant conditionals. (Contributed by NM, 28-Feb-1999.)

Assertion

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

Proof of Theorem i1i2con2

Step	Hyp	Ref	Expression
1		i1i2 266	. 2 (a⊥ →1 b) = (b⊥ →2 a⊥ ⊥ )
2		ax-a1 30	. . . 4 a = a⊥ ⊥
3	2	ax-r1 35	. . 3 a⊥ ⊥ = a
4	3	ud2lem0a 258	. 2 (b⊥ →2 a⊥ ⊥ ) = (b⊥ →2 a)
5	1, 4	ax-r2 36	1 (a⊥ →1 b) = (b⊥ →2 a)

Syntax hints:   = wb 1  ^⊥ wn 4   →₁ wi1 12   →₂ wi2 13

This theorem was proved from axioms:  ax-a1 30  ax-a2 31  ax-r1 35  ax-r2 36  ax-r4 37  ax-r5 38

This theorem depends on definitions:  df-a 40  df-i1 44  df-i2 45

This theorem is referenced by:  2oai1u  822  1oath1i1u  828