so
(a->a) = (a->a)
(a->a) = (b->b)
(a->a) != (a->b)
Jay
On Wed, Dec 1, 2010 at 6:06 PM, Samuel Stephens <almo...@gmail.com> wrote:
> Jay:
>
> What is the behavior of the Type=? function? I don't understand what
> the modulo renaming mentioned in its helper function is. Is it
> supposed to return true for two t-vars that have different names, or
> error?
>
> Samuel Stephens
> --"Your reputation is what others know about you. Your honor is what
> you know about yourself."
> --Aral Vorkosigan
> from "Memory" by Lois McMaster Bujold
>
--
Jay McCarthy <j...@cs.byu.edu>
Assistant Professor / Brigham Young University
http://faculty.cs.byu.edu/~jay
"The glory of God is Intelligence" - D&C 93