On 8 Dec 2008, at 23:15, Joachim Breitner wrote:
Am Montag, den 08.12.2008, 15:59 -0600 schrieb Nathan Bloomfield:
Slightly off topic, but the A^B notation for hom-sets also makes the
natural isomorphism we call currying expressable as A^(BxC) = (A^B)
^C.
So A^(B+C) = A^B × A^C ?
Oh, right, I guess that’s actually true:...
I posted some of those relations for lambda-calculus two days ago
(this thread).
It is so very off-topic, because one can reverse the process, take
those operators and some relations, and show it contains the lambda-
calculus (or so I remember, don't recall details). Then one problem
is that these operators are not so intuitive for all practical purposes.
Hans
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