9 Dec
2008
9 Dec
'08
4:38 p.m.
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