On Mon, 2008-12-08 at 23:15 +0100, Joachim Breitner wrote:
Hi,
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
On 15 Dec 2008, at 12:52 pm, Derek Elkins wrote:
I want to point out a quick categorical way of proving this (and
almost
all the other arithmetic laws follow similarly.) This is just
continuity of right adjoints. The interesting thing is the
adjunction,
one that is commonly neglected in
On Mon, Dec 8, 2008 at 23:10, Dan Piponi [EMAIL PROTECTED] wrote:
More generally, all of Tarski's high school algebra axioms carry
over to types. You can see the axioms here:
http://math.bu.edu/people/kayeats/papers/saga_paper4.ps That proves
type theory is child's play :-)
Ah, the power of a
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,
Don Stewart wrote:
Which suggests that $ was already in the 1.0 report going to SIGPLAN.
Perhaps Paul or Simon could shed light on it? Anyone have the 1.0 report
lying around to check if it was in earlier?
As far as Haskell is concerned, the first report-ed occurrence
of the $ operator was in
On Mon, 2008-12-08 at 09:58 -0800, Mark P. Jones wrote:
Don Stewart wrote:
Which suggests that $ was already in the 1.0 report going to SIGPLAN.
Perhaps Paul or Simon could shed light on it? Anyone have the 1.0 report
lying around to check if it was in earlier?
As far as Haskell is
On Sun, Dec 7, 2008 at 2:05 AM, Hans Aberg [EMAIL PROTECTED] wrote:
As for the operator itself, it appears in Alonzo Church, The Calculi of
Lambda-Conversion, where it is written as exponentiation, like x^f
That's reminiscent of the notation in Lambek and Scott where (roughly
speaking) the
In set theory, and sometimes in category theory, A^B is just another
notation for Hom(B, A), and the latter might be given the alternate notation
B - A. And th reason is that for finite sets, computing cardinalities
result in the usual power function of natural numbers - same as Church,
Hi,
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:
uncurry either
2008/12/8 Joachim Breitner [EMAIL PROTECTED]:
So A^(B+C) = A^B × A^C ?
That's part of the basis for Hinze's paper on memoization:
http://www.informatik.uni-bonn.de/~ralf/publications/WGP00b.ps.gz
It's always nice to see that I havn't learned the elementary power
calculation rules for nothing
On Sun, Dec 7, 2008 at 3:05 AM, Hans Aberg [EMAIL PROTECTED] wrote:
One can define operators
a ^ b := b(a) -- Application in inverse.
(a * b)(x) := b(a(x)) -- Function composition in inverse.
(a + b)(x) := a(x) * b(x)
O(x) := I -- Constant function returning
On 7 Dec 2008, at 11:34, Luke Palmer wrote:
On Sun, Dec 7, 2008 at 3:05 AM, Hans Aberg [EMAIL PROTECTED] wrote:
One can define operators
a ^ b := b(a) -- Application in inverse.
(a * b)(x) := b(a(x)) -- Function composition in inverse.
(a + b)(x) := a(x) * b(x)
O(x) := I
On 7 Dec 2008, at 04:30, George Pollard wrote:
This is a little bit random, but I was just wondering if anyone knew
where the $ low-precedence parenthesis-eliminating application
operator
originated. The Haskell Report doesn't mention anything, and I can't
search for $ on Google. So... who
Hello Haskell-Café:
This is a little bit random, but I was just wondering if anyone knew
where the $ low-precedence parenthesis-eliminating application operator
originated. The Haskell Report doesn't mention anything, and I can't
search for $ on Google. So... who thought it up? Does it originate
porges:
Hello Haskell-Café:
This is a little bit random, but I was just wondering if anyone knew
where the $ low-precedence parenthesis-eliminating application operator
originated. The Haskell Report doesn't mention anything, and I can't
search for $ on Google. So... who thought it up? Does
Paul Hudak [EMAIL PROTECTED] writes:
As for x:xs, the xs is meant to be the plural of x, and is
pronounced exs (I guess...).
Similarly, n:ns is one n followed by many more ens. Make sense?
I think this convention is often used in the Prolog community as well,
as in X|Xs.
--
Doug Quale
Pattern matching goes back to Burstall and Darlington's work in the 1970's.
As for x:xs, the xs is meant to be the plural of x, and is
pronounced exs (I guess...).
Similarly, n:ns is one n followed by many more ens. Make sense?
(By the way, : is often pronounced followed by.)
-Paul
Quoth Toby Hutton, nevermore:
Does anyone know why (x:xs)? Is xs meant to be a synonym for 'excess'?
I've seen it said somewhere (possibly Hudak's _Haskell School of
Expression_) that xs should be read as the plural of x. Although
personally I always gravitate towards the excess notion,
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