On Fri, Dec 30, 2011 at 9:43 AM, Gregg Reynolds <d...@mobileink.com> wrote:
> On Dec 30, 2011, at 11:20 AM, Colin Adams wrote: > On 30 December 2011 17:17, Gregg Reynolds <d...@mobileink.com> wrote: > > On Dec 30, 2011, at 11:04 AM, Colin Adams wrote: >> On 30 December 2011 16:59, Gregg Reynolds <d...@mobileink.com> wrote: >> >> On Fri, Dec 30, 2011 at 12:49 AM, Heinrich Apfelmus < >>> apfel...@quantentunnel.de> wrote: >>> >>> The function >>>> >>>> f :: Int -> IO Int >>>> f x = getAnIntFromTheUser >>= \i -> return (i+x) >>>> >>>> is pure according to the common definition of "pure" in the context of >>>> purely functional programming. That's because >>>> >>>> f 42 = f (43-1) = etc. >>>> >>>> Conclusion: f 42 != f 42 >>> >>> (This seems so extraordinarily obvious that maybe Heinrich has something >>> else in mind.) >>> >>> This seems such an obviously incorrect conclusion. >> >> f42 is a funtion for returning a program for returning an int, not a >> function for returning an int. >> >> >> My conclusion holds: f 42 != f 42. Obviously, so I won't burden you >> with an explanation. ;) >> >> -Gregg >> > Your conclusion is clearly erroneous. > > proof: f is a function, and it is taking the same argument each time. > Therefore the result is the same each time. > > > That's called begging the question. f is not a function, so I guess your > proof is flawed. > > It seems pretty clear that we're working with different ideas of what > constitutes a function. When I use the term, I intend what I take to be > the standard notion of a function in computation: not just a unique mapping > from one input to one output, but one where the output is computable from > the input. Any "function" that depends on a non-computable component is by > that definition not a true function. For clarity let's call such critters > quasi-functions, so we can retain the notion of application. Equality > cannot be defined for quasi-functions, for obvious reasons. > > f is a quasi-function because it depends on getAnIntFromUser, which is not > definable and is obviously not a function. When applied to an argument > like 42, it yields another quasi-function, and therefore "f 42 = f 42" is > false, or at least unknown, and the same goes for f 42 != f 42 I suppose. > > -Gregg > Please don't redefine "function" to mean "computable function". Besides distancing yourself from math, I don't think doing so really helps your case. And on what do you base your claim that getAnIntFromUser is not definable? Or that applying it (what?) to 42 gives a quasi-function?
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