Re: [Haskell-beginners] Simple function comparison

25 Nov 2015

      On Wed, Nov 25, 2015 at 03:48:08PM +0000, MJ Williams wrote:
...
[snip]
...
I don't think there is a way to /prove/ f (g a) == g (f a) if their domain
is not finite inside Haskell (you could do it with pen and paper).
[snip]
        Just out of interest, could you demonstrate the proof without a
finite domain?
Sincerely, Matthew
Say we have:

    f(x) = x + 1
    g(x) = x + 7

Then we can substitute:

    f(g(x))
    f(x+7)
    (x+7) + 1
    x+8

and

    g(f(x))
    g(x+1)
    (x+1) + 7
    x+8

which shows f(g(x)) = g(f(x))

For anything more, my lawyer suggested I say: I have discovered a
truly marvellous proof for it, but the margin of this email is to
narrow to contain it. :P

Re: [Haskell-beginners] Simple function comparison

Francesco Ariis