[Haskell-beginners] Simple function comparison
Francesco Ariis
fa-ml at ariis.it
Wed Nov 25 16:22:50 UTC 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
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