[Haskell-cafe] Proof of induction case of simple foldl identity.
Daniel Fischer
daniel.is.fischer at web.de
Sat Mar 14 17:25:21 EDT 2009
Am Samstag, 14. März 2009 21:45 schrieb R J:
> Can someone provide the induction-case proof of the following identity:
>
> foldl (-) ((-) x y) ys = (foldl (-) x ys) - y
>
> If foldl is defined as usual:
>
> foldl :: (b -> a -> b) -> b -> [a] -> b
> foldl f e [] = e
(R) foldl f e (x : xs) = foldl f (f e x) xs
>
> The first two cases, _|_ and [], are trivial.
>
> Here's my best attempt at the (y : ys) case (the left and right sides
> reduce to expressions that are obviously equal, but I don't know how to
> show it):
>
> Case (y : ys).
>
> Left-side reduction:
>
> foldl (-) ((-) x y) (y : ys)
> = {second equation of "foldl"}
> foldl (-) ((-) ((-) x y) y) ys
> = {notation}
> foldl (-) ((-) (x - y) y) ys
> = {notation}
> foldl (-) ((x - y) - y) ys
> = {arithmetic}
> foldl (-) (x - 2 * y) ys
>
> Right-side reduction:
>
> (foldl (-) x (y : ys)) - y
> = {second equation of "foldl"}
> (foldl (-) ((-) x y) ys) - y
> = {induction hypothesis: foldl (-) ((-) x y) ys = (foldl (-) x ys) -
> y} ((foldl (-) x ys) - y) - y
> = {arithmetic}
> (foldl (-) x ys) - 2 * y
>
> Thanks as always.
>
Consider a one-element list.
foldl (-) (x-y) [z] = (x-y)-z
(foldl (-) x [z]) - y = (x-z)-y
So a necessary condition for the identity to generally hold is that the Num
instance obeys the law
(L) forall u v w. (u - v) - w = (u - v) - w
Then take as inductive hypothesis that zs is a list such that
forall a b. foldl (-) (a-b) zs = (foldl (-) a zs) - b
Let z be an arbitrary value of appropriate type, x and y too.
Then
foldl (-) (x - y) (z:zs)
= foldl (-) ((x-y)-z) zs (R)
= (foldl (-) (x-y) zs) - z (IH, a = x-y, b = z)
= ((foldl (-) x zs) - y) - z (IH, a = x, b = y)
= ((foldl (-) x zs) - z) - y (L, u = foldl (-) x zs, v = y, w = z)
= (foldl (-) (x-z) zs) - y (IH, a = x, b = z)
= (foldl (-) x (z:zs)) - y (R)
The trick is to formulate the inductive hypothesis with enough generality, so
you have strong foundations to build on.
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