[Haskell-cafe] Proof of induction case of simple foldl identity.

[Dean Herington]

At 8:45 PM +0000 3/14/09, R J wrote:
>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
>    foldl f e (x : xs)     =  myFoldl 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).

Careful.  Your introduced variables y and ys already appear in the 
equation to be proved.  Try introducing fresh variables.

>
>       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.
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