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.