[Haskell-cafe] Re: A non-inductive Haskell proof?

[Heinrich Apfelmus]

R J wrote:
> The following theorem is obviously true, but how is it proved (most cleanly and simply)
> in Haskell?
> 
> Theorem:  (nondecreasing xs) => nondecreasing (insert x xs), where:
> 
>    nondecreasing               :: (Ord a) => [a] -> Bool
>    nondecreasing []            =  True
>    nondecreasing xxs@(x : xs)  =  and [a <= b | (a, b) <- zip xxs xs]
> 
>    insert                      :: (Ord a) => a -> [a] -> [a]
>    insert x xs                 =  takeWhile (<= x) xs ++ [x] ++ dropWhile (<= x) xs

Since  insert  involves list concatenation at the outermost level, the
first step is to prove a lemma along the lines of

     nondecreasing xs && nondecreasing ys && (last xs <= head ys)
 =>
     nondecreasing (xs ++ ys)

from which the wanted theorem follows immediately. The lemma itself is
proved readily by noting/proving

      and (xs ++ ys)
    = and xs && and ys

      zip (xs ++ ys) (tail (xs ++ ys))
   ~= zip xs (tail xs) ++ [(last xs, head ys)] ++ zip ys (tail ys)


Regards,
apfelmus

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