How to set up an inductive proof in haskell?

Question

I need to prove

f (g xs) == g (f xs)

whenver xs is a finite list of Ints.

Assume both f and g are of type [Int]->[Int]

This is for specific `f` and `g`, right? (For arbitrary `f` and `g` it does not hold.) — Joachim Breitner, Apr 24 '17 at 22:47
They are both of type [Int]->[Int] and both take a finite integer list. — user3358850, Apr 24 '17 at 22:50

amalloy · Answer 1 · 2017-04-24T23:05:39.940

5

Disproof by counterexample:

f xs = []
g xs = [1]

If you want this property to hold, you need more specific constraints on what f and g are.

You may be thinking of the law that

(map f . map g) == map (f . g)

which can indeed be proven.

edited Apr 24 '17 at 23:05

answered Apr 24 '17 at 22:55

amalloy

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score -1 · Answer 2 · answered Apr 25 '17 at 00:16

Found this from somewhere

Theorem: (map f . map g) xs = map (f . g) xs
       - Proof is by induction on xs
    *** Base Case, xs = []
        - Left side: (map f . map g) [] = map f (map g []) = map f [] = []
        - Right side: map (f . g) [] = []
    *** Inductive Case, xs = k:ks
        - Inductive Hypothesis: (map f . map g) ks = map (f . g) ks
        - Left Side
          + (map f . map g) xs
          + map f (map g (k:ks))
          + map f ((g k) : (map g ks))
          + (f (g k)) : (map f (map g ks))
          + ((f . g) k) : ((map f . map g) ks)  i.e. change from bracket form back to point form
          + ((f . g) k) : (map (f . g) ks) by inductive hypothesis
          + map (f . g) (k:ks) by definition of map
          + map (f . g) xs

This is very different than saying `f (g xs) == g (f xs)` which, as others have pointed out, is false in general. — David Young, Apr 25 '17 at 00:39

How to set up an inductive proof in haskell?

2 Answers2