Functional proofs (Haskell)

Question

I failed at reading RWH; and not one to quit, I ordered Haskell: The Craft of Functional Programming. Now I'm curious about these functional proofs on page 146. Specifically I'm trying to prove 8.5.1 sum (reverse xs) = sum xs. I can do some of the induction proof but then I get stuck..

HYP:

sum ( reverse xs ) = sum xs

BASE:

sum ( reverse [] ) = sum []

Left  = sum ( [] ) (reverse.1)
      = 0          (sum.1)

Right = 0          (sum.1)

INDUCTION:

sum ( reverse (x:xs) ) = sum (x:xs) 

Left = sum ( reverse xs ++ [x] )    (reverse.2)

Right = sum (x:xs)   
      = x + sum xs                  (sum.2)

So now I'm just trying ot prove that Left sum ( reverse xs ++ [x] ) is equal to Right x + sum xs, but that isn't too far off from where I started sum ( reverse (x:xs) ) = sum (x:xs).

I'm not quite sure why this needs to be proved, it seems totally reasonable to use the symbolic proof of reverse x:y:z = z:y:x (by defn), and because + is commutative (arth) then reverse 1+2+3 = 3+2+1,

Answer 1

sum (reverse [])     = sum []                     -- def reverse
sum (reverse (x:xs)) = sum (reverse xs ++ [x])    -- def reverse
                     = sum (reverse xs) + sum [x] -- sum lemma below
                     = sum (reverse xs) + x       -- def sum
                     = x + sum (reverse xs)       -- commutativity assumption!
                     = x + sum xs                 -- inductive hypothesis
                     = sum (x:xs)                 -- definition of sum

However, there are underlying assumptions of associativity and commutativity that are not strictly warranted and this will not work properly for a number of numerical types such as Float and Double where those assumptions are violated.

Lemma: sum (xs ++ ys) == sum xs + sum ys given the associativity of (+)

Proof:

sum ([] ++ ys)     = sum ys           -- def (++)
                   = 0 + sum ys       -- identity of addition
                   = sum [] ++ sum ys -- def sum

sum ((x:xs) ++ ys) = sum (x : (xs ++ ys))  -- def (++)
                   = x + sum (xs ++ ys)    -- def sum 
                   = x + (sum xs + sum ys) -- inductive hypothesis
                   = (x + sum xs) + sum ys -- associativity assumption!
                   = sum (x:xs) + sum ys   -- def sum

Answer 2

Basically you need to show that

sum (reverse xs ++ [x]) = sum (reverse xs) + sum [x]

which then easily leads to

                        = x + sum (reverse xs)
                        = x + sum xs  -- by inductive hyp.

The problem is to show that sum distributes over list concatenation.

Answer 3

Use the definition of a sum to break up (sum reverse xs ++[x]) into x + sum(reverse(xs)), and using your inductive hypothesis you know sum(reverse(xs)) = sum(xs). But I agree, induction is overkill for a problem like this.