fold on Monoids

I was looking at Wikipedia’s article regarding Monoids, and decided that it was interesting to make a post regarding the section “Monoids in Computer Science”

So there they have posted this formula:

And let’s also take a look at Haskell’s definition for fold:

(>>>) = (++) 
fold []     = []
fold (m:l') = m >>> (fold l')

Notice the relation between the formulas? Just replace >>> with *, and there you have it!

Regarding the types, in Haskell we have:

*Main> :t fold
fold :: [[a]] -> [a]

And in Mathematics, we have:
fold: M* -> M

The star here represents Kleene’s star, and it is defined to be all concatenations on the elements of the set M, including the empty string (). Example, if M = {a, b} then M* = {(), (a), (b), (a,a), (a,b), (b,b), (b,a), (a,b,a), …}. So what this means is, that if we have M defined as list of integers, then M* would be list of list of integers (which is the result of list concatenation of the elements of M).

And what is fold? Well, fold basically takes a Monoid and applies its operation. Now that we have these two formulas, we can show how fold operates on a Monoid.

Let’s consider the Monoid M ([Int], ++, []). First, to prove that this triple makes a monoid, we must first take a look at the definition of ++:

[]     ++ ys = ys
(x:xs) ++ ys = x : (xs ++ ys) (*)

1. Closure
Namely, the (++) is the list concatenation operator, and when we concatenate two lists, we get list as an output, according to ++ definition.

2. Associativity
(xs ++ ys) ++ zs = xs ++ (ys ++ zs), order of evaluation will not change the resulting list
To prove this, we again need to look at ++ definition.

We can use Structural Induction in order to prove associativity.

Base case: If we let x = ys + zs, then [] ++ x = x by definition, LHS equals to x. and for the RHS, we get “” + ys = ys by definition, so ys ++ zs == x, and so LHS = RHS.

Inductive step: Assume that (xs ++ ys) ++ zs = xs ++ (ys ++ zs)

We need to prove that ((x:xs) ++ ys) ++ zs = (x:xs) ++ (ys ++ zs)

LHS: by (*) = (x : (xs ++ ys)) ++ zs => let xs ++ ys = p => x : p ++ zs = x : ((xs ++ ys) ++ zs)
RHS: by (*) = x : (xs ++ (ys ++ zs)) => by assumption => x : ((xs ++ ys) ++ zs)

So we get that LHS = RHS.

3. Identity
The empty list [] is the identity here, because [] ++ ys or ys ++ [] always equals to ys, according to ++ definition (already proved by base case in 2).

Now, we can start playing around with our fold:

*Main> fold [[1,2], [3]] -- per Monoid M
*Main> fold [['a'],['b'],['c']] -- same as Monoid M, but operates on set [Char] instead of [Int]

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