MonadFix in F#

mfix.fsx

let force (value : Lazy<_>) = value.Force()

let fix : (Lazy<'T> -> 'T) -> Lazy<'T> = fun f -> 
    let rec x = lazy (f x) in x

// Maybe MonadFix
let mfix : (Lazy<'T> -> option<Lazy<'T>>) -> Lazy<option<Lazy<'T>>> = fun f -> 
    let rec x = lazy ( match force x with | Some v -> f v | None -> None ) in x

Hmm... Looking at this more closely, isn't the body of mfix above equivalent to

let rec x = lazy (force x) in x

and undefined?

Hmmm maybe you are correct...
I was trying to mimic the Haskell definition

mfix :: (α → Maybe α) → Maybe α
mfix f = let x = (f · unJust) x in x
where unJust (Just x ) = x

According to an old snippet of mine
http://www.fssnip.net/36
the trick is to push force inside the f

Maybe something like this ... ???

mfix :: (α → Maybe α) → Maybe α
mfix f = let x = (f · unJust) x in x
where unJust (Just x ) = x

let mfix : (Lazy<'T> -> option<'T>) -> Lazy<option<'T>> = fun f -> 
    let rec x = lazy ( f (lazy (Option.get (force x) )) ) in x

I think maybe option<'T> is not a good example for mfix and something more lazy like

let mfix : (Lazy<'T> -> seq<'T>) -> Lazy<seq<'T>>

is more easily demonstrable