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Functionalize this function by using fold

I am looking to rewrite this function using foldr (in the spirit of this question).
I am new to functional programming and don’t know how to work with foldr. I am hoping for a code review and a general direction.
import Control.Monad

append :: a -> [a] -> [a]
append x [] = x : []
append x (y:ys) = (y : x) : append x ys

appendUndup :: [a] -> [a]
appendUndup [] = []
appendUndup (x:xs) = (x:xs’) : xs”
(xs’,xs”) = (append x xs, xs)
(y:ys’,ys”) = (append y ys’, ys)


You want to use foldr instead of foldl’ which is for lists. It takes a function with 2 arguments and a starting state:
a `foldr` f z = f a z

There are a few things to note about foldr:

Lines of equalities are not permitted; foldr must be written as a function definition.
A helper function is supplied that you can use to accumulate items into a tuple. This function takes an arbitrary number of arguments, each of which is the output of the function being applied.
You need to specify the starting state for the function. The starting state is a placeholder for any number of intermediate results. The state is what the function operates on and gets back. You don’t have to specify the state in the helper function.
The output of the function must be the same as the input. So each function iteration produces the same number of items as the previous iteration.

There is a function called flip that reverses a function. It takes a function of 2 arguments and flips the first argument of the function (and returns the result of the flipped function). Here is an example:
flip id :: a -> b -> a
flip 384a16bd22

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