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-Hints for reverse.
-
-*	If left and right are two v3 lists, what is the result of:
-
-		left make_list right
-
-*	What is `reverse []`? What is `reverse [c]`? What is `reverse [b;c]`? How is it related to `reverse [c]`?
-
-*	How is `reverse [a;b;c]` related to `reverse [b;c]`?
-
-*	Getting any ideas?
-
-
-Another strategy.
-
-*	Our version 3 lists are _right_-folds. That is, `[a;b;c]` is implemented as:
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-		\f z. f a (f b (f c z))
-
-	which is the result of first combining the base value `z` with the rightmost element of the list, using `f`, then combining the result of that with the next element to the left, and so on.
-
-	A _left_-fold on the same list would be:
-
-		\f z. f (f (f z a) b) c
-
-	which is the result of first combining the base value `z` with the leftmost element of the list, then combining the result of that with the next element to the right, and so on.
-
-	It's conventional for `f` to take the accumulated value so far as its second argument when doing a right-fold, and for it to take it as its first argument when doing a left-fold. However, this convention could be ignored. We could also call this the left-fold of `[a;b;c]`:
-
-		\f z. f c (f b (f a z))
-
-*	Getting any ideas?
-
-*	Our `make_list` function for taking an existing, right-fold-based list, and a new element, and returning a new right-fold-based list, looks like this:
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-		let make_list = \hd tl. \f z. f hd (tl f z)
-
-	How would you write a `make_left_list` function, that takes an existing, left-fold-based list, like `\f z. f c (f b z)`, and a new element, `a`, and returned the new, left-fold based list:
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-		\f z. f c (f b (f a z))
-
-