+ let b = bool in
+ let y = yes in
+ let n = no in
+ match b with true -> y | false -> n
+
+ This almost works. For instance,
+
+ if true then 1 else 2;;
+
+ evaluates to 1, and
+
+ let b = true in let y = 1 in let n = 2 in
+ match b with true -> y | false -> n;;
+
+ also evaluates to 1. Likewise,
+
+ if false then 1 else 2;;
+
+ and
+
+ let b = false in let y = 1 in let n = 2 in
+ match b with true -> y | false -> n;;
+
+ both evaluate to 2.
+
+ However,
+
+ let rec blackhole x = blackhole x in
+ if true then blackhole else blackhole ();;
+
+ terminates, but
+
+ let rec blackhole x = blackhole x in
+ let b = true in
+ let y = blackhole in
+ let n = blackhole () in
+ match b with true -> y | false -> n;;
+
+ does not terminate. Incidentally, `match bool with true -> yes |
+ false -> no;;` works as desired, but your assignment is to solve it
+ without using the magical evaluation order properties of either `if`
+ or of `match`. That is, you must keep the `let` statements, though
+ you're allowed to adjust what `b`, `y`, and `n` get assigned to.
+
+ [[Hint assignment 5 problem 3]]
+
+Booleans, Church numbers, and Church lists in OCaml
+---------------------------------------------------
+
+(These questions adapted from web materials by Umut Acar. See <http://www.mpi-sws.org/~umut/>.)
+
+The idea is to get booleans, Church numbers, v3 lists, and
+binary trees working in OCaml.
+
+Recall from class System F, or the polymorphic λ-calculus.
+
+ τ ::= α | τ1 → τ2 | ∀α. τ
+ e ::= x | λx:τ. e | e1 e2 | Λα. e | e [τ ]
+
+Recall that bool may be encoded as follows:
+
+ bool := ∀α. α → α → α
+ true := Λα. λt:α. λf :α. t
+ false := Λα. λt:α. λf :α. f
+
+(where τ indicates the type of e1 and e2)
+
+Note that each of the following terms, when applied to the
+appropriate arguments, return a result of type bool.
+
+(a) the term not that takes an argument of type bool and computes its negation;
+(b) the term and that takes two arguments of type bool and computes their conjunction;
+(c) the term or that takes two arguments of type bool and computes their disjunction.
+
+The type nat (for "natural number") may be encoded as follows:
+
+ nat := ∀α. α → (α → α) → α
+ zero := Λα. λz:α. λs:α → α. z
+ succ := λn:nat. Λα. λz:α. λs:α → α. s (n [α] z s)
+
+A nat n is defined by what it can do, which is to compute a function iterated n times. In the polymorphic
+encoding above, the result of that iteration can be any type α, as long as you have a base element z : α and
+a function s : α → α.
+
+**Excercise**: get booleans and Church numbers working in OCaml,
+including OCaml versions of bool, true, false, zero, succ, add.
+
+Consider the following list type:
+
+ type ’a list = Nil | Cons of ’a * ’a list
+
+We can encode τ lists, lists of elements of type τ as follows:
+
+ τ list := ∀α. α → (τ → α → α) → α
+ nilτ := Λα. λn:α. λc:τ → α → α. n
+ makeListτ := λh:τ. λt:τ list. Λα. λn:α. λc:τ → α → α. c h (t [α] n c)