PL: Lecture #13  Tuesday, March 2nd
(text)

More Encodings

Our choice of encoding numbers makes sense — the idea is that the main feature of a natural number is repeating something a number of times. For booleans, the main property we’re looking for is choosing between two values. So we can encode true and false by functions of two arguments that return either the first or the second argument:

(define #t (lambda (x y) x))
(define #f (lambda (x y) y))

Note that this encoding of #f is really the same as the encoding of 0, so we have to know what type to expect an use the proper operations (this is similar to C, where everything is just integers). Now that we have these two, we can define if:

(define if (lambda (c t e) (c t e)))

it expects a boolean which is a function of two arguments, and passes it the two expressions. The #t boolean will simply return the first, and the #f boolean will return the second. Strictly speaking, we don’t really need this definition, since instead of writing (if c t e), we can simply write (c t e). In any case, we need the language to be lazy for this to work. To demonstrate this, we’ll intentionally use the quote back-door to use a non-functional value, using this will normally result in an error:

(+ '1 '2)

But testing our if definition, things work just fine:

(if #t (+ 4 5) (+ 1 2))

and we see that DrRacket leaves the second addition expression in red, which indicates that it was not executed. We can also make sure that even when it is defined as a function, it is still working fine because the language is lazy:

(if #f ((lambda (x) (x x)) (lambda (x) (x x))) 3)

What about and and or? Simple, or takes two arguments, and returns either true or false if one of the inputs is true:

(define or (lambda (a b) (if a #t (if b #t #f))))

but (if b #t #f) is really the same as just b because it must be a boolean (we cannot use more than one “truty” or “falsy” values):

(define or (lambda (a b) (if a #t b)))

also, if a is true, we want to return #t, but that is exactly the value of a, so:

(define or (lambda (a b) (if a a b)))

and finally, we can get rid of the if (which is actually breaking the if abstraction, if we encode booleans in some other way):

(define or (lambda (a b) (a a b)))

Similarly, you can convince yourself that the definition of and is:

(define and (lambda (a b) (a b a)))

Schlac has to-Racket conversion functions for booleans too:

(->bool (or #f #f))
(->bool (or #f #t))
(->bool (or #t #f))
(->bool (or #t #t))

and

(->bool (and #f #f))
(->bool (and #f #t))
(->bool (and #t #f))
(->bool (and #t #t))

A not function is quite simple — one alternative is to choose from true and false in the usual way:

(define not (lambda (a) (a #f #t)))

and another is to return a function that switches the inputs to an input boolean:

(define not (lambda (a) (lambda (x y) (a y x))))

which is the same as

(define not (lambda (a x y) (a y x)))

We can now put numbers and booleans together: we define a zero? function.

(define zero? (lambda (n) (n (lambda (x) #f) #t)))
(test (->bool (and (zero? 0) (not (zero? 3)))) => '#t)

(Good question: is this fast?)

(Note that it is better to test that the value is explicitly #t, if we just use (test (->bool ...)) then the test will work even if the expression in question evaluated to some bogus value.)

The idea is simple — if n is the encoding of zero, it will return it’s second argument which is #t:

(zero? 0) --> ((lambda (f n) n) (lambda (x) #f) #t) -> #t

if n is an encoding of a bigger number, then it is a self-composition, and the function that we give it is one that always returns #f, no matter how many times it is self-composed. Try 2 for example:

(zero? 2) --> ((lambda (f n) (f (f n))) (lambda (x) #f) #t)
          --> ((lambda (x) #f) ((lambda (x) #f) #t))
          --> #f

Now, how about an encoding for compound values? A minimal approach is what we use in Racket — a way to generate pairs (cons), and encode lists as chains of pairs with a special value at the end (null). There is a natural encoding for pairs that we have previously seen — a pair is a function that expects a selector, and will apply that on the two values:

(define cons (lambda (x y) (lambda (s) (s x y))))

Or, equivalently:

(define cons (lambda (x y s) (s x y)))

To extract the two values from a pair, we need to pass a selector that consumes two values and returns one of them. In our framework, this is exactly what the two boolean values do, so we get:

(define car (lambda (x) (x #t)))
(define cdr (lambda (x) (x #f)))

(->nat (+ (car (cons 2 3)) (cdr (cons 2 3))))

We can even do this:

(define 1st car)
(define 2nd (lambda (l) (car (cdr l))))
(define 3rd (lambda (l) (car (cdr (cdr l)))))
(define 4th (lambda (l) (car (cdr (cdr (cdr l))))))
(define 5th (lambda (l) (car (cdr (cdr (cdr (cdr l)))))))

or write a list-ref function:

(define list-ref (lambda (l n) (car (n cdr l))))

Note that we don’t need a recursive function for this: our encoding of natural numbers makes it easy to “iterate N times”. What we get with this encoding is essentially free natural-number recursion.

We now need a special null value to mark list ends. This value should have the same number of arguments as a cons value (one: a selector/boolean function), and it should be possible to distinguish it from other values. We choose

(define null (lambda (s) #t))

Testing the list encoding:

(define l123 (cons 1 (cons 2 (cons 3 null))))
(->nat (2nd l123))

And as with natural numbers and booleans, Schlac has built-in facility to convert encoded lists to Racket values, except that this requires specifying the type of values in a list so it’s a higher-order function:

((->listof ->nat) l123)

which (“as usual”) can be written as

(->listof ->nat l123)

We can even do this:

(->listof (->listof ->nat) (cons l123 (cons l123 null)))

Defining null? is now relatively easy (and it’s actually already used by the above ->listof conversion). The following definition

(define null? (lambda (x) (x (lambda (x y) #f))))

works because if x is null, then it simply ignores its argument and returns #t, and if it’s a pair, then it uses the input selector, which always returns #f in its turn. Using some arbitrary A and B:

(null? (cons A B))
  --> ((lambda (x) (x (lambda (x y) #f))) (lambda (s) (s A B)))
  --> ((lambda (s) (s A B)) (lambda (x y) #f))
  --> ((lambda (x y) #f) A B)
  --> #f
(null? null)
  --> ((lambda (x) (x (lambda (x y) #f))) (lambda (s) #t))
  --> ((lambda (s) #t) (lambda (x y) #f))
  --> #t

We can use the Y combinator to create recursive functions — we can even use the rewrite rules facility that Schlac contains (the same one that we have previously seen):

(define Y
  (lambda (f)
    ((lambda (x) (x x)) (lambda (x) (f (x x))))))
(rewrite (define/rec f E) => (define f (Y (lambda (f) E))))

and using it:

(define/rec length
  (lambda (l)
    (if (null? l)
      0
      (add1 (length (cdr l))))))
(->nat (length l123))

And to complete this, um, journey — we’re still missing subtraction. There are many ways to solve the problem of subtraction, and for a challenge try to come up with a solution yourself. One of the clearer solutions uses a simple idea — begin with a pair of two zeroes <0,0>, and repeat this transformation n times: <a,b> -> <b,b+1>. After n steps, we will have <n-1,n> — so we get:

(define inccons (lambda (p) (cons (cdr p) (add1 (cdr p)))))
(define sub1 (lambda (n) (car (n inccons (cons 0 0)))))
(->nat (sub1 5))

And from this the road is short to general subtraction, m-n is simply n applications of sub1 on m:

(define - (lambda (m n) (n sub1 m)))
(test (->nat (- 3 2)) => '1)
(test (->nat (- (* 4 (* 5 5)) 5)) => '95)

We now have a normal-looking language, and we’re ready to do anything we want. Here are two popular examples:

(define/rec fact
  (lambda (x)
    (if (zero? x) 1 (* x (fact (sub1 x))))))
(test (->nat (fact 5)) => '120)

(define/rec fib
  (lambda (x)
    (if (or (zero? x) (zero? (sub1 x)))
      1
      (+ (fib (- x 1)) (fib (- x 2))))))
(test (->nat (fib (* 5 2))) => '89)

To get generalized arithmetic capability, Schlac has yet another built-in facility for translating Racket natural numbers into Church numerals:

(->nat (fib (nat-> '10)))

… and to get to that frightening expression in the beginning, all you need to do is replace all definitions in the fib definition over and over again until you’re left with nothing but lambda expressions and applications, then reformat the result into some cute shape. For extra fun, you can look for immediate applications of lambda expressions and reduce them manually.

All of this is in the following code:

;; Making Schlac into a practical language (not an interpreter)

#lang pl schlac

(define identity (lambda (x) x))

;; Natural numbers

(define 0 (lambda (f x) x))

(define add1 (lambda (n) (lambda (f x) (f (n f x)))))
;; same as:
;;  (define add1 (lambda (n) (lambda (f x) (n f (f x)))))

(define 1 (add1 0))
(define 2 (add1 1))
(define 3 (add1 2))
(define 4 (add1 3))
(define 5 (add1 4))
(test (->nat (add1 (add1 5))) => '7)

(define + (lambda (m n) (m add1 n)))
(test (->nat (+ 4 5)) => '9)

;; (define * (lambda (m n) (m (+ n) 0)))
(define * (lambda (m n f) (m (n f))))
(test (->nat (* 4 5)) => '20)
(test (->nat (+ 4 (* (+ 2 5) 5))) => '39)

;; (define ^ (lambda (m n) (n (* m) 1)))
(define ^ (lambda (m n) (n m)))
(test (->nat (^ 3 4)) => '81)

;; Booleans

(define #t (lambda (x y) x))
(define #f (lambda (x y) y))

(define if (lambda (c t e) (c t e))) ; not really needed

(test (->nat (if #t 1 2)) => '1)
(test (->nat (if #t (+ 4 5) (+ '1 '2))) => '9)

(define and (lambda (a b) (a b a)))
(define or  (lambda (a b) (a a b)))
;; (define not (lambda (a) (a #f #t)))
(define not (lambda (a x y) (a y x)))

(test (->bool (and #f #f)) => '#f)
(test (->bool (and #t #f)) => '#f)
(test (->bool (and #f #t)) => '#f)
(test (->bool (and #t #t)) => '#t)
(test (->bool (or  #f #f)) => '#f)
(test (->bool (or  #t #f)) => '#t)
(test (->bool (or  #f #t)) => '#t)
(test (->bool (or  #t #t)) => '#t)
(test (->bool (not #f)) => '#t)
(test (->bool (not #t)) => '#f)

(define zero? (lambda (n) (n (lambda (x) #f) #t)))
(test (->bool (and (zero? 0) (not (zero? 3)))) => '#t)

;; Lists

(define cons (lambda (x y s) (s x y)))

(define car (lambda (x) (x #t)))
(define cdr (lambda (x) (x #f)))

(test (->nat (+ (car (cons 2 3)) (cdr (cons 2 3)))) => '5)

(define 1st car)
(define 2nd (lambda (l) (car (cdr l))))
(define 3rd (lambda (l) (car (cdr (cdr l)))))
(define 4th (lambda (l) (car (cdr (cdr (cdr l))))))
(define 5th (lambda (l) (car (cdr (cdr (cdr (cdr l)))))))

(define null (lambda (s) #t))
(define null? (lambda (x) (x (lambda (x y) #f))))

(define l123 (cons 1 (cons 2 (cons 3 null))))
;; Note that `->listof' is a H.O. converter
(test ((->listof ->nat) l123) => '(1 2 3))
(test (->listof ->nat l123) => '(1 2 3))  ; same as the above
(test (->listof (->listof ->nat) (cons l123 (cons l123 null)))
      => '((1 2 3) (1 2 3)))

;; Subtraction is tricky

(define inccons (lambda (p) (cons (cdr p) (add1 (cdr p)))))
(define sub1 (lambda (n) (car (n inccons (cons 0 0)))))

(test (->nat (sub1 5)) => '4)

(define - (lambda (a b) (b sub1 a)))

(test (->nat (- 3 2)) => '1)
(test (->nat (- (* 4 (* 5 5)) 5)) => '95)
(test (->nat (- 2 4)) => '0) ; this is "natural subtraction"

;; Recursive functions

(define Y
  (lambda (f)
    ((lambda (x) (x x)) (lambda (x) (f (x x))))))
(rewrite (define/rec f E) => (define f (Y (lambda (f) E))))

(define/rec length
  (lambda (l)
    (if (null? l)
      0
      (add1 (length (cdr l))))))
(test (->nat (length l123)) => '3)

(define/rec fact
  (lambda (x)
    (if (zero? x) 1 (* x (fact (sub1 x))))))
(test (->nat (fact 5)) => '120)

(define/rec fib
  (lambda (x)
    (if (or (zero? x) (zero? (sub1 x)))
      1
      (+ (fib (sub1 x)) (fib (sub1 (sub1 x)))))))
(test (->nat (fib (* 5 2))) => '89)

#|
;; Fully-expanded Fibonacci
(define fib
  ((lambda (f)
    ((lambda (x) (x x)) (lambda (x) (f (x x)))))
  (lambda (f)
    (lambda (x)
      ((lambda (c t e) (c t e))
        ((lambda (a b) (a a b))
        ((lambda (n)
            (n (lambda (x) (lambda (x y) y)) (lambda (x y) x)))
          x)
        ((lambda (n)
            (n (lambda (x) (lambda (x y) y)) (lambda (x y) x)))
          ((lambda (n)
            ((lambda (x) (x (lambda (x y) x)))
              (n (lambda (p)
                  ((lambda (x y s) (s x y))
                    ((lambda (x) (x (lambda (x y) y))) p)
                    ((lambda (n) (lambda (f x) (f (n f x))))
                    ((lambda (x) (x (lambda (x y) y))) p))))
                ((lambda (x y s) (s x y))
                  (lambda (f x) x)
                  (lambda (f x) x)))))
          x)))
        ((lambda (n) (lambda (f x) (f (n f x)))) (lambda (f x) x))
        ((lambda (x y)
          (x (lambda (n) (lambda (f x) (f (n f x)))) y))
        (f ((lambda (n)
              ((lambda (x) (x (lambda (x y) x)))
                (n (lambda (p)
                    ((lambda (x y s) (s x y))
                      ((lambda (x) (x (lambda (x y) y))) p)
                      ((lambda (n) (lambda (f x) (f (n f x))))
                      ((lambda (x) (x (lambda (x y) y))) p))))
                  ((lambda (x y s) (s x y))
                    (lambda (f x) x)
                    (lambda (f x) x)))))
            x))
        (f ((lambda (n)
              ((lambda (x) (x (lambda (x y) x)))
                (n (lambda (p)
                    ((lambda (x y s) (s x y))
                      ((lambda (x) (x (lambda (x y) y))) p)
                      ((lambda (n) (lambda (f x) (f (n f x))))
                      ((lambda (x) (x (lambda (x y) y))) p))))
                  ((lambda (x y s) (s x y))
                    (lambda (f x) x)
                    (lambda (f x) x)))))
            ((lambda (n)
                ((lambda (x) (x (lambda (x y) x)))
                (n (lambda (p)
                      ((lambda (x y s) (s x y))
                      ((lambda (x) (x (lambda (x y) y))) p)
                      ((lambda (n) (lambda (f x) (f (n f x))))
                        ((lambda (x) (x (lambda (x y) y))) p))))
                    ((lambda (x y s) (s x y))
                    (lambda (f x) x)
                    (lambda (f x) x)))))
              x)))))))))

;; The same after reducing all immediate function applications
(define fib
  ((lambda (f)
    ((lambda (x) (x x)) (lambda (x) (f (x x)))))
  (lambda (f)
    (lambda (x)
      (((x (lambda (x) (lambda (x y) y)) (lambda (x y) x))
        (x (lambda (x) (lambda (x y) y)) (lambda (x y) x))
        (((x (lambda (p)
                (lambda (s)
                  (s (p (lambda (x y) y))
                    (lambda (f x)
                      (f ((p (lambda (x y) y)) f x))))))
              (lambda (s)
                (s (lambda (f x) x) (lambda (f x) x))))
          (lambda (x y) x))
          (lambda (x) (lambda (x y) y))
          (lambda (x y) x)))
        (lambda (f x) (f x))
        ((f ((x (lambda (p)
                  (lambda (s)
                    (s (p (lambda (x y) y))
                      (lambda (f x)
                        (f ((p (lambda (x y) y)) f x))))))
                (lambda (y s)
                  (s (lambda (f x) x) (lambda (f x) x))))
            (lambda (x y) x)))
        (lambda (n) (lambda (f x) (f (n f x))))
        (f ((((x (lambda (p)
                    (lambda (s)
                      (s (p (lambda (x y) y))
                        (lambda (f x)
                          (f ((p (lambda (x y) y)) f x))))))
                  (lambda (s)
                    (s (lambda (f x) x) (lambda (f x) x))))
              (lambda (x y) x))
              (lambda (p)
                (lambda (s)
                  (s (p (lambda (x y) y))
                    (lambda (f x)
                      (f ((p (lambda (x y) y)) f x))))))
              (lambda (s)
                (s (lambda (f x) x) (lambda (f x) x))))
            (lambda (x y) x)))))))))

;; Cute reformatting of the above:
(define fib((lambda(f)((lambda(x)(x x))(lambda(x)(f(x x)))))(lambda(
f)(lambda(x)(((x(lambda(x)(lambda(x y)y))(lambda(x y)x))(x(lambda(x)
(lambda(x y)y))(lambda(x y) x))(((x(lambda(p)(lambda(s)(s(p(lambda(x
y)y))(lambda(f x)(f((p(lambda(x y)y))f x))))))(lambda(s) (s(lambda(f
x)x)(lambda(f x)x))))(lambda(x y)x))(lambda(x)(lambda(x y)y))(lambda
(x y)x)))(lambda(f x)(f x))((f((x(lambda(p)(lambda(s)(s(p(lambda(x y
)y))(lambda(f x)(f((p(lambda(x y)y))f x))))))(lambda(y s)(s(lambda(f
x)x)(lambda(f x)x))))(lambda(x y)x)))(lambda(n)(lambda(f x)(f(n f x)
)))(f((((x(lambda(p)(lambda(s)(s(p (lambda(x y)y))(lambda(f x)(f((p(
lambda(x y) y))f x))))))(lambda(s)(s(lambda(f x)x)(lambda(f x)x))))(
;;                      ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
;;                      `---------------(cons 0 0)---------------'
lambda(x y)x))(lambda(p)(lambda(s)(s(p(lambda(x y)y))(lambda(f x)(f(
(p(lambda(x y)y))f x))))))(lambda(s)(s(lambda(f x)x)(lambda(f x)x)))
)(lambda(x y)x)))))))))

And for extra fun:

        (λ(f)(λ
        (x)(((x(λ(
        x)(λ(x y)y)
        )(λ(x y)x))(
        x(λ(x)(λ(x y)
            y))(λ(x y
              )x))(((
                x(λ(p)(
                λ(s)(s
                (p (λ(
                x y)y))
                  (λ(f x
                  )(f((p(
                  λ(x y)
                  y))f x
                  ))))))(
                  λ(s)(s(
                  λ(f x)x)
                  (λ(f x)x)
                )))(λ(x y)
                x))(λ(x)(λ(
                x y)y)) (λ(
                x y) x)))(λ(
                f x)(f x))((f
              ((x(λ(p )(λ (s
              )(s(p(  λ(x y)
              y))(λ (  f x)(f(
              (p (λ(    x y)y)
            )f x)))    )))(λ(
            y s)(s    (λ (f x
            )x)(λ(      f x)x)
            )))(λ(      x y)x))
            )(λ(n)        (λ (f
          x)(f (n        f x)))
          )(f(((        (x(λ(p)
          (λ(s)(s          (p( λ(
          x y )y          ))(λ(f
        x) (f((          p(λ(x y
        )y)) f            x)))))
        )(λ(s)(            s(λ(f x
        )x)(λ(              f x)x)
        ))) (λ              (x y)x
      ))(λ(p                )(λ(s)(
      s(p(λ(                x y)y)
      )(λ (f                  x)(f((
      p(λ (x                  y)y)) f
    x))))))                  (λ(s)(
    s(λ (f                    x)x)(λ
    (f x)x)                    )))(λ(
    x y)x)                      ))))))

|#

Alternative Church Encoding

Finally, note that this is just one way to encode things — other encodings are possible. One alternative encoding is in the following code — it uses a list of N falses as the encoding for N. This encoding makes it easier to add1 (just cons another #f), and to sub1 (simply cdr). The tradeoff is that some arithmetics operations becomes more complicated, for example, the definition of + requires the fixpoint combinator. (As expected, some people want to see what can we do with a language without recursion, so they don’t like jumping to Y too fast.)

;; An alternative "Church" encoding: use lists to encode numbers

#lang pl schlac

(define identity (lambda (x) x))

;; Booleans (same as before)

(define #t (lambda (x y) x))
(define #f (lambda (x y) y))

(define if (lambda (c t e) (c t e))) ; not really needed

(test (->bool (if #t #f #t))
      => '#f)
(test (->bool (if #f ((lambda (x) (x x)) (lambda (x) (x x))) #t))
      => '#t)

(define and (lambda (a b) (a b a)))
(define or  (lambda (a b) (a a b)))
(define not (lambda (a x y) (a y x)))

(test (->bool (and #f #f)) => '#f)
(test (->bool (and #t #f)) => '#f)
(test (->bool (and #f #t)) => '#f)
(test (->bool (and #t #t)) => '#t)
(test (->bool (or  #f #f)) => '#f)
(test (->bool (or  #t #f)) => '#t)
(test (->bool (or  #f #t)) => '#t)
(test (->bool (or  #t #t)) => '#t)
(test (->bool (not #f)) => '#t)
(test (->bool (not #t)) => '#f)

;; Lists (same as before)

(define cons (lambda (x y s) (s x y)))
(define car (lambda (x) (x #t)))
(define cdr (lambda (x) (x #f)))

(define 1st car)
(define 2nd (lambda (l) (car (cdr l))))
(define 3rd (lambda (l) (car (cdr (cdr l)))))
(define 4th (lambda (l) (car (cdr (cdr (cdr l))))))
(define 5th (lambda (l) (car (cdr (cdr (cdr (cdr l)))))))

(define null (lambda (s) #t))
(define null? (lambda (x) (x (lambda (x y) #f))))

;; Natural numbers (alternate encoding)

(define 0 identity)

(define add1 (lambda (n) (cons #f n)))

(define zero? car) ; tricky

(define sub1 cdr) ; this becomes very simple

;; Note that we could have used something more straightforward:
;; (define 0 null)
;; (define add1 (lambda (n) (cons #t n))) ; cons anything
;; (define zero? null?)
;; (define sub1 (lambda (l) (if (zero? l) l (cdr l))))

(define 1 (add1 0))
(define 2 (add1 1))
(define 3 (add1 2))
(define 4 (add1 3))
(define 5 (add1 4))

(test (->nat* (add1 (add1 5))) => '7)
(test (->nat* (sub1 (sub1 (add1 (add1 5))))) => '5)
(test (->bool (and (zero? 0) (not (zero? 3)))) => '#t)
(test (->bool (zero? (sub1 (sub1 (sub1 3))))) => '#t)

;; list-of-numbers tests
(define l123 (cons 1 (cons 2 (cons 3 null))))
(test (->listof ->nat* l123) => '(1 2 3))
(test (->listof (->listof ->nat*) (cons l123 (cons l123 null)))
      => '((1 2 3) (1 2 3)))

;; Recursive functions

(define Y
  (lambda (f)
    ((lambda (x) (x x)) (lambda (x) (f (x x))))))
(rewrite (define/rec f E) => (define f (Y (lambda (f) E))))

;; note that this example is doing something silly now
(define/rec length
  (lambda (l)
    (if (null? l)
      0
      (add1 (length (cdr l))))))
(test (->nat* (length l123)) => '3)

;; addition becomes hard since it requires a recursive definition
;; (define/rec +
;;  (lambda (m n) (if (zero? n) m (+ (add1 m) (sub1 n)))))
;; (test (->nat* (+ 4 5)) => '9)

;; faster alternative:
(define/rec +
  (lambda (m n)
    (if (zero? m) n
        (if (zero? n) m
            (add1 (add1 (+ (sub1 m) (sub1 n))))))))
(test (->nat* (+ 4 5)) => '9)

;; subtraction is similar to addition
;; (define/rec -
;;  (lambda (m n) (if (zero? n) m (- (sub1 m) (sub1 n)))))
;; (test (->nat* (- (+ 4 5) 4)) => '5)
;; but this is not "natural subtraction": doesn't work when n>m,
;; because (sub1 0) does not return 0.

;; a solution is like alternative form of +:
(define/rec -
  (lambda (m n)
    (if (zero? m) 0
        (if (zero? n) m
            (- (sub1 m) (sub1 n))))))
(test (->nat* (- (+ 4 5) 4)) => '5)
(test (->nat* (- 2 5)) => '0)
;; alternatively, could change sub1 above:
;; (define sub1 (lambda (n) (if (zero? n) n (cdr n))))

;; we can do multiplication in a similar way
(define/rec *
  (lambda (m n)
    (if (zero? m) 0
        (+ n (* (sub1 m) n)))))
(test (->nat* (* 4 5)) => '20)
(test (->nat* (+ 4 (* (+ 2 5) 5))) => '39)

;; and the rest of the examples

(define/rec fact
  (lambda (x)
    (if (zero? x) 1 (* x (fact (sub1 x))))))
(test (->nat* (fact 5)) => '120)

(define/rec fib
  (lambda (x)
    (if (or (zero? x) (zero? (sub1 x)))
      1
      (+ (fib (sub1 x)) (fib (sub1 (sub1 x)))))))
(test (->nat* (fib (* 5 2))) => '89)

#|
;; Fully-expanded Fibonacci (note: much shorter than the previous
;; encoding, but see how Y appears twice -- two "((lambda" pairs)
(define fib((lambda(f)((lambda(x)(x x))(lambda(x)(f(x x)))))(lambda(
f)(lambda(x)(((((x(lambda(x y)x))(x(lambda(x y)x)))((x(lambda(x y)y)
)(lambda(x y)x)))(lambda(s)(s(lambda(x y)y)(lambda(x)x))))((((lambda
(f)((lambda(x)(x x))(lambda(x)(f(x x))))) (lambda(f)(lambda(m n)((m(
lambda(x y)x))n (((n(lambda(x y)x)) m)(lambda(s)((s (lambda(x y)y))(
lambda(s)((s (lambda(x y)y))((f(m(lambda(x y)y)))(n(lambda(x y)y))))
))))))))(f(x(lambda(x y)y))))(f((x(lambda(x y)y))(lambda(x y)y))))))
)))
|#