[racket] missing solution 20.1.2 ex:syn-funcs

On Mon, Sep 8, 2014 at 9:51 PM,
Matthias Felleisen <matthias at ccs.neu.edu> wrote:

On Sep 2, 2014, at 12:05 PM, Daniel Bastos wrote:
>
> > Exercise 20.1.2. Argue why the following sentences are legal
> > definitions:
> >
> > (define (f x) (x 10))
> >
> > (define (f x) f)
> >
> > (define (f x y) (x 'a y 'b))
> >
> > Solution. The relevant part of the grammar is the following.
> >
> >  <def> = (define (<var> <var> ...<var>) <exp>)
> >        | (define <var> <exp>)
> >        | (define-struct <var> (<var> <var> ...<var>))
> >
> > (*) First definition
> >
> > The LHS is a list of <var>, since we find f and x as members of the
> > list and they're both variables. The RHS is a list consisting of a
> > <var> and a <num>.
>
> Small correction. Let's not call (x 10) a list. It's an application. '(x
> 10) would be a list and this one character is critical.
>

Indeed. Here's a full rewrite for completeness. (Thanks!)

Exercise 20.1.2. Argue why the following sentences are legal
definitions:

(define (f x) (x 10))

(define (f x) f)

(define (f x y) (x 'a y 'b))

Solution. The relevant part of the grammar is the following.

  <def> = (define (<var> <var> ...<var>) <exp>)
        | (define <var> <exp>)
        | (define-struct <var> (<var> <var> ...<var>))

(*) First definition

The LHS is a list of <var>, since we find f and x as members of the
list and they're both variables. The RHS is an application of the form
(<var> <num>), which makes up an <exp>, satisfying the first form
of <def>. In other words,

  (define (<var> <var>) (<var> <num)) =
  (define (<var> <var>) (<exp> <exp>) =
  (define (<var> <var>) <exp>).

Therefore it's a legal <def>.

(*) Second definition

Same LHS as the previous, so we need only check the RHS which is a
<var>. <var> is a valid form of <exp>, so we have

  (define (<var> <var>) <var>) =
  (define (<var> <var>) <exp>).

Therefore it's a legal <def>.

(*) Third definition

The LHS is (<var> <var> <var>), while the RHS is (<var> <sym> <var>
<sym>), so we have

  (define (<var> <var> <var>) (<var> <sym> <var> <sym>)) =
  (define (<var> <var> <var>) (<exp> <exp> <exp> <exp>)) =
  (define (<var> <var> <var>) <exp>).

Therefore it's a legal <def>.
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