SICP Exercise 4.15 halts?

Exercise 4.15.  Given a one-argument procedure p and an object a, p is said to ``halt'' on a if evaluating the expression (p a) returns a value (as opposed to terminating with an error message or running forever). Show that it is impossible to write a procedure halts? that correctly determines whether p halts on a for any procedure p and object a. Use the following reasoning: If you had such a procedure halts?, you could implement the following program:

(define (run-forever) (run-forever))

(define (try p)
  (if (halts? p p)
      (run-forever)
      'halted))

Now consider evaluating the expression (try try) and show that any possible outcome (either halting or running forever) violates the intended behavior of halts?.²³

R E A S O N I N G

The following is a simple zero-argument procedure that runs forever:

(define (run-forever) (run-forever))

The following is a hypothetical procedure that checks if procedure p halts on input a:

(define (halts? p a)

(if <p halts on a>

true

false

)

)

Assuming that the "halts?" procedure above exists and works, we should be able to write 

another procedure:

(define (try p)

(if (halts? p p)

(run-forever)

'halted

)

)

The "try" procedure takes a procedure as its argument and checks if that procedure halts

on itself. If true, then try runs forever. If false, then try returns.

If p halts on itself, then try runs forever.

If p runs forever, then try halts on p.

So "try" behaves oppositely to the procedure supplied. 

Now if we evaluate (try try), then the logic would work like this:

If try halts on itself, try runs forever.

If try runs forever, try halts on itself.

In both cases, the behaviour of "try" violates the intended behavior of "halts?"