1 2 3 (define (double f) (lambda (x) (f (f x))) )
Now evaluating the statement given:
1 > (((double (double double)) inc) 5)
double double returns a procedure, lets call this procedure X, that applies double twice to an argument passed to it.
And call to
double (double double), return a procedure, lets call this procedure Y, that applies X twice to the argument passed to it.
And the invocation
Y inc, returns a procedure, lets call it Z, that applies ‘X’ twice to
(X (X inc)).
Thus now finally the call to
Z 5 is evaluated as follows:
1 2 3 (Z 5) ((X (X inc)) 5) ((double (double (double (double inc))) 5))
Now we can see that in the innermost call, inc is called twice, than it is doubled thus called 4 times, again it is doubled thus called 8 times and again it is double thus called 16 times.
Thus thus function will return after calling
inc 16 times starting with argument 5.
Thus the output is $16+5 = 21$.