Problem 1
a)
set-equalp = (and (set-equalp x x)
		          ((set-equalp x y) => (set-equalp y x))
                                  ((and (set-equalp x y) (set-equalp y z)) => (set-equalp x z)))
 
Reflexive (set-equalp x x) = T
Proof -
(set-equalp x x)
|| by def. of setequalp
    (and (set-subsetp x x) 
            (set-subsetp x x)))
|| by def. of set-subsetp
(and 
           (if (endp x) 
               t 
              (and (set-memberp (car x) x) 
                      (set-subsetp (cdr x) x))))
           (if (endp x) 
              t 
             (and (set-memberp (car x) x) 
                     (set-subsetp (cdr x) x))))
|| I chose not to expand this any further (by def. of set-memberp), just because it would become too messy and unreadable.  However, (set-subsetp x x) always returns T.  This is because (set-memberp (car x) x) obviously always returns true because an element of set “x” is always going to be in set “x”.  Eventually (set-subsetp x x) will run (set-memberp) on every element of x and the function will return T
(and (T
        (T))
|| and axiom
T.. Therefore, set-equalp is reflexive