Prolog Unification Proof

gcd(U, 0, U) .
gcd(U, V, W) :-
V>0,
U1 is  U mod V,
gcd(V, U1, W).

% z:/prog lang/prolog/test compiled 0.00 sec, 7 clauses

1 ?- gcd(29, 7, X).

X = 1 .

Unification

gcd(u, 0, u)
gcd(u, v, w) <- not zero(v), gcd(v, u mod v, w)

<-gcd(29, 7, X)

**********************************************
gcd(29, 7, X) <- not zero(7), gcd(7,1,X), gcd(29, 7, X)
gcd(1, 7, X) <- not zero(7), gcd(7,0,X), gcd(1, 7, X)
gcd(1,0,1) <- gcd(1,0,1)

Scheme Tail Recursion

(define (listlen2);I am lazy so i made a built in list for testing
  (listlen1 (list 1 2 3 4 5)))
(define (listlen1 tlist);procedure listlen without the need to define starting count
  (listlen tlist 0))
(define (listlen tlist count)
           (if (null? tlist);check if reached end of list
             count;write count if reached end
             (listlen (cdr tlist) (+ 1 count))));recurse with remainder of list and increment count
 
(define (trprod2);I am lazy so i made a built in list for testing
  (trprod1 (list 1 2 3 4 5)))
(define (trprod1 tlist);procedure tprod without need to define start prod
  (trprod tlist 1))
(define (trprod tlist prod)
  (if (null? tlist);check is reached end of list
      prod;write prod if reached end
      (trprod (cdr tlist) (* (car tlist) prod))));recurse with remainder of list and product of car and current product

**************************************************************
> (listlen2)
5
> (listlen1 (list 1 2 3 4 5 6 7 8 9))
9
> (trprod2)
120
> (trprod1 (list 1 2 3))
6
>