We consider universal computability of the LCTRS with only rule scheme Calc:

  Signature: gcd :: Int -> Int -> Int

  Rules: gcd(x, 0) -> x
         gcd(0, y) -> y
         gcd(x, y) -> gcd(x - y, y) | x >= y /\ y > 0
         gcd(x, y) -> gcd(y - x, x) | y > x /\ x > 0

The system is accessible function passing by a sort ordering that equates all sorts.

We start by computing the initial DP problem D1 = (P1, R UNION R_?, f, c), where:

  P1. (1) gcd#(x, y) => gcd#(x - y, y) | x >= y /\ y > 0
      (2) gcd#(x, y) => gcd#(y - x, x) | y > x /\ x > 0

***** We apply the Integer Function Processor on D1 = (P1, R UNION R_?, f, c).

We use the following integer mapping:

  J(gcd#) = arg_2

We thus have:

  (1) x >= y /\ y > 0 |= y >= y
  (2) y > x /\ x > 0  |= y >  x (and y >= 0)

We may remove the strictly oriented DPs, which yields:

Processor output: { D2 = (P2, R UNION R_?, f, c) }, where:

  P2. (1) gcd#(x, y) => gcd#(x - y, y) | x >= y /\ y > 0

***** We apply the Integer Function Processor on D2 = (P2, R UNION R_?, f, c).

We use the following integer mapping:

  J(gcd#) = arg_1 - arg_2

We thus have:

  (1) x >= y /\ y > 0 |= x - y > x - y - y (and x - y >= 0)

All DPs are strictly oriented, and may be removed.  Hence, this DP problem is finite.

Processor output: { }.