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

  Signature: eval :: Int -> Int -> o

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

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) eval#(x, y) => eval#(x - y, y) | x > y /\ x > 0 /\ y > 0
      (2) eval#(x, y) => eval#(x - y, y) | y > x /\ x > 0 /\ y > 0 /\ x > y
      (3) eval#(x, y) => eval#(x, y - x) | x > y /\ x > 0 /\ y > 0 /\ y >= x
      (4) eval#(x, y) => eval#(x, y - x) | y > x /\ x > 0 /\ y > 0 /\ y >= x

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

We compute a graph approximation with the following edges:

  1: 1 4 
  2:
  3:
  4: 1 4 

There is only one SCC, so all DPs not inside the SCC can be removed.

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

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

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

We use the following integer mapping:

  J(eval#) = arg_1

We thus have:

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

We may remove the strictly oriented DPs, which yields:

Processor output: { D3 = (P3, R UNION R_?, f, c) }, where:

  P3. (1) eval#(x, y) => eval#(x, y - x) | y > x /\ x > 0 /\ y > 0 /\ y >= x

***** We apply the Integer Function Processor on D3 = (P3, R UNION R_?, f, c).

We use the following integer mapping:

  J(eval#) = arg_2 - arg_1

We thus have:

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

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

Processor output: { }.