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

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

  Rules: eval(x, y) -> eval(x - 1, y) | x + y > 0 /\ x > y
         eval(x, y) -> eval(x - 1, y) | x + y > 0 /\ y >= x /\ x = y
         eval(x, y) -> eval(x, y - 1) | x + y > 0 /\ y >= x /\ y > x
         eval(x, y) -> eval(x, y - 1) | x + y > 0 /\ y >= x /\ x > y

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_?, i, c), where:

  P1. (1) eval#(x, y) => eval#(x - 1, y) | x + y > 0 /\ x > y
      (2) eval#(x, y) => eval#(x - 1, y) | x + y > 0 /\ y >= x /\ x = y
      (3) eval#(x, y) => eval#(x, y - 1) | x + y > 0 /\ y >= x /\ y > x
      (4) eval#(x, y) => eval#(x, y - 1) | x + y > 0 /\ y >= x /\ x > y

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

We compute a graph approximation with the following edges:

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

There are 2 SCCs.

Processor output: { D2 = (P2, R UNION R_?, i, c) ; D3 = (P3, R UNION R_?, i, c) }, where:

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

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

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

We use the following integer mapping:

  J(eval#) = arg_1

We thus have:

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

We may remove the strictly oriented DPs, which yields:

Processor output: { D4 = (P4, R UNION R_?, i, c) }, where:

  P4. (1) eval#(x, y) => eval#(x, y - 1) | x + y > 0 /\ y >= x /\ y > x

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

We use the following integer mapping:

  J(eval#) = arg_1 - 1

We thus have:

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

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

Processor output: { }.

***** We apply the Integer Function Processor on D4 = (P4, R UNION R_?, i, c).

We use the following integer mapping:

  J(eval#) = arg_2

We thus have:

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

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

Processor output: { }.