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

  Signature: eval_1 :: Int -> Int -> o
             eval_2 :: Int -> Int -> o

  Rules: eval_1(x, y) -> eval_2(x, y) | x > 0 /\ y = y
         eval_2(x, y) -> eval_2(x, y - 1) | x > 0 /\ y > 0
         eval_2(x, y) -> eval_1(x - 1, y) | x > 0 /\ 0 >= 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_1#(x, y) => eval_2#(x, y) | x > 0 /\ y = y
      (2) eval_2#(x, y) => eval_2#(x, y - 1) | x > 0 /\ y > 0
      (3) eval_2#(x, y) => eval_1#(x - 1, y) | x > 0 /\ 0 >= y

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

We use the following integer mapping:

  J(eval_1#) = arg_1
  J(eval_2#) = arg_1 - 1

We thus have:

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

We may remove the strictly oriented DPs, which yields:

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

  P2. (1) eval_2#(x, y) => eval_2#(x, y - 1) | x > 0 /\ y > 0
      (2) eval_2#(x, y) => eval_1#(x - 1, y) | x > 0 /\ 0 >= y

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

We compute a graph approximation with the following edges:

  1: 1 2 
  2:

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

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

  P3. (1) eval_2#(x, y) => eval_2#(x, y - 1) | x > 0 /\ y > 0

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

We use the following integer mapping:

  J(eval_2#) = arg_2 - 1

We thus have:

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

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

Processor output: { }.