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

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

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

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, z) => eval#(x + 1, y, z) | y > x /\ z > x
      (2) eval#(x, y, z) => eval#(x, y, z + 1) | y > x /\ x >= z

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

We use the following integer mapping:

  J(eval#) = arg_2 - arg_1 - 1

We thus have:

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

We may remove the strictly oriented DPs, which yields:

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

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

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

We use the following integer mapping:

  J(eval#) = arg_1 - arg_3

We thus have:

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

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

Processor output: { }.