We consider termination of the LCTRS with only rule scheme Calc:

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

  Rules: eval(x, y) -> eval(x - 1, z) | x >= 0 /\ y = y /\ z = z
         eval(x, y) -> eval(x, y - 1) | y >= 0 /\ x = 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, i, c), where:

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

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

We use the following integer mapping:

  J(eval#) = arg_1 + 1

We thus have:

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

We may remove the strictly oriented DPs, which yields:

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

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

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

We use the following integer mapping:

  J(eval#) = arg_2

We thus have:

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

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

Processor output: { }.