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 > 0 /\ z = z
         eval(x, y) -> eval(x, y - 1) | x > 0 /\ y > 0

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

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

We use the following integer mapping:

  J(eval#) = arg_1

We thus have:

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

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) | x > 0 /\ y > 0

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

We use the following integer mapping:

  J(eval#) = 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: { }.