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

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

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

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

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

We compute a graph approximation with the following edges:

  1: 1 2 
  2: 2 
  3:

There are 2 SCCs.

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

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

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

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

We use the following integer mapping:

  J(eval#) = arg_1 + arg_2 - arg_3 - 1

We thus have:

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

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

Processor output: { }.