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

  Signature: eval_1 :: Int -> Int -> o
             eval_2 :: Int -> Int -> o
             eval_3 :: Int -> Int -> o

  Rules: eval_1(x, y) -> eval_2(x, y) | x > 0 /\ y > 0 /\ x > y
         eval_1(x, y) -> eval_3(x, y) | x > 0 /\ y > 0 /\ y >= x
         eval_2(x, y) -> eval_2(x - 1, y) | x > 0
         eval_2(x, y) -> eval_1(x, y) | 0 >= x
         eval_3(x, y) -> eval_3(x, y - 1) | y > 0
         eval_3(x, y) -> eval_1(x, y) | 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, f, c), where:

  P1. (1) eval_1#(x, y) => eval_2#(x, y) | x > 0 /\ y > 0 /\ x > y
      (2) eval_1#(x, y) => eval_3#(x, y) | x > 0 /\ y > 0 /\ y >= x
      (3) eval_2#(x, y) => eval_2#(x - 1, y) | x > 0
      (4) eval_2#(x, y) => eval_1#(x, y) | 0 >= x
      (5) eval_3#(x, y) => eval_3#(x, y - 1) | y > 0
      (6) eval_3#(x, y) => eval_1#(x, y) | 0 >= y

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

We compute a graph approximation with the following edges:

  1: 3 
  2: 5 
  3: 3 4 
  4:
  5: 5 6 
  6:

There are 2 SCCs.

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

  P2. (1) eval_2#(x, y) => eval_2#(x - 1, y) | x > 0

  P3. (1) eval_3#(x, y) => eval_3#(x, y - 1) | y > 0

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

We use the following integer mapping:

  J(eval_2#) = arg_1

We thus have:

  (1) x > 0 |= x > x - 1 (and x >= 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, f, c).

We use the following integer mapping:

  J(eval_3#) = arg_2

We thus have:

  (1) y > 0 |= y > y - 1 (and y >= 0)

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

Processor output: { }.