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

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

  Rules: eval_1(x, y) -> eval_2(x, 0) | x > 0
         eval_2(x, y) -> eval_2(x, y + 1) | x > 0 /\ y >= 0 /\ x > y
         eval_2(x, y) -> eval_1(x - 1, y) | x > 0 /\ y >= 0 /\ y >= 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_1#(x, y) => eval_2#(x, 0) | x > 0
      (2) eval_2#(x, y) => eval_2#(x, y + 1) | x > 0 /\ y >= 0 /\ x > y
      (3) eval_2#(x, y) => eval_1#(x - 1, y) | x > 0 /\ y >= 0 /\ y >= x

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

We chain DPs according to the following mapping:


  eval_2#(x, y) => eval_2#(x - 1, 0) | x > 0 /\ y >= 0 /\ y >= x /\ x - 1 > 0  is obtained by chaining  eval_2#(x, y) => eval_1#(x - 1, y) | x > 0 /\ y >= 0 /\ y >= x and eval_1#(x', y') => eval_2#(x', 0) | x' > 0


The following DPs were deleted:

eval_2#(x, y) => eval_1#(x - 1, y) | x > 0 /\ y >= 0 /\ y >= x

eval_1#(x, y) => eval_2#(x, 0) | x > 0


By chaining, we added 1 DPs and removed 2 DPs.

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

  P2. (1) eval_2#(x, y) => eval_2#(x, y + 1) | x > 0 /\ y >= 0 /\ x > y
      (2) eval_2#(x, y) => eval_2#(x - 1, 0) | x > 0 /\ y >= 0 /\ y >= x /\ x - 1 > 0

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

We use the following integer mapping:

  J(eval_2#) = arg_1 - 1 - 1

We thus have:

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

We may remove the strictly oriented DPs, which yields:

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

  P3. (1) eval_2#(x, y) => eval_2#(x, y + 1) | x > 0 /\ y >= 0 /\ x > y

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

We use the following integer mapping:

  J(eval_2#) = arg_1 - arg_2 - 1

We thus have:

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

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

Processor output: { }.