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

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

  Rules: eval(x, y) -> eval(x, y + 1) | x > y
         eval(x, y) -> eval(x, y + 1) | y > x /\ x > y
         eval(x, y) -> eval(x + 1, y) | x > y /\ y >= x
         eval(x, y) -> eval(x + 1, y) | y > x /\ 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#(x, y) => eval#(x, y + 1) | x > y
      (2) eval#(x, y) => eval#(x, y + 1) | y > x /\ x > y
      (3) eval#(x, y) => eval#(x + 1, y) | x > y /\ y >= x
      (4) eval#(x, y) => eval#(x + 1, y) | y > x /\ y >= x

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

We compute a graph approximation with the following edges:

  1: 1 
  2:
  3:
  4: 4 

There are 2 SCCs.

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

  P2. (1) eval#(x, y) => eval#(x, y + 1) | x > y

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

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

We use the following integer mapping:

  J(eval#) = arg_1 - arg_2

We thus have:

  (1) x > y |= x - y > x - (y + 1) (and x - y >= 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_2 - arg_1

We thus have:

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

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

Processor output: { }.