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

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

  Rules: eval_1(x, y, z) -> eval_2(x, y, z) | x > y
         eval_2(x, y, z) -> eval_1(x, y + 1, z) | x > z
         eval_2(x, y, z) -> eval_1(x, y, z + 1) | x > z
         eval_2(x, y, z) -> eval_1(x - 1, y, z) | z >= 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, f, c), where:

  P1. (1) eval_1#(x, y, z) => eval_2#(x, y, z) | x > y
      (2) eval_2#(x, y, z) => eval_1#(x, y + 1, z) | x > z
      (3) eval_2#(x, y, z) => eval_1#(x, y, z + 1) | x > z
      (4) eval_2#(x, y, z) => eval_1#(x - 1, y, z) | z >= x

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

We use the following theory arguments function:

  eval_1# : [1, 2, 3]
  eval_2# : [1, 2, 3]

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

  P2. (1) eval_1#(x, y, z) => eval_2#(x, y, z) | x > y
      (2) eval_2#(x, y, z) => eval_1#(x, y + 1, z) | x > z { x, y, z }
      (3) eval_2#(x, y, z) => eval_1#(x, y, z + 1) | x > z { x, y, z }
      (4) eval_2#(x, y, z) => eval_1#(x - 1, y, z) | z >= x { x, y, z }

  P3. (1) eval_2#(x, y, z) => eval_1#(x, y + 1, z) | x > z
      (2) eval_2#(x, y, z) => eval_1#(x, y, z + 1) | x > z
      (3) eval_2#(x, y, z) => eval_1#(x - 1, y, z) | z >= x

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

We use the following theory arguments function:

  eval_1# : [1, 2, 3]
  eval_2# : [1, 2, 3]

Processor output: { D4 = (P4, R, f, c) ; D5 = (P5, R, f, c) }, where:

  P4. (1) eval_1#(x, y, z) => eval_2#(x, y, z) | x > y { x, y, z }
      (2) eval_2#(x, y, z) => eval_1#(x, y + 1, z) | x > z { x, y, z }
      (3) eval_2#(x, y, z) => eval_1#(x, y, z + 1) | x > z { x, y, z }
      (4) eval_2#(x, y, z) => eval_1#(x - 1, y, z) | z >= x { x, y, z }

  P5. (1) eval_1#(x, y, z) => eval_2#(x, y, z) | x > y

***** We apply the Graph Processor on D3 = (P3, R, f, c).

We compute a graph approximation with the following edges:

  1:
  2:
  3:

As there are no SCCs, this DP problem is removed.

Processor output: { }.

***** No progress could be made on DP problem D4 = (P4, R, f, c).