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

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

We use the following integer mapping:

  J(eval#) = arg_1

We thus have:

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

We may remove the strictly oriented DPs, which yields:

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

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

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

We obtain 0 usable rules (out of 2 rules in the input problem).

Processor output: { D3 = (P2, {}, i, c) }.

***** No progress could be made on DP problem D3 = (P2, {}, i, c).