The system is accessible function passing by a sort ordering that equates all sorts.

We start by computing the following initial DP problem:

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

***** We apply the Integer Function Processor on P1.

We use the following integer mapping:

  J(eval_1#) = arg_1 - arg_3 - 1
  J(eval_2#) = arg_1 - arg_3 - 1

We thus have:

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

We may remove the strictly oriented DPs, which yields:

  P2. (1) eval_1#(x, y, z) => eval_2#(x, y, z) | x > z
      (2) eval_2#(x, y, z) => eval_2#(x, y - 1, z) | x > z /\ y > z

***** We apply the Graph Processor on P2.

There is only one SCC, so all DPs not inside the SCC can be removed:

  P3. (1) eval_2#(x, y, z) => eval_2#(x, y - 1, z) | x > z /\ y > z

***** We apply the Integer Function Processor on P3.

We use the following integer mapping:

  J(eval_2#) = arg_2 - arg_3

We thus have:

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

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