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

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

We use the following integer mapping:

  J(eval_1#) = arg_1 + 1
  J(eval_2#) = arg_1

We thus have:

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

We may remove the strictly oriented DPs, which yields:

  P2. (1) eval_1#(x, y) => eval_2#(x + 1, 1) | x >= 0
      (2) eval_2#(x, y) => eval_2#(x, y + 1) | x >= 0 /\ y > 0 /\ x >= y

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

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

We use the following integer mapping:

  J(eval_2#) = arg_1 - arg_2

We thus have:

  (1) x >= 0 /\ y > 0 /\ 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.