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

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

  Rules: eval(x, y) -> eval(y, y) | x > 0 /\ not (x = 0) /\ x > y
         eval(x, y) -> eval(x - 1, y) | x > 0 /\ not (x = 0) /\ x <= 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, f, c), where:

  P1. (1) eval#(x, y) => eval#(y, y) | x > 0 /\ not (x = 0) /\ x > y
      (2) eval#(x, y) => eval#(x - 1, y) | x > 0 /\ not (x = 0) /\ x <= y

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

We compute a graph approximation with the following edges:

  1: 2 
  2: 2 

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

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

  P2. (1) eval#(x, y) => eval#(x - 1, y) | x > 0 /\ not (x = 0) /\ x <= y

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

We use the following integer mapping:

  J(eval#) = arg_1 + 1

We thus have:

  (1) x > 0 /\ not (x = 0) /\ x <= y |= x + 1 > x - 1 + 1 (and x + 1 >= 0)

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

Processor output: { }.