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

  Signature: f :: Int -> A (private)
             g :: Int -> A

  Rules: f(x) -> g(x - 1) | x = x
         g(x) -> f(x) | x > 0

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) f#(x) => g#(x - 1) | x = x (private)
      (2) g#(x) => f#(x) | x > 0

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

We use the following integer mapping:

  J(f#) = arg_1
  J(g#) = arg_1 + 1

We thus have:

  (1) x = x |= x     >= x - 1 + 1
  (2) x > 0 |= x + 1 >  x         (and x + 1 >= 0)

We may remove the strictly oriented DPs, which yields:

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

  P2. (1) f#(x) => g#(x - 1) | x = x

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

We compute a graph approximation with the following edges:

  1:

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

Processor output: { }.