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

  Signature: f :: Int -> Int -> A
             g :: Int -> Int -> A

  Rules: f(x, y) -> g(x, y - 1) | x = x /\ y >= 0
         g(x, y) -> f(x, y) | x = x /\ y = 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) f#(x, y) => g#(x, y - 1) | x = x /\ y >= 0
      (2) g#(x, y) => f#(x, y) | x = x /\ y = y

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

We chain DPs according to the following mapping:


  g#(x, y) => g#(x, y - 1) | x = x /\ y = y /\ (x = x /\ y >= 0)  is obtained by chaining  g#(x, y) => f#(x, y) | x = x /\ y = y and f#(x', y') => g#(x', y' - 1) | x' = x' /\ y' >= 0


The following DPs were deleted:

g#(x, y) => f#(x, y) | x = x /\ y = y

f#(x, y) => g#(x, y - 1) | x = x /\ y >= 0


By chaining, we added 1 DPs and removed 2 DPs.

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

  P2. (1) g#(x, y) => g#(x, y - 1) | x = x /\ y = y /\ (x = x /\ y >= 0)

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

We use the following integer mapping:

  J(g#) = arg_2

We thus have:

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

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

Processor output: { }.