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

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

  Rules: f(x) -> f(x - 1) | x >= 0
         g(y) -> h(y) | y = y
         h(y) -> g(y + 1) | y <= 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) => f#(x - 1) | x >= 0
      (2) g#(y) => h#(y) | y = y
      (3) h#(y) => g#(y + 1) | y <= 0

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

We compute a graph approximation with the following edges:

  1: 1 
  2: 3 
  3: 2 

There are 2 SCCs.

Processor output: { D2 = (P2, R, i, c) ; D3 = (P3, R, i, c) }, where:

  P2. (1) f#(x) => f#(x - 1) | x >= 0

  P3. (1) g#(y) => h#(y) | y = y
      (2) h#(y) => g#(y + 1) | y <= 0

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

We use the following integer mapping:

  J(f#) = arg_1 + 1

We thus have:

  (1) x >= 0 |= 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: { }.

***** We apply the Chaining Processor Processor on D3 = (P3, R, i, c).

We chain DPs according to the following mapping:


  h#(y) => h#(y + 1) | y <= 0 /\ y + 1 = y + 1  is obtained by chaining  h#(y) => g#(y + 1) | y <= 0 and g#(y') => h#(y') | y' = y'


The following DPs were deleted:

h#(y) => g#(y + 1) | y <= 0

g#(y) => h#(y) | y = y


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

Processor output: { D4 = (P4, R, i, c) }, where:

  P4. (1) h#(y) => h#(y + 1) | y <= 0 /\ y + 1 = y + 1

***** We apply the Integer Function Processor on D4 = (P4, R, i, c).

We use the following integer mapping:

  J(h#) = 0 - arg_1

We thus have:

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

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

Processor output: { }.