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

  Signature: sum :: Int -> Int

  Rules: sum(x) -> 0 | x <= 0
         sum(x) -> x + sum(x - 1) | 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) sum#(x) => sum#(x - 1) | x != 0

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

We replace sum#(x) => sum#(x - 1) | x != 0 by:

  sum#(x) => sum#(x - 1) | x > 0
  sum#(x) => sum#(x - 1) | x < 0

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

  P2. (1) sum#(x) => sum#(x - 1) | x > 0
      (2) sum#(x) => sum#(x - 1) | x < 0

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

We compute a graph approximation with the following edges:

  1: 1 
  2: 2 

There are 2 SCCs.

Processor output: { D3 = (P3, R, i, c) ; D4 = (P4, R, i, c) }, where:

  P3. (1) sum#(x) => sum#(x - 1) | x > 0

  P4. (1) sum#(x) => sum#(x - 1) | x < 0

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

We use the following integer mapping:

  J(sum#) = arg_1

We thus have:

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

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

Processor output: { }.

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

We obtain 0 usable rules (out of 2 rules in the input problem).

Processor output: { D5 = (P4, {}, i, c) }.

***** No progress could be made on DP problem D5 = (P4, {}, i, c).