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

  Signature: if    :: o -> Int -> Int -> o
             sumto :: Int -> Int -> o
             wrap  :: Int -> o

  Rules: sumto(x, y) -> wrap(0) | x > y
         sumto(x, y) -> if(sumto(x + 1, y), x, y) | y >= x
         if(wrap(z), x, y) -> wrap(x + z) | z = z /\ 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) sumto#(x, y) => sumto#(x + 1, y) | y >= x
      (2) sumto#(x, y) => if#(sumto(x + 1, y), x, y) | y >= x

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

We compute a graph approximation with the following edges:

  1: 1 2 
  2:

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

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

  P2. (1) sumto#(x, y) => sumto#(x + 1, y) | y >= x

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

We use the following integer mapping:

  J(sumto#) = arg_2 - arg_1

We thus have:

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

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

Processor output: { }.