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

  Signature: comp   :: (Int -> ret) -> (Int -> Int) -> Int -> ret
             error  :: ret
             fact   :: Int -> (Int -> ret) -> ret
             return :: Int -> ret

  Rules: comp(f, g, x) -> f(g(x))
         fact(n, k) -> error | n < 0
         fact(n, k) -> k(1) | n = 0
         fact(n, k) -> fact(n - 1, comp(k, [*](n))) | n > 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) fact#(n, k) => comp#(k, [*](n), fresh1) | n > 0
      (2) fact#(n, k) => fact#(n - 1, comp(k, [*](n))) | n > 0

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

We compute a graph approximation with the following edges:

  1:
  2: 1 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) fact#(n, k) => fact#(n - 1, comp(k, [*](n))) | n > 0

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

We use the following integer mapping:

  J(fact#) = arg_1

We thus have:

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

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

Processor output: { }.