The system is accessible function passing by a sort ordering that equates all sorts.

We start by computing the following initial DP problem:

  P1. (1) init#(F, X{8}) => fsum#(F, 10)
      (2) init#(F, X{8}) => map#([+](fsum(F, 10)), X{8})
      (3) map#(F, cons(H, T)) => map#(F, T)
      (4) fsum#(F, N) => fsum#(F, N - 1) | N != 0

***** We apply the Constraint Modification Processor on P1.

We replace fsum#(F, N) => fsum#(F, N - 1) | N != 0 by:

  fsum#(F, N) => fsum#(F, N - 1) | N > 0
  fsum#(F, N) => fsum#(F, N - 1) | N < 0

This yields:

  P2. (1) init#(F, X{8}) => fsum#(F, 10)
      (2) init#(F, X{8}) => map#([+](fsum(F, 10)), X{8})
      (3) map#(F, cons(H, T)) => map#(F, T)
      (4) fsum#(F, N) => fsum#(F, N - 1) | N > 0
      (5) fsum#(F, N) => fsum#(F, N - 1) | N < 0

***** We apply the Reachability Processor on P2.

There is one unreachable dependency pair, which is removed.

  P3. (1) init#(F, X{8}) => fsum#(F, 10)
      (2) init#(F, X{8}) => map#([+](fsum(F, 10)), X{8})
      (3) map#(F, cons(H, T)) => map#(F, T)
      (4) fsum#(F, N) => fsum#(F, N - 1) | N > 0

***** We apply the Graph Processor on P3.

Considering the 2 SCCs, this DP problem is split into the following new problems.

  P4. (1) fsum#(F, N) => fsum#(F, N - 1) | N > 0

  P5. (1) map#(F, cons(H, T)) => map#(F, T)

***** We apply the Integer Function Processor on P4.

We use the following integer mapping:

  J(fsum#) = arg_2

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.

***** We apply the Subterm Criterion Processor on P5.

We use the following projection function:

  nu(map#) = 2

We thus have:

  (1) cons(H, T) |>| T

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