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) neq#(s(U), s(V)) => neq#(U, V)
      (2) filter#(J, cons(X1, Y1)) => filtersub#(J(X1), J, cons(X1, Y1))
      (3) filtersub#(true, G1, cons(V1, W1)) => filter#(G1, W1)
      (4) filtersub#(false, J1, cons(X2, Y2)) => filter#(J1, Y2)
      (5) nonzero#(X{14}) => neq#(0, X{15})
      (6) nonzero#(X{14}) => filter#(neq(0), X{14})

***** We apply the Graph Processor on P1.

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

  P2. (1) neq#(s(U), s(V)) => neq#(U, V)

  P3. (1) filter#(J, cons(X1, Y1)) => filtersub#(J(X1), J, cons(X1, Y1))
      (2) filtersub#(true, G1, cons(V1, W1)) => filter#(G1, W1)
      (3) filtersub#(false, J1, cons(X2, Y2)) => filter#(J1, Y2)

***** We apply the Subterm Criterion Processor on P2.

We use the following projection function:

  nu(neq#) = 1

We thus have:

  (1) s(U) |>| U

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

***** We apply the Subterm Criterion Processor on P3.

We use the following projection function:

  nu(filter#)    = 2
  nu(filtersub#) = 3

We thus have:

  (1) cons(X1, Y1) |>=| cons(X1, Y1)
  (2) cons(V1, W1) |>|  W1
  (3) cons(X2, Y2) |>|  Y2

We may remove the strictly oriented DPs, which yields:

  P4. (1) filter#(J, cons(X1, Y1)) => filtersub#(J(X1), J, cons(X1, Y1))

***** We apply the Graph Processor on P4.

As there are no SCCs, this DP problem is removed.