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) f#(0, 1, X) => f#(s(X), X, X)
      (2) f#(Y, U, s(V)) => f#(0, 1, V)
      (3) map#(J, cons(X1, Y1)) => map#(J, Y1)
      (4) filter#(H1, cons(W1, P1)) => filter2#(H1(W1), H1, W1, P1)
      (5) filter2#(true, F2, Y2, U2) => filter#(F2, U2)
      (6) filter2#(false, H2, W2, P2) => filter#(H2, P2)

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

Considering the 3 SCCs, this DP problem is split into the following new problems.

  P2. (1) f#(0, 1, X) => f#(s(X), X, X)
      (2) f#(Y, U, s(V)) => f#(0, 1, V)

  P3. (1) map#(J, cons(X1, Y1)) => map#(J, Y1)

  P4. (1) filter#(H1, cons(W1, P1)) => filter2#(H1(W1), H1, W1, P1)
      (2) filter2#(true, F2, Y2, U2) => filter#(F2, U2)
      (3) filter2#(false, H2, W2, P2) => filter#(H2, P2)

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

We use the following projection function:

  nu(f#) = 3

We thus have:

  (1) X    |>=| X
  (2) s(V) |>|  V

We may remove the strictly oriented DPs, which yields:

  P5. (1) f#(0, 1, X) => f#(s(X), X, X)

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

We use the following projection function:

  nu(map#) = 2

We thus have:

  (1) cons(X1, Y1) |>| Y1

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

***** We apply the Subterm Criterion Processor on P4.

We use the following projection function:

  nu(filter#)  = 2
  nu(filter2#) = 4

We thus have:

  (1) cons(W1, P1) |>|  P1
  (2) U2           |>=| U2
  (3) P2           |>=| P2

We may remove the strictly oriented DPs, which yields:

  P6. (1) filter2#(true, F2, Y2, U2) => filter#(F2, U2)
      (2) filter2#(false, H2, W2, P2) => filter#(H2, P2)

***** We apply the Graph Processor on P5.

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

***** We apply the Graph Processor on P6.

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