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) intlist#(cons(X, Y)) => intlist#(Y)
      (2) int#(0, s(U)) => int#(s(0), s(U))
      (3) int#(s(W), s(P)) => int#(W, P)
      (4) int#(s(W), s(P)) => intlist#(int(W, P))
      (5) map#(Z1, cons(U1, V1)) => map#(Z1, V1)
      (6) filter#(J1, cons(X2, Y2)) => filter2#(J1(X2), J1, X2, Y2)
      (7) filter2#(true, G2, V2, W2) => filter#(G2, W2)
      (8) filter2#(false, J2, X3, Y3) => filter#(J2, Y3)

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

Considering the 4 SCCs, this DP problem is split into the following new problems.

  P2. (1) intlist#(cons(X, Y)) => intlist#(Y)

  P3. (1) int#(0, s(U)) => int#(s(0), s(U))
      (2) int#(s(W), s(P)) => int#(W, P)

  P4. (1) map#(Z1, cons(U1, V1)) => map#(Z1, V1)

  P5. (1) filter#(J1, cons(X2, Y2)) => filter2#(J1(X2), J1, X2, Y2)
      (2) filter2#(true, G2, V2, W2) => filter#(G2, W2)
      (3) filter2#(false, J2, X3, Y3) => filter#(J2, Y3)

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

We use the following projection function:

  nu(intlist#) = 1

We thus have:

  (1) cons(X, Y) |>| Y

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(int#) = 2

We thus have:

  (1) s(U) |>=| s(U)
  (2) s(P) |>|  P

We may remove the strictly oriented DPs, which yields:

  P6. (1) int#(0, s(U)) => int#(s(0), s(U))

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

We use the following projection function:

  nu(map#) = 2

We thus have:

  (1) cons(U1, V1) |>| V1

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

We thus have:

  (1) cons(X2, Y2) |>|  Y2
  (2) W2           |>=| W2
  (3) Y3           |>=| Y3

We may remove the strictly oriented DPs, which yields:

  P7. (1) filter2#(true, G2, V2, W2) => filter#(G2, W2)
      (2) filter2#(false, J2, X3, Y3) => filter#(J2, Y3)

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

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

***** We apply the Graph Processor on P7.

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