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) times#(Y, s(U)) => times#(Y, U)
      (2) times#(Y, s(U)) => plus#(times(Y, U), Y)
      (3) plus#(P, s(X1)) => plus#(P, X1)
      (4) plus#(s(Y1), U1) => plus#(Y1, U1)
      (5) map#(I1, cons(P1, X2)) => map#(I1, X2)
      (6) filter#(G2, cons(V2, W2)) => filter2#(G2(V2), G2, V2, W2)
      (7) filter2#(true, J2, X3, Y3) => filter#(J2, Y3)
      (8) filter2#(false, G3, V3, W3) => filter#(G3, W3)

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

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

  P2. (1) plus#(P, s(X1)) => plus#(P, X1)
      (2) plus#(s(Y1), U1) => plus#(Y1, U1)

  P3. (1) times#(Y, s(U)) => times#(Y, U)

  P4. (1) map#(I1, cons(P1, X2)) => map#(I1, X2)

  P5. (1) filter#(G2, cons(V2, W2)) => filter2#(G2(V2), G2, V2, W2)
      (2) filter2#(true, J2, X3, Y3) => filter#(J2, Y3)
      (3) filter2#(false, G3, V3, W3) => filter#(G3, W3)

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

We use the following projection function:

  nu(plus#) = 1

We thus have:

  (1) P     |>=| P
  (2) s(Y1) |>|  Y1

We may remove the strictly oriented DPs, which yields:

  P6. (1) plus#(P, s(X1)) => plus#(P, X1)

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

We use the following projection function:

  nu(times#) = 2

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 P4.

We use the following projection function:

  nu(map#) = 2

We thus have:

  (1) cons(P1, X2) |>| X2

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(V2, W2) |>|  W2
  (2) Y3           |>=| Y3
  (3) W3           |>=| W3

We may remove the strictly oriented DPs, which yields:

  P7. (1) filter2#(true, J2, X3, Y3) => filter#(J2, Y3)
      (2) filter2#(false, G3, V3, W3) => filter#(G3, W3)

***** We apply the Subterm Criterion Processor on P6.

We use the following projection function:

  nu(plus#) = 2

We thus have:

  (1) s(X1) |>| X1

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

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

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