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) *#(X, +(Y, U)) => *#(X, Y)
      (2) *#(X, +(Y, U)) => *#(X, U)
      (3) map#(I, cons(P, X1)) => map#(I, X1)
      (4) filter#(G1, cons(V1, W1)) => filter2#(G1(V1), G1, V1, W1)
      (5) filter2#(true, J1, X2, Y2) => filter#(J1, Y2)
      (6) filter2#(false, G2, V2, W2) => filter#(G2, W2)

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

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

  P2. (1) *#(X, +(Y, U)) => *#(X, Y)
      (2) *#(X, +(Y, U)) => *#(X, U)

  P3. (1) map#(I, cons(P, X1)) => map#(I, X1)

  P4. (1) filter#(G1, cons(V1, W1)) => filter2#(G1(V1), G1, V1, W1)
      (2) filter2#(true, J1, X2, Y2) => filter#(J1, Y2)
      (3) filter2#(false, G2, V2, W2) => filter#(G2, W2)

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

We use the following projection function:

  nu(*#) = 2

We thus have:

  (1) +(Y, U) |>| Y
  (2) +(Y, 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(map#) = 2

We thus have:

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

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(V1, W1) |>|  W1
  (2) Y2           |>=| Y2
  (3) W2           |>=| W2

We may remove the strictly oriented DPs, which yields:

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

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

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