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#(g(X)) => f#(X)
      (2) map#(H, cons(W, P)) => map#(H, P)
      (3) filter#(Z1, cons(U1, V1)) => filter2#(Z1(U1), Z1, U1, V1)
      (4) filter2#(true, I1, P1, X2) => filter#(I1, X2)
      (5) filter2#(false, Z2, U2, V2) => filter#(Z2, V2)

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

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

  P2. (1) f#(g(X)) => f#(X)

  P3. (1) map#(H, cons(W, P)) => map#(H, P)

  P4. (1) filter#(Z1, cons(U1, V1)) => filter2#(Z1(U1), Z1, U1, V1)
      (2) filter2#(true, I1, P1, X2) => filter#(I1, X2)
      (3) filter2#(false, Z2, U2, V2) => filter#(Z2, V2)

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

We use the following projection function:

  nu(f#) = 1

We thus have:

  (1) g(X) |>| X

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(W, P) |>| P

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(U1, V1) |>|  V1
  (2) X2           |>=| X2
  (3) V2           |>=| V2

We may remove the strictly oriented DPs, which yields:

  P5. (1) filter2#(true, I1, P1, X2) => filter#(I1, X2)
      (2) filter2#(false, Z2, U2, V2) => filter#(Z2, V2)

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

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