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#(cons(f(cons(nil, Y)), U)) => copy#(n, Y, U)
      (2) copy#(0, V, W) => f#(W)
      (3) copy#(s(P), X1, Y1) => f#(X1)
      (4) copy#(s(P), X1, Y1) => copy#(P, X1, cons(f(X1), Y1))
      (5) map#(H1, cons(W1, P1)) => map#(H1, P1)
      (6) filter#(Z2, cons(U2, V2)) => filter2#(Z2(U2), Z2, U2, V2)
      (7) filter2#(true, I2, P2, X3) => filter#(I2, X3)
      (8) filter2#(false, Z3, U3, V3) => filter#(Z3, V3)

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

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

  P2. (1) copy#(s(P), X1, Y1) => copy#(P, X1, cons(f(X1), Y1))

  P3. (1) map#(H1, cons(W1, P1)) => map#(H1, P1)

  P4. (1) filter#(Z2, cons(U2, V2)) => filter2#(Z2(U2), Z2, U2, V2)
      (2) filter2#(true, I2, P2, X3) => filter#(I2, X3)
      (3) filter2#(false, Z3, U3, V3) => filter#(Z3, V3)

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

We use the following projection function:

  nu(copy#) = 1

We thus have:

  (1) s(P) |>| P

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(W1, P1) |>| P1

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(U2, V2) |>|  V2
  (2) X3           |>=| X3
  (3) V3           |>=| V3

We may remove the strictly oriented DPs, which yields:

  P5. (1) filter2#(true, I2, P2, X3) => filter#(I2, X3)
      (2) filter2#(false, Z3, U3, V3) => filter#(Z3, V3)

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

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