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#(s(X)) => f#(X)
      (2) g#(cons(0, Y)) => g#(Y)
      (3) h#(cons(W, P)) => g#(cons(W, P))
      (4) h#(cons(W, P)) => h#(g(cons(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 5 SCCs, this DP problem is split into the following new problems.

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

  P3. (1) g#(cons(0, Y)) => g#(Y)

  P4. (1) h#(cons(W, P)) => h#(g(cons(W, P)))

  P5. (1) map#(Z1, cons(U1, V1)) => map#(Z1, V1)

  P6. (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(f#) = 1

We thus have:

  (1) s(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(g#) = 1

We thus have:

  (1) cons(0, Y) |>| Y

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

***** No progress could be made on DP problem P4.