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)  g#(h(g(X))) => g#(X)
      (2)  g#(g(Y)) => g#(Y)
      (3)  g#(g(Y)) => h#(g(Y))
      (4)  g#(g(Y)) => g#(h(g(Y)))
      (5)  h#(h(U)) => h#(U)
      (6)  h#(h(U)) => h#(f(h(U), U))
      (7)  map#(I, cons(P, X1)) => map#(I, X1)
      (8)  filter#(G1, cons(V1, W1)) => filter2#(G1(V1), G1, V1, W1)
      (9)  filter2#(true, J1, X2, Y2) => filter#(J1, Y2)
      (10) filter2#(false, G2, V2, W2) => filter#(G2, W2)

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

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

  P2. (1) h#(h(U)) => h#(U)

  P3. (1) g#(h(g(X))) => g#(X)
      (2) g#(g(Y)) => g#(Y)
      (3) g#(g(Y)) => g#(h(g(Y)))

  P4. (1) map#(I, cons(P, X1)) => map#(I, X1)

  P5. (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(h#) = 1

We thus have:

  (1) h(U) |>| U

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

***** No progress could be made on DP problem P3.