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) rev#(cons(Y, X)) => rev1#(Y, X)
      (2) rev#(cons(Y, X)) => rev2#(Y, X)
      (3) rev1#(W, cons(P, V)) => rev1#(P, V)
      (4) rev2#(U1, cons(V1, Y1)) => rev2#(V1, Y1)
      (5) rev2#(U1, cons(V1, Y1)) => rev#(cons(U1, rev2(V1, Y1)))
      (6) map#(J1, cons(X2, Y2)) => map#(J1, Y2)
      (7) filter#(H2, cons(W2, P2)) => filter2#(H2(W2), H2, W2, P2)
      (8) filter2#(true, F3, Y3, U3) => filter#(F3, U3)
      (9) filter2#(false, H3, W3, P3) => filter#(H3, P3)

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

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

  P2. (1) rev1#(W, cons(P, V)) => rev1#(P, V)

  P3. (1) rev#(cons(Y, X)) => rev2#(Y, X)
      (2) rev2#(U1, cons(V1, Y1)) => rev2#(V1, Y1)
      (3) rev2#(U1, cons(V1, Y1)) => rev#(cons(U1, rev2(V1, Y1)))

  P4. (1) map#(J1, cons(X2, Y2)) => map#(J1, Y2)

  P5. (1) filter#(H2, cons(W2, P2)) => filter2#(H2(W2), H2, W2, P2)
      (2) filter2#(true, F3, Y3, U3) => filter#(F3, U3)
      (3) filter2#(false, H3, W3, P3) => filter#(H3, P3)

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

We use the following projection function:

  nu(rev1#) = 2

We thus have:

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

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

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