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) minus#(U, s(V)) => minus#(U, V)
      (2) minus#(U, s(V)) => pred#(minus(U, V))
      (3) quot#(s(P), s(X1)) => minus#(P, X1)
      (4) quot#(s(P), s(X1)) => quot#(minus(P, X1), s(X1))
      (5) map#(G1, cons(V1, W1)) => map#(G1, W1)
      (6) filter#(F2, cons(Y2, U2)) => filter2#(F2(Y2), F2, Y2, U2)
      (7) filter2#(true, H2, W2, P2) => filter#(H2, P2)
      (8) filter2#(false, F3, Y3, U3) => filter#(F3, U3)

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

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

  P2. (1) minus#(U, s(V)) => minus#(U, V)

  P3. (1) quot#(s(P), s(X1)) => quot#(minus(P, X1), s(X1))

  P4. (1) map#(G1, cons(V1, W1)) => map#(G1, W1)

  P5. (1) filter#(F2, cons(Y2, U2)) => filter2#(F2(Y2), F2, Y2, U2)
      (2) filter2#(true, H2, W2, P2) => filter#(H2, P2)
      (3) filter2#(false, F3, Y3, U3) => filter#(F3, U3)

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

We use the following projection function:

  nu(minus#) = 2

We thus have:

  (1) s(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.