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#(s(Y), s(U)) => minus#(Y, U)
      (2) quot#(s(W), s(P)) => minus#(W, P)
      (3) quot#(s(W), s(P)) => quot#(minus(W, P), s(P))
      (4) log#(s(s(X1))) => quot#(X1, s(s(0)))
      (5) log#(s(s(X1))) => log#(s(quot(X1, s(s(0)))))
      (6) map#(G1, cons(V1, W1)) => map#(G1, W1)
      (7) filter#(F2, cons(Y2, U2)) => filter2#(F2(Y2), F2, Y2, U2)
      (8) filter2#(true, H2, W2, P2) => filter#(H2, P2)
      (9) filter2#(false, F3, Y3, U3) => filter#(F3, U3)

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

Considering the 5 SCCs, this DP problem is split into the following new problems.

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

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

  P4. (1) log#(s(s(X1))) => log#(s(quot(X1, s(s(0)))))

  P5. (1) map#(G1, cons(V1, W1)) => map#(G1, W1)

  P6. (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#) = 1

We thus have:

  (1) s(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 P3.