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)  double#(s(V)) => double#(V)
      (3)  plus#(s(P), X1) => plus#(P, X1)
      (4)  plus#(s(Y1), U1) => plus#(Y1, s(U1))
      (5)  plus#(s(V1), W1) => minus#(V1, W1)
      (6)  plus#(s(V1), W1) => double#(W1)
      (7)  plus#(s(V1), W1) => plus#(minus(V1, W1), double(W1))
      (8)  map#(F2, cons(Y2, U2)) => map#(F2, U2)
      (9)  filter#(I2, cons(P2, X3)) => filter2#(I2(P2), I2, P2, X3)
      (10) filter2#(true, Z3, U3, V3) => filter#(Z3, V3)
      (11) filter2#(false, I3, P3, X4) => filter#(I3, X4)

***** 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) double#(s(V)) => double#(V)

  P4. (1) plus#(s(P), X1) => plus#(P, X1)
      (2) plus#(s(Y1), U1) => plus#(Y1, s(U1))
      (3) plus#(s(V1), W1) => plus#(minus(V1, W1), double(W1))

  P5. (1) map#(F2, cons(Y2, U2)) => map#(F2, U2)

  P6. (1) filter#(I2, cons(P2, X3)) => filter2#(I2(P2), I2, P2, X3)
      (2) filter2#(true, Z3, U3, V3) => filter#(Z3, V3)
      (3) filter2#(false, I3, P3, X4) => filter#(I3, X4)

***** 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.

***** We apply the Subterm Criterion Processor on P3.

We use the following projection function:

  nu(double#) = 1

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 P4.