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)  fact#(s(Y)) => p#(s(Y))
      (2)  fact#(s(Y)) => fact#(p(s(Y)))
      (3)  fact#(s(Y)) => *#(s(Y), fact(p(s(Y))))
      (4)  *#(s(V), W) => *#(V, W)
      (5)  *#(s(V), W) => +#(*(V, W), W)
      (6)  +#(X1, s(Y1)) => +#(X1, Y1)
      (7)  map#(H1, cons(W1, P1)) => map#(H1, P1)
      (8)  filter#(Z2, cons(U2, V2)) => filter2#(Z2(U2), Z2, U2, V2)
      (9)  filter2#(true, I2, P2, X3) => filter#(I2, X3)
      (10) filter2#(false, Z3, U3, V3) => filter#(Z3, V3)

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

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

  P2. (1) +#(X1, s(Y1)) => +#(X1, Y1)

  P3. (1) *#(s(V), W) => *#(V, W)

  P4. (1) fact#(s(Y)) => fact#(p(s(Y)))

  P5. (1) map#(H1, cons(W1, P1)) => map#(H1, P1)

  P6. (1) filter#(Z2, cons(U2, V2)) => filter2#(Z2(U2), Z2, U2, V2)
      (2) filter2#(true, I2, P2, X3) => filter#(I2, X3)
      (3) filter2#(false, Z3, U3, V3) => filter#(Z3, V3)

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

We use the following projection function:

  nu(+#) = 2

We thus have:

  (1) s(Y1) |>| Y1

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(*#) = 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.