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)  f#(s(V)) => f#(V)
      (3)  f#(s(V)) => g#(f(V))
      (4)  f#(s(V)) => minus#(s(V), g(f(V)))
      (5)  g#(s(W)) => g#(W)
      (6)  g#(s(W)) => f#(g(W))
      (7)  g#(s(W)) => minus#(s(W), f(g(W)))
      (8)  map#(F1, cons(Y1, U1)) => map#(F1, U1)
      (9)  filter#(I1, cons(P1, X2)) => filter2#(I1(P1), I1, P1, X2)
      (10) filter2#(true, Z2, U2, V2) => filter#(Z2, V2)
      (11) filter2#(false, I2, P2, X3) => filter#(I2, X3)

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

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

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

  P3. (1) f#(s(V)) => f#(V)
      (2) f#(s(V)) => g#(f(V))
      (3) g#(s(W)) => g#(W)
      (4) g#(s(W)) => f#(g(W))

  P4. (1) map#(F1, cons(Y1, U1)) => map#(F1, U1)

  P5. (1) filter#(I1, cons(P1, X2)) => filter2#(I1(P1), I1, P1, X2)
      (2) filter2#(true, Z2, U2, V2) => filter#(Z2, V2)
      (3) filter2#(false, I2, P2, X3) => filter#(I2, X3)

***** 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.