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)  f#(s(X)) => f#(X)
      (2)  g#(P, c(X1)) => g#(P, X1)
      (3)  g#(Y1, c(U1)) => f#(Y1)
      (4)  g#(Y1, c(U1)) => g#(s(Y1), U1)
      (5)  g#(Y1, c(U1)) => if#(f(Y1), c(g(s(Y1), U1)), c(U1))
      (6)  g#(Y1, c(U1)) => g#(Y1, if(f(Y1), c(g(s(Y1), U1)), c(U1)))
      (7)  map#(I1, cons(P1, X2)) => map#(I1, X2)
      (8)  filter#(G2, cons(V2, W2)) => filter2#(G2(V2), G2, V2, W2)
      (9)  filter2#(true, J2, X3, Y3) => filter#(J2, Y3)
      (10) filter2#(false, G3, V3, W3) => filter#(G3, W3)

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

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

  P2. (1) f#(s(X)) => f#(X)

  P3. (1) g#(P, c(X1)) => g#(P, X1)
      (2) g#(Y1, c(U1)) => g#(s(Y1), U1)
      (3) g#(Y1, c(U1)) => g#(Y1, if(f(Y1), c(g(s(Y1), U1)), c(U1)))

  P4. (1) map#(I1, cons(P1, X2)) => map#(I1, X2)

  P5. (1) filter#(G2, cons(V2, W2)) => filter2#(G2(V2), G2, V2, W2)
      (2) filter2#(true, J2, X3, Y3) => filter#(J2, Y3)
      (3) filter2#(false, G3, V3, W3) => filter#(G3, W3)

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

We use the following projection function:

  nu(f#) = 1

We thus have:

  (1) s(X) |>| X

All DPs are strictly oriented, and may be removed.  Hence, this DP problem is finite.

***** No progress could be made on DP problem P3.