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#(X, c(X), c(Y)) => f#(Y, X, Y)
      (2) f#(X, c(X), c(Y)) => f#(Y, Y, f(Y, X, Y))
      (3) f#(s(U), V, W) => f#(U, s(c(V)), c(W))
      (4) map#(F2, cons(Y2, U2)) => map#(F2, U2)
      (5) filter#(I2, cons(P2, X3)) => filter2#(I2(P2), I2, P2, X3)
      (6) filter2#(true, Z3, U3, V3) => filter#(Z3, V3)
      (7) filter2#(false, I3, P3, X4) => filter#(I3, X4)

***** 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(U), V, W) => f#(U, s(c(V)), c(W))

  P3. (1) f#(X, c(X), c(Y)) => f#(Y, X, Y)
      (2) f#(X, c(X), c(Y)) => f#(Y, Y, f(Y, X, Y))

  P4. (1) map#(F2, cons(Y2, U2)) => map#(F2, U2)

  P5. (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(f#) = 1

We thus have:

  (1) s(U) |>| U

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

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