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#(a) => f#(b)
      (2) g#(b) => g#(a)
      (3) map#(Z, cons(U, V)) => map#(Z, V)
      (4) filter#(J, cons(X1, Y1)) => filter2#(J(X1), J, X1, Y1)
      (5) filter2#(true, G1, V1, W1) => filter#(G1, W1)
      (6) filter2#(false, J1, X2, Y2) => filter#(J1, Y2)

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

Considering the 2 SCCs, this DP problem is split into the following new problems.

  P2. (1) map#(Z, cons(U, V)) => map#(Z, V)

  P3. (1) filter#(J, cons(X1, Y1)) => filter2#(J(X1), J, X1, Y1)
      (2) filter2#(true, G1, V1, W1) => filter#(G1, W1)
      (3) filter2#(false, J1, X2, Y2) => filter#(J1, Y2)

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

We use the following projection function:

  nu(map#) = 2

We thus have:

  (1) cons(U, V) |>| V

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(filter#)  = 2
  nu(filter2#) = 4

We thus have:

  (1) cons(X1, Y1) |>|  Y1
  (2) W1           |>=| W1
  (3) Y2           |>=| Y2

We may remove the strictly oriented DPs, which yields:

  P4. (1) filter2#(true, G1, V1, W1) => filter#(G1, W1)
      (2) filter2#(false, J1, X2, Y2) => filter#(J1, Y2)

***** We apply the Graph Processor on P4.

As there are no SCCs, this DP problem is removed.