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) map#(Z3, cons(U3, V3)) => map#(Z3, V3)
      (2) filter#(J3, cons(X4, Y4)) => filter2#(J3(X4), J3, X4, Y4)
      (3) filter2#(true, G4, V4, W4) => filter#(G4, W4)
      (4) filter2#(false, J4, X5, Y5) => filter#(J4, Y5)

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

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

  P2. (1) map#(Z3, cons(U3, V3)) => map#(Z3, V3)

  P3. (1) filter#(J3, cons(X4, Y4)) => filter2#(J3(X4), J3, X4, Y4)
      (2) filter2#(true, G4, V4, W4) => filter#(G4, W4)
      (3) filter2#(false, J4, X5, Y5) => filter#(J4, Y5)

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

We use the following projection function:

  nu(map#) = 2

We thus have:

  (1) cons(U3, V3) |>| V3

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(X4, Y4) |>|  Y4
  (2) W4           |>=| W4
  (3) Y5           |>=| Y5

We may remove the strictly oriented DPs, which yields:

  P4. (1) filter2#(true, G4, V4, W4) => filter#(G4, W4)
      (2) filter2#(false, J4, X5, Y5) => filter#(J4, Y5)

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

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