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#(I1, cons(P1, X2)) => map#(I1, X2)
      (2) filter#(G2, cons(V2, W2)) => filter2#(G2(V2), G2, V2, W2)
      (3) filter2#(true, J2, X3, Y3) => filter#(J2, Y3)
      (4) filter2#(false, G3, V3, W3) => filter#(G3, W3)

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

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

  P2. (1) map#(I1, cons(P1, X2)) => map#(I1, X2)

  P3. (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(map#) = 2

We thus have:

  (1) cons(P1, X2) |>| X2

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(V2, W2) |>|  W2
  (2) Y3           |>=| Y3
  (3) W3           |>=| W3

We may remove the strictly oriented DPs, which yields:

  P4. (1) filter2#(true, J2, X3, Y3) => filter#(J2, Y3)
      (2) filter2#(false, G3, V3, W3) => filter#(G3, W3)

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

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