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)  sub#(s(P), s(X1)) => sub#(P, X1)
      (2)  gtr#(s(V1), s(W1)) => gtr#(V1, W1)
      (3)  d#(s(X2), s(Y2)) => gtr#(X2, Y2)
      (4)  d#(s(X2), s(Y2)) => sub#(Y2, X2)
      (5)  d#(s(X2), s(Y2)) => d#(s(X2), sub(Y2, X2))
      (6)  d#(s(X2), s(Y2)) => if#(gtr(X2, Y2), false, d(s(X2), sub(Y2, X2)))
      (7)  len#(cons(U2, V2)) => len#(V2)
      (8)  filter#(J2, cons(X3, Y3)) => filter#(J2, Y3)
      (9)  filter#(J2, cons(X3, Y3)) => filter#(J2, Y3)
      (10) filter#(J2, cons(X3, Y3)) => if#(J2(X3), cons(X3, filter(J2, Y3)), filter(J2, Y3))

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

Considering the 5 SCCs, this DP problem is split into the following new problems.

  P2. (1) sub#(s(P), s(X1)) => sub#(P, X1)

  P3. (1) gtr#(s(V1), s(W1)) => gtr#(V1, W1)

  P4. (1) d#(s(X2), s(Y2)) => d#(s(X2), sub(Y2, X2))

  P5. (1) len#(cons(U2, V2)) => len#(V2)

  P6. (1) filter#(J2, cons(X3, Y3)) => filter#(J2, Y3)
      (2) filter#(J2, cons(X3, Y3)) => filter#(J2, Y3)

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

We use the following projection function:

  nu(sub#) = 1

We thus have:

  (1) s(P) |>| P

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(gtr#) = 1

We thus have:

  (1) s(V1) |>| V1

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

***** No progress could be made on DP problem P4.