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)  le#(s(U), s(V)) => le#(U, V)
      (2)  eq#(s(X1), s(Y1)) => eq#(X1, Y1)
      (3)  minsort#(cons(X2, Y2)) => min#(X2, Y2)
      (4)  minsort#(cons(X2, Y2)) => min#(X2, Y2)
      (5)  minsort#(cons(X2, Y2)) => del#(min(X2, Y2), cons(X2, Y2))
      (6)  minsort#(cons(X2, Y2)) => minsort#(del(min(X2, Y2), cons(X2, Y2)))
      (7)  min#(V2, cons(W2, P2)) => le#(V2, W2)
      (8)  min#(V2, cons(W2, P2)) => min#(V2, P2)
      (9)  min#(V2, cons(W2, P2)) => min#(W2, P2)
      (10) min#(V2, cons(W2, P2)) => if#(le(V2, W2), min(V2, P2), min(W2, P2))
      (11) del#(Y3, cons(U3, V3)) => eq#(Y3, U3)
      (12) del#(Y3, cons(U3, V3)) => del#(Y3, V3)
      (13) del#(Y3, cons(U3, V3)) => if#(eq(Y3, U3), V3, cons(U3, del(Y3, V3)))
      (14) map#(J3, cons(X4, Y4)) => map#(J3, Y4)
      (15) filter#(H4, cons(W4, P4)) => filter2#(H4(W4), H4, W4, P4)
      (16) filter2#(true, F5, Y5, U5) => filter#(F5, U5)
      (17) filter2#(false, H5, W5, P5) => filter#(H5, P5)

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

Considering the 7 SCCs, this DP problem is split into the following new problems.

  P2. (1) le#(s(U), s(V)) => le#(U, V)

  P3. (1) eq#(s(X1), s(Y1)) => eq#(X1, Y1)

  P4. (1) min#(V2, cons(W2, P2)) => min#(V2, P2)
      (2) min#(V2, cons(W2, P2)) => min#(W2, P2)

  P5. (1) del#(Y3, cons(U3, V3)) => del#(Y3, V3)

  P6. (1) minsort#(cons(X2, Y2)) => minsort#(del(min(X2, Y2), cons(X2, Y2)))

  P7. (1) map#(J3, cons(X4, Y4)) => map#(J3, Y4)

  P8. (1) filter#(H4, cons(W4, P4)) => filter2#(H4(W4), H4, W4, P4)
      (2) filter2#(true, F5, Y5, U5) => filter#(F5, U5)
      (3) filter2#(false, H5, W5, P5) => filter#(H5, P5)

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

We use the following projection function:

  nu(le#) = 1

We thus have:

  (1) s(U) |>| U

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(eq#) = 1

We thus have:

  (1) s(X1) |>| X1

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

***** We apply the Subterm Criterion Processor on P4.

We use the following projection function:

  nu(min#) = 2

We thus have:

  (1) cons(W2, P2) |>| P2
  (2) cons(W2, P2) |>| P2

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

***** We apply the Subterm Criterion Processor on P5.

We use the following projection function:

  nu(del#) = 2

We thus have:

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

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

***** No progress could be made on DP problem P6.