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#(Z, cons(U, V)) => map#(Z, V)
      (2) append#(cons(X1, Y1), U1) => append#(Y1, U1)
      (3) zip#(cons(P1, X2), cons(Y2, U2)) => append#(P1, Y2)
      (4) zip#(cons(P1, X2), cons(Y2, U2)) => zip#(X2, U2)
      (5) combine#(W2, cons(P2, X3)) => zip#(W2, P2)
      (6) combine#(W2, cons(P2, X3)) => combine#(zip(W2, P2), X3)
      (7) levels#(node(Y3, U3)) => levels#(X{21})
      (8) levels#(node(Y3, U3)) => map#(levels, U3)
      (9) levels#(node(Y3, U3)) => combine#(nil, map(levels, U3))

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

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

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

  P3. (1) append#(cons(X1, Y1), U1) => append#(Y1, U1)

  P4. (1) zip#(cons(P1, X2), cons(Y2, U2)) => zip#(X2, U2)

  P5. (1) combine#(W2, cons(P2, X3)) => combine#(zip(W2, P2), X3)

  P6. (1) levels#(node(Y3, U3)) => levels#(X{21})

***** 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(append#) = 1

We thus have:

  (1) cons(X1, Y1) |>| Y1

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(zip#) = 1

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 P5.

We use the following projection function:

  nu(combine#) = 2

We thus have:

  (1) cons(P2, X3) |>| X3

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

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