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) append#(cons(Y, V), U) => append#(V, U)
      (2) map#(J, cons(X1, Y1)) => map#(J, Y1)
      (3) append#(append(U1, V1), W1) => append#(V1, W1)
      (4) append#(append(U1, V1), W1) => append#(U1, append(V1, W1))
      (5) map#(J1, append(X2, Y2)) => map#(J1, X2)
      (6) map#(J1, append(X2, Y2)) => map#(J1, Y2)
      (7) map#(J1, append(X2, Y2)) => append#(map(J1, X2), map(J1, Y2))

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

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

  P2. (1) append#(cons(Y, V), U) => append#(V, U)
      (2) append#(append(U1, V1), W1) => append#(V1, W1)
      (3) append#(append(U1, V1), W1) => append#(U1, append(V1, W1))

  P3. (1) map#(J, cons(X1, Y1)) => map#(J, Y1)
      (2) map#(J1, append(X2, Y2)) => map#(J1, X2)
      (3) map#(J1, append(X2, Y2)) => map#(J1, Y2)

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

We use the following projection function:

  nu(append#) = 1

We thus have:

  (1) cons(Y, V)     |>| V
  (2) append(U1, V1) |>| V1
  (3) append(U1, V1) |>| U1

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(map#) = 2

We thus have:

  (1) cons(X1, Y1)   |>| Y1
  (2) append(X2, Y2) |>| X2
  (3) append(X2, Y2) |>| Y2

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