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) ack#(succ(Y), U) => ack#(Y, succ(0))
      (2) ack#(succ(V), succ(W)) => ack#(succ(V), W)
      (3) ack#(succ(V), succ(W)) => ack#(V, ack(succ(V), W))
      (4) map#(F1, cons(Y1, U1)) => map#(F1, U1)
      (5) filter#(I1, cons(P1, X2)) => filter2#(I1(P1), I1, P1, X2)
      (6) filter2#(true, Z2, U2, V2) => filter#(Z2, V2)
      (7) filter2#(false, I2, P2, X3) => filter#(I2, X3)

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

Considering the 3 SCCs, this DP problem is split into the following new problems.

  P2. (1) ack#(succ(Y), U) => ack#(Y, succ(0))
      (2) ack#(succ(V), succ(W)) => ack#(succ(V), W)
      (3) ack#(succ(V), succ(W)) => ack#(V, ack(succ(V), W))

  P3. (1) map#(F1, cons(Y1, U1)) => map#(F1, U1)

  P4. (1) filter#(I1, cons(P1, X2)) => filter2#(I1(P1), I1, P1, X2)
      (2) filter2#(true, Z2, U2, V2) => filter#(Z2, V2)
      (3) filter2#(false, I2, P2, X3) => filter#(I2, X3)

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

We use the following projection function:

  nu(ack#) = 1

We thus have:

  (1) succ(Y) |>|  Y
  (2) succ(V) |>=| succ(V)
  (3) succ(V) |>|  V

We may remove the strictly oriented DPs, which yields:

  P5. (1) ack#(succ(V), succ(W)) => ack#(succ(V), W)

***** We apply the Subterm Criterion Processor on P3.

We use the following projection function:

  nu(map#) = 2

We thus have:

  (1) cons(Y1, U1) |>| U1

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(filter#)  = 2
  nu(filter2#) = 4

We thus have:

  (1) cons(P1, X2) |>|  X2
  (2) V2           |>=| V2
  (3) X3           |>=| X3

We may remove the strictly oriented DPs, which yields:

  P6. (1) filter2#(true, Z2, U2, V2) => filter#(Z2, V2)
      (2) filter2#(false, I2, P2, X3) => filter#(I2, X3)

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

We use the following projection function:

  nu(ack#) = 2

We thus have:

  (1) succ(W) |>| W

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

***** We apply the Graph Processor on P6.

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