We consider termination of the STRS with no additional rule schemes:

  Signature: ack :: N -> N -> N
             s   :: N -> N
             z   :: N

  Rules: ack(z, X) -> s(X)
         ack(s(Y), z) -> ack(Y, s(z))
         ack(s(U), s(V)) -> ack(U, ack(s(U), V))

The system is accessible function passing by a sort ordering that equates all sorts.

We start by computing the initial DP problem D1 = (P1, R, i, c), where:

  P1. (1) ack#(s(Y), z) => ack#(Y, s(z))
      (2) ack#(s(U), s(V)) => ack#(s(U), V)
      (3) ack#(s(U), s(V)) => ack#(U, ack(s(U), V))

***** We apply the Subterm Criterion Processor on D1 = (P1, R, i, c).

We use the following projection function:

  nu(ack#) = 1

We thus have:

  (1) s(Y) |>|  Y
  (2) s(U) |>=| s(U)
  (3) s(U) |>|  U

We may remove the strictly oriented DPs.

Processor output: { D2 = (P2, R, i, c) }, where:

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

***** We apply the Subterm Criterion Processor on D2 = (P2, R, i, c).

We use the following projection function:

  nu(ack#) = 2

We thus have:

  (1) s(V) |>| V

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

Processor output: { }.