We consider termination of the LCTRS with only rule scheme Calc:

  Signature: cons :: Int -> list -> list
             nil  :: list
             take :: Int -> list -> list

  Rules: take(n, nil) -> nil | n = n
         take(n, l) -> nil | n <= 0
         take(n, cons(x, l)) -> cons(x, take(n - 1, l)) | n > 0 /\ x = x

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) take#(n, cons(x, l)) => take#(n - 1, l) | n > 0 /\ x = x

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

We use the following projection function:

  nu(take#) = 2

We thus have:

  (1) cons(x, l) |>| l

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

Processor output: { }.