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

  Signature: cond    :: Bool -> Int -> o -> o
             cons    :: o -> o -> o
             length  :: o -> Int
             nil     :: o
             nthtail :: Int -> o -> o
             tail    :: o -> o

  Rules: nthtail(n, l) -> cond(n >= length(l), n, l) | n = n
         cond(true, n, l) -> l | n = n
         cond(false, n, l) -> tail(nthtail(n + 1, l)) | n = n
         tail(nil) -> nil
         tail(cons(x, l)) -> l
         length(nil) -> 0
         length(cons(x, l)) -> 1 + length(l)

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) nthtail#(n, l) => length#(l) | n = n
      (2) nthtail#(n, l) => cond#(n >= length(l), n, l) | n = n
      (3) cond#(false, n, l) => nthtail#(n + 1, l) | n = n
      (4) cond#(false, n, l) => tail#(nthtail(n + 1, l)) | n = n
      (5) length#(cons(x, l)) => length#(l)

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

We compute a graph approximation with the following edges:

  1: 5 
  2:
  3: 1 2 
  4:
  5: 5 

There is only one SCC, so all DPs not inside the SCC can be removed.

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

  P2. (1) length#(cons(x, l)) => length#(l)

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

We use the following projection function:

  nu(length#) = 1

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: { }.