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

  Signature: comp  :: (c -> c) -> (c -> c) -> c -> c
             cons  :: a -> b -> b
             map   :: (a -> a) -> b -> b
             nil   :: b
             twice :: (c -> c) -> c -> c

  Rules: map(F, nil) -> nil
         map(Z, cons(U, V)) -> cons(Z(U), map(Z, V))
         comp(I, J, X1) -> I(J(X1))
         twice(Z1) -> comp(Z1, Z1)

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) map#(Z, cons(U, V)) => map#(Z, V)
      (2) twice#(Z1, arg2) => comp#(Z1, Z1, arg2)

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

We compute a graph approximation with the following edges:

  1: 1 
  2:

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) map#(Z, cons(U, V)) => map#(Z, V)

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

We use the following projection function:

  nu(map#) = 2

We thus have:

  (1) cons(U, V) |>| V

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

Processor output: { }.