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

  Signature: append :: b -> b -> b
             cons   :: a -> b -> b
             map    :: (a -> a) -> b -> b
             nil    :: b

  Rules: append(nil, X) -> X
         append(cons(Y, V), U) -> cons(Y, append(V, U))
         map(I, nil) -> nil
         map(J, cons(X1, Y1)) -> cons(J(X1), map(J, Y1))
         append(append(U1, V1), W1) -> append(U1, append(V1, W1))
         map(J1, append(X2, Y2)) -> append(map(J1, X2), map(J1, Y2))

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) append#(cons(Y, V), U) => append#(V, U)
      (2) map#(J, cons(X1, Y1)) => map#(J, Y1)
      (3) append#(append(U1, V1), W1) => append#(V1, W1)
      (4) append#(append(U1, V1), W1) => append#(U1, append(V1, W1))
      (5) map#(J1, append(X2, Y2)) => map#(J1, X2)
      (6) map#(J1, append(X2, Y2)) => map#(J1, Y2)
      (7) map#(J1, append(X2, Y2)) => append#(map(J1, X2), map(J1, Y2))

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

We compute a graph approximation with the following edges:

  1: 1 3 4 
  2: 2 5 6 7 
  3: 1 3 4 
  4: 1 3 4 
  5: 2 5 6 7 
  6: 2 5 6 7 
  7: 1 3 4 

There are 2 SCCs.

Processor output: { D2 = (P2, R, i, c) ; D3 = (P3, R, i, c) }, where:

  P2. (1) append#(cons(Y, V), U) => append#(V, U)
      (2) append#(append(U1, V1), W1) => append#(V1, W1)
      (3) append#(append(U1, V1), W1) => append#(U1, append(V1, W1))

  P3. (1) map#(J, cons(X1, Y1)) => map#(J, Y1)
      (2) map#(J1, append(X2, Y2)) => map#(J1, X2)
      (3) map#(J1, append(X2, Y2)) => map#(J1, Y2)

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

We use the following projection function:

  nu(append#) = 1

We thus have:

  (1) cons(Y, V)     |>| V
  (2) append(U1, V1) |>| V1
  (3) append(U1, V1) |>| U1

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

Processor output: { }.

***** We apply the Subterm Criterion Processor on D3 = (P3, R, i, c).

We use the following projection function:

  nu(map#) = 2

We thus have:

  (1) cons(X1, Y1)   |>| Y1
  (2) append(X2, Y2) |>| X2
  (3) append(X2, Y2) |>| Y2

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

Processor output: { }.