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

  Signature: 0       :: a
             1       :: a
             cons    :: c -> d -> d
             f       :: a -> a -> a -> a
             false   :: b
             filter  :: (c -> b) -> d -> d
             filter2 :: b -> (c -> b) -> c -> d -> d
             map     :: (c -> c) -> d -> d
             nil     :: d
             s       :: a -> a
             true    :: b

  Rules: f(0, 1, X) -> f(s(X), X, X)
         f(Y, U, s(V)) -> s(f(0, 1, V))
         map(I, nil) -> nil
         map(J, cons(X1, Y1)) -> cons(J(X1), map(J, Y1))
         filter(G1, nil) -> nil
         filter(H1, cons(W1, P1)) -> filter2(H1(W1), H1, W1, P1)
         filter2(true, F2, Y2, U2) -> cons(Y2, filter(F2, U2))
         filter2(false, H2, W2, P2) -> filter(H2, P2)

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) f#(0, 1, X) => f#(s(X), X, X)
      (2) f#(Y, U, s(V)) => f#(0, 1, V)
      (3) map#(J, cons(X1, Y1)) => map#(J, Y1)
      (4) filter#(H1, cons(W1, P1)) => filter2#(H1(W1), H1, W1, P1)
      (5) filter2#(true, F2, Y2, U2) => filter#(F2, U2)
      (6) filter2#(false, H2, W2, P2) => filter#(H2, P2)

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

We compute a graph approximation with the following edges:

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

There are 3 SCCs.

Processor output: { D2 = (P2, R, i, c) ; D3 = (P3, R, i, c) ; D4 = (P4, R, i, c) }, where:

  P2. (1) f#(0, 1, X) => f#(s(X), X, X)
      (2) f#(Y, U, s(V)) => f#(0, 1, V)

  P3. (1) map#(J, cons(X1, Y1)) => map#(J, Y1)

  P4. (1) filter#(H1, cons(W1, P1)) => filter2#(H1(W1), H1, W1, P1)
      (2) filter2#(true, F2, Y2, U2) => filter#(F2, U2)
      (3) filter2#(false, H2, W2, P2) => filter#(H2, P2)

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

We use the following projection function:

  nu(f#) = 3

We thus have:

  (1) X    |>=| X
  (2) s(V) |>|  V

We may remove the strictly oriented DPs.

Processor output: { D5 = (P5, R, i, c) }, where:

  P5. (1) f#(0, 1, X) => f#(s(X), X, X)

***** 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

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

Processor output: { }.

***** We apply the Subterm Criterion Processor on D4 = (P4, R, i, c).

We use the following projection function:

  nu(filter#)  = 2
  nu(filter2#) = 4

We thus have:

  (1) cons(W1, P1) |>|  P1
  (2) U2           |>=| U2
  (3) P2           |>=| P2

We may remove the strictly oriented DPs.

Processor output: { D6 = (P6, R, i, c) }, where:

  P6. (1) filter2#(true, F2, Y2, U2) => filter#(F2, U2)
      (2) filter2#(false, H2, W2, P2) => filter#(H2, P2)

***** We apply the Graph Processor on D5 = (P5, R, i, c).

We compute a graph approximation with the following edges:

  1:

As there are no SCCs, this DP problem is removed.

Processor output: { }.

***** We apply the Graph Processor on D6 = (P6, R, i, c).

We compute a graph approximation with the following edges:

  1:
  2:

As there are no SCCs, this DP problem is removed.

Processor output: { }.