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

  Signature: 0       :: b
             cons    :: b -> c -> c
             false   :: a
             filter  :: (b -> a) -> c -> c
             filter2 :: a -> (b -> a) -> b -> c -> c
             int     :: b -> b -> c
             intlist :: c -> c
             map     :: (b -> b) -> c -> c
             nil     :: c
             s       :: b -> b
             true    :: a

  Rules: intlist(nil) -> nil
         intlist(cons(X, Y)) -> cons(s(X), intlist(Y))
         int(0, 0) -> cons(0, nil)
         int(0, s(U)) -> cons(0, int(s(0), s(U)))
         int(s(V), 0) -> nil
         int(s(W), s(P)) -> intlist(int(W, P))
         map(F1, nil) -> nil
         map(Z1, cons(U1, V1)) -> cons(Z1(U1), map(Z1, V1))
         filter(I1, nil) -> nil
         filter(J1, cons(X2, Y2)) -> filter2(J1(X2), J1, X2, Y2)
         filter2(true, G2, V2, W2) -> cons(V2, filter(G2, W2))
         filter2(false, J2, X3, Y3) -> filter(J2, Y3)

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) intlist#(cons(X, Y)) => intlist#(Y)
      (2) int#(0, s(U)) => int#(s(0), s(U))
      (3) int#(s(W), s(P)) => int#(W, P)
      (4) int#(s(W), s(P)) => intlist#(int(W, P))
      (5) map#(Z1, cons(U1, V1)) => map#(Z1, V1)
      (6) filter#(J1, cons(X2, Y2)) => filter2#(J1(X2), J1, X2, Y2)
      (7) filter2#(true, G2, V2, W2) => filter#(G2, W2)
      (8) filter2#(false, J2, X3, Y3) => filter#(J2, Y3)

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

We compute a graph approximation with the following edges:

  1: 1 
  2: 3 4 
  3: 2 3 4 
  4: 1 
  5: 5 
  6: 7 8 
  7: 6 
  8: 6 

There are 4 SCCs.

Processor output: { D2 = (P2, R, i, c) ; D3 = (P3, R, i, c) ; D4 = (P4, R, i, c) ; D5 = (P5, R, i, c) }, where:

  P2. (1) intlist#(cons(X, Y)) => intlist#(Y)

  P3. (1) int#(0, s(U)) => int#(s(0), s(U))
      (2) int#(s(W), s(P)) => int#(W, P)

  P4. (1) map#(Z1, cons(U1, V1)) => map#(Z1, V1)

  P5. (1) filter#(J1, cons(X2, Y2)) => filter2#(J1(X2), J1, X2, Y2)
      (2) filter2#(true, G2, V2, W2) => filter#(G2, W2)
      (3) filter2#(false, J2, X3, Y3) => filter#(J2, Y3)

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

We use the following projection function:

  nu(intlist#) = 1

We thus have:

  (1) cons(X, Y) |>| Y

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(int#) = 2

We thus have:

  (1) s(U) |>=| s(U)
  (2) s(P) |>|  P

We may remove the strictly oriented DPs.

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

  P6. (1) int#(0, s(U)) => int#(s(0), s(U))

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

We use the following projection function:

  nu(map#) = 2

We thus have:

  (1) cons(U1, V1) |>| V1

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

Processor output: { }.

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

We use the following projection function:

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

We thus have:

  (1) cons(X2, Y2) |>|  Y2
  (2) W2           |>=| W2
  (3) Y3           |>=| Y3

We may remove the strictly oriented DPs.

Processor output: { D7 = (P7, R, i, c) }, where:

  P7. (1) filter2#(true, G2, V2, W2) => filter#(G2, W2)
      (2) filter2#(false, J2, X3, Y3) => filter#(J2, Y3)

***** We apply the Graph Processor on D6 = (P6, 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 D7 = (P7, 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: { }.