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

  Signature: *        :: a -> a -> a
             +        :: a -> a -> a
             -        :: a -> a -> a
             0        :: a
             1        :: a
             D        :: a -> a
             cons     :: c -> d -> d
             constant :: a
             false    :: b
             filter   :: (c -> b) -> d -> d
             filter2  :: b -> (c -> b) -> c -> d -> d
             map      :: (c -> c) -> d -> d
             nil      :: d
             t        :: a
             true     :: b

  Rules: D(t) -> 1
         D(constant) -> 0
         D(+(X, Y)) -> +(D(X), D(Y))
         D(*(U, V)) -> +(*(V, D(U)), *(U, D(V)))
         D(-(W, P)) -> -(D(W), D(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)  D#(+(X, Y)) => D#(X)
      (2)  D#(+(X, Y)) => D#(Y)
      (3)  D#(*(U, V)) => D#(U)
      (4)  D#(*(U, V)) => D#(V)
      (5)  D#(-(W, P)) => D#(W)
      (6)  D#(-(W, P)) => D#(P)
      (7)  map#(Z1, cons(U1, V1)) => map#(Z1, V1)
      (8)  filter#(J1, cons(X2, Y2)) => filter2#(J1(X2), J1, X2, Y2)
      (9)  filter2#(true, G2, V2, W2) => filter#(G2, W2)
      (10) 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 5 6 
  2:  1 2 3 4 5 6 
  3:  1 2 3 4 5 6 
  4:  1 2 3 4 5 6 
  5:  1 2 3 4 5 6 
  6:  1 2 3 4 5 6 
  7:  7 
  8:  9 10 
  9:  8 
  10: 8 

There are 3 SCCs.

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

  P2. (1) D#(+(X, Y)) => D#(X)
      (2) D#(+(X, Y)) => D#(Y)
      (3) D#(*(U, V)) => D#(U)
      (4) D#(*(U, V)) => D#(V)
      (5) D#(-(W, P)) => D#(W)
      (6) D#(-(W, P)) => D#(P)

  P3. (1) map#(Z1, cons(U1, V1)) => map#(Z1, V1)

  P4. (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(D#) = 1

We thus have:

  (1) +(X, Y) |>| X
  (2) +(X, Y) |>| Y
  (3) *(U, V) |>| U
  (4) *(U, V) |>| V
  (5) -(W, P) |>| W
  (6) -(W, P) |>| P

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(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 D4 = (P4, 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: { D5 = (P5, R, i, c) }, where:

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

***** We apply the Graph Processor on D5 = (P5, 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: { }.