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

  Signature: *       :: a -> b -> b
             +       :: b -> b -> b
             cons    :: d -> e -> e
             false   :: c
             filter  :: (d -> c) -> e -> e
             filter2 :: c -> (d -> c) -> d -> e -> e
             map     :: (d -> d) -> e -> e
             nil     :: e
             true    :: c

  Rules: *(X, +(Y, U)) -> +(*(X, Y), *(X, U))
         map(H, nil) -> nil
         map(I, cons(P, X1)) -> cons(I(P), map(I, X1))
         filter(Z1, nil) -> nil
         filter(G1, cons(V1, W1)) -> filter2(G1(V1), G1, V1, W1)
         filter2(true, J1, X2, Y2) -> cons(X2, filter(J1, Y2))
         filter2(false, G2, V2, W2) -> filter(G2, W2)

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) *#(X, +(Y, U)) => *#(X, Y)
      (2) *#(X, +(Y, U)) => *#(X, U)
      (3) map#(I, cons(P, X1)) => map#(I, X1)
      (4) filter#(G1, cons(V1, W1)) => filter2#(G1(V1), G1, V1, W1)
      (5) filter2#(true, J1, X2, Y2) => filter#(J1, Y2)
      (6) filter2#(false, G2, V2, W2) => filter#(G2, W2)

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

We compute a graph approximation with the following edges:

  1: 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) *#(X, +(Y, U)) => *#(X, Y)
      (2) *#(X, +(Y, U)) => *#(X, U)

  P3. (1) map#(I, cons(P, X1)) => map#(I, X1)

  P4. (1) filter#(G1, cons(V1, W1)) => filter2#(G1(V1), G1, V1, W1)
      (2) filter2#(true, J1, X2, Y2) => filter#(J1, Y2)
      (3) filter2#(false, G2, V2, W2) => filter#(G2, W2)

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

We use the following projection function:

  nu(*#) = 2

We thus have:

  (1) +(Y, U) |>| Y
  (2) +(Y, U) |>| U

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(P, X1) |>| X1

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(V1, W1) |>|  W1
  (2) Y2           |>=| Y2
  (3) W2           |>=| W2

We may remove the strictly oriented DPs.

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

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

***** 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: { }.