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

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

  Rules: f(0, 1, g(X, Y), U) -> f(g(X, Y), g(X, Y), g(X, Y), h(X))
         g(0, 1) -> 0
         g(0, 1) -> 1
         h(g(V, W)) -> h(V)
         map(J, nil) -> nil
         map(F1, cons(Y1, U1)) -> cons(F1(Y1), map(F1, U1))
         filter(H1, nil) -> nil
         filter(I1, cons(P1, X2)) -> filter2(I1(P1), I1, P1, X2)
         filter2(true, Z2, U2, V2) -> cons(U2, filter(Z2, V2))
         filter2(false, I2, P2, X3) -> filter(I2, X3)

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, g(X, Y), U) => g#(X, Y)
      (2)  f#(0, 1, g(X, Y), U) => g#(X, Y)
      (3)  f#(0, 1, g(X, Y), U) => g#(X, Y)
      (4)  f#(0, 1, g(X, Y), U) => h#(X)
      (5)  f#(0, 1, g(X, Y), U) => f#(g(X, Y), g(X, Y), g(X, Y), h(X))
      (6)  h#(g(V, W)) => h#(V)
      (7)  map#(F1, cons(Y1, U1)) => map#(F1, U1)
      (8)  filter#(I1, cons(P1, X2)) => filter2#(I1(P1), I1, P1, X2)
      (9)  filter2#(true, Z2, U2, V2) => filter#(Z2, V2)
      (10) filter2#(false, I2, P2, X3) => filter#(I2, X3)

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

We compute a graph approximation with the following edges:

  1:
  2:
  3:
  4:  6 
  5:  1 2 3 4 5 
  6:  6 
  7:  7 
  8:  9 10 
  9:  8 
  10: 8 

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) h#(g(V, W)) => h#(V)

  P3. (1) f#(0, 1, g(X, Y), U) => f#(g(X, Y), g(X, Y), g(X, Y), h(X))

  P4. (1) map#(F1, cons(Y1, U1)) => map#(F1, U1)

  P5. (1) filter#(I1, cons(P1, X2)) => filter2#(I1(P1), I1, P1, X2)
      (2) filter2#(true, Z2, U2, V2) => filter#(Z2, V2)
      (3) filter2#(false, I2, P2, X3) => filter#(I2, X3)

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

We use the following projection function:

  nu(h#) = 1

We thus have:

  (1) g(V, W) |>| V

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

Processor output: { }.

***** No progress could be made on DP problem D3 = (P3, R, i, c).