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

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

  Rules: f(X, g(X)) -> f(1, g(X))
         g(1) -> g(0)
         map(Z, nil) -> nil
         map(G, cons(V, W)) -> cons(G(V), map(G, W))
         filter(J, nil) -> nil
         filter(F1, cons(Y1, U1)) -> filter2(F1(Y1), F1, Y1, U1)
         filter2(true, H1, W1, P1) -> cons(W1, filter(H1, P1))
         filter2(false, F2, Y2, U2) -> filter(F2, U2)

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#(X, g(X)) => g#(X)
      (2) f#(X, g(X)) => f#(1, g(X))
      (3) g#(1) => g#(0)
      (4) map#(G, cons(V, W)) => map#(G, W)
      (5) filter#(F1, cons(Y1, U1)) => filter2#(F1(Y1), F1, Y1, U1)
      (6) filter2#(true, H1, W1, P1) => filter#(H1, P1)
      (7) filter2#(false, F2, Y2, U2) => filter#(F2, U2)

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

We compute a graph approximation with the following edges:

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

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#(X, g(X)) => f#(1, g(X))

  P3. (1) map#(G, cons(V, W)) => map#(G, W)

  P4. (1) filter#(F1, cons(Y1, U1)) => filter2#(F1(Y1), F1, Y1, U1)
      (2) filter2#(true, H1, W1, P1) => filter#(H1, P1)
      (3) filter2#(false, F2, Y2, U2) => filter#(F2, U2)

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

We obtain 1 usable rules (out of 8 rules in the input problem).

Processor output: { D5 = (P2, R2, i, c) }, where:

  R2. (1) g(1) -> g(0)

***** 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(V, W) |>| W

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(Y1, U1) |>|  U1
  (2) P1           |>=| P1
  (3) U2           |>=| U2

We may remove the strictly oriented DPs.

Processor output: { D6 = (P5, R, i, c) }, where:

  P5. (1) filter2#(true, H1, W1, P1) => filter#(H1, P1)
      (2) filter2#(false, F2, Y2, U2) => filter#(F2, U2)

***** No progress could be made on DP problem D5 = (P2, R2, i, c).