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

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

  Rules: f(0, 1, X) -> f(X, X, X)
         f(Y, U, V) -> 2
         0 -> 2
         1 -> 2
         g(W, W, P) -> P
         g(X1, Y1, Y1) -> X1
         map(G1, nil) -> nil
         map(H1, cons(W1, P1)) -> cons(H1(W1), map(H1, P1))
         filter(F2, nil) -> nil
         filter(Z2, cons(U2, V2)) -> filter2(Z2(U2), Z2, U2, V2)
         filter2(true, I2, P2, X3) -> cons(P2, filter(I2, X3))
         filter2(false, Z3, U3, V3) -> filter(Z3, V3)

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, X) => f#(X, X, X)
      (2) map#(H1, cons(W1, P1)) => map#(H1, P1)
      (3) filter#(Z2, cons(U2, V2)) => filter2#(Z2(U2), Z2, U2, V2)
      (4) filter2#(true, I2, P2, X3) => filter#(I2, X3)
      (5) filter2#(false, Z3, U3, V3) => filter#(Z3, V3)

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

We compute a graph approximation with the following edges:

  1: 1 
  2: 2 
  3: 4 5 
  4: 3 
  5: 3 

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#(0, 1, X) => f#(X, X, X)

  P3. (1) map#(H1, cons(W1, P1)) => map#(H1, P1)

  P4. (1) filter#(Z2, cons(U2, V2)) => filter2#(Z2(U2), Z2, U2, V2)
      (2) filter2#(true, I2, P2, X3) => filter#(I2, X3)
      (3) filter2#(false, Z3, U3, V3) => filter#(Z3, V3)

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

We obtain 0 usable rules (out of 12 rules in the input problem).

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

***** 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(W1, P1) |>| P1

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(U2, V2) |>|  V2
  (2) X3           |>=| X3
  (3) V3           |>=| V3

We may remove the strictly oriented DPs.

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

  P5. (1) filter2#(true, I2, P2, X3) => filter#(I2, X3)
      (2) filter2#(false, Z3, U3, V3) => filter#(Z3, V3)

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