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

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

  Rules: f(g(X), s(0), Y) -> f(Y, Y, g(X))
         g(s(U)) -> s(g(U))
         g(0) -> 0
         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) f#(g(X), s(0), Y) => g#(X)
      (2) f#(g(X), s(0), Y) => f#(Y, Y, g(X))
      (3) g#(s(U)) => g#(U)
      (4) map#(I, cons(P, X1)) => map#(I, X1)
      (5) filter#(G1, cons(V1, W1)) => filter2#(G1(V1), G1, V1, W1)
      (6) filter2#(true, J1, X2, Y2) => filter#(J1, Y2)
      (7) 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: 3 
  2: 1 2 
  3: 3 
  4: 4 
  5: 6 7 
  6: 5 
  7: 5 

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) g#(s(U)) => g#(U)

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

  P4. (1) map#(I, cons(P, X1)) => map#(I, X1)

  P5. (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(g#) = 1

We thus have:

  (1) s(U) |>| U

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

Processor output: { }.

***** We apply the Usable Rules Processor on D3 = (P3, R, i, c).

We obtain 2 usable rules (out of 9 rules in the input problem).

Processor output: { D6 = (P3, R2, i, c) }, where:

  R2. (1) g(s(U)) -> s(g(U))
      (2) g(0) -> 0

***** We apply the Subterm Criterion Processor on D4 = (P4, 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 D5 = (P5, 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: { D7 = (P6, R, i, c) }, where:

  P6. (1) filter2#(true, J1, X2, Y2) => filter#(J1, Y2)
      (2) filter2#(false, G2, V2, W2) => filter#(G2, W2)

***** No progress could be made on DP problem D6 = (P3, R2, i, c).