We consider universal computability of the STRS with no additional rule schemes:

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

  Rules: f(s(X)) -> f(X)
         g(cons(0, Y)) -> g(Y)
         g(cons(s(U), V)) -> s(U)
         h(cons(W, P)) -> h(g(cons(W, P)))
         map(F1, nil) -> nil
         map(Z1, cons(U1, V1)) -> cons(Z1(U1), map(Z1, V1))
         filter(I1, nil) -> nil
         filter(J1, cons(X2, Y2)) -> filter2(J1(X2), J1, X2, Y2)
         filter2(true, G2, V2, W2) -> cons(V2, filter(G2, W2))
         filter2(false, J2, X3, Y3) -> filter(J2, Y3)

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 UNION R_?, f, c), where:

  P1. (1) f#(s(X)) => f#(X)
      (2) g#(cons(0, Y)) => g#(Y)
      (3) h#(cons(W, P)) => g#(cons(W, P))
      (4) h#(cons(W, P)) => h#(g(cons(W, P)))
      (5) map#(Z1, cons(U1, V1)) => map#(Z1, V1)
      (6) filter#(J1, cons(X2, Y2)) => filter2#(J1(X2), J1, X2, Y2)
      (7) filter2#(true, G2, V2, W2) => filter#(G2, W2)
      (8) filter2#(false, J2, X3, Y3) => filter#(J2, Y3)

***** We apply the Graph Processor on D1 = (P1, R UNION R_?, f, c).

We compute a graph approximation with the following edges:

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

There are 5 SCCs.

Processor output: { D2 = (P2, R UNION R_?, f, c) ; D3 = (P3, R UNION R_?, f, c) ; D4 = (P4, R UNION R_?, f, c) ; D5 = (P5, R UNION R_?, f, c) ; D6 = (P6, R UNION R_?, f, c) }, where:

  P2. (1) f#(s(X)) => f#(X)

  P3. (1) g#(cons(0, Y)) => g#(Y)

  P4. (1) h#(cons(W, P)) => h#(g(cons(W, P)))

  P5. (1) map#(Z1, cons(U1, V1)) => map#(Z1, V1)

  P6. (1) filter#(J1, cons(X2, Y2)) => filter2#(J1(X2), J1, X2, Y2)
      (2) filter2#(true, G2, V2, W2) => filter#(G2, W2)
      (3) filter2#(false, J2, X3, Y3) => filter#(J2, Y3)

***** We apply the Subterm Criterion Processor on D2 = (P2, R UNION R_?, f, c).

We use the following projection function:

  nu(f#) = 1

We thus have:

  (1) s(X) |>| X

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 UNION R_?, f, c).

We use the following projection function:

  nu(g#) = 1

We thus have:

  (1) cons(0, Y) |>| Y

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 D4 = (P4, R UNION R_?, f, c).