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

  Signature: 0       :: a
             bits    :: a -> a
             cons    :: c -> d -> d
             false   :: b
             filter  :: (c -> b) -> d -> d
             filter2 :: b -> (c -> b) -> c -> d -> d
             half    :: a -> a
             map     :: (c -> c) -> d -> d
             nil     :: d
             s       :: a -> a
             true    :: b

  Rules: half(0) -> 0
         half(s(0)) -> 0
         half(s(s(X))) -> s(half(X))
         bits(0) -> 0
         bits(s(Y)) -> s(bits(half(s(Y))))
         map(G, nil) -> nil
         map(H, cons(W, P)) -> cons(H(W), map(H, P))
         filter(F1, nil) -> nil
         filter(Z1, cons(U1, V1)) -> filter2(Z1(U1), Z1, U1, V1)
         filter2(true, I1, P1, X2) -> cons(P1, filter(I1, X2))
         filter2(false, Z2, U2, V2) -> filter(Z2, V2)

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_?, i, c), where:

  P1. (1) half#(s(s(X))) => half#(X)
      (2) bits#(s(Y)) => half#(s(Y))
      (3) bits#(s(Y)) => bits#(half(s(Y)))
      (4) map#(H, cons(W, P)) => map#(H, P)
      (5) filter#(Z1, cons(U1, V1)) => filter2#(Z1(U1), Z1, U1, V1)
      (6) filter2#(true, I1, P1, X2) => filter#(I1, X2)
      (7) filter2#(false, Z2, U2, V2) => filter#(Z2, V2)

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

We compute a graph approximation with the following edges:

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

There are 4 SCCs.

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

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

  P3. (1) bits#(s(Y)) => bits#(half(s(Y)))

  P4. (1) map#(H, cons(W, P)) => map#(H, P)

  P5. (1) filter#(Z1, cons(U1, V1)) => filter2#(Z1(U1), Z1, U1, V1)
      (2) filter2#(true, I1, P1, X2) => filter#(I1, X2)
      (3) filter2#(false, Z2, U2, V2) => filter#(Z2, V2)

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

We use the following projection function:

  nu(half#) = 1

We thus have:

  (1) s(s(X)) |>| X

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 UNION R_?, i, c).

We obtain 3 usable rules (out of 11 rules in the input problem).

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

  R2. (1) half(0) -> 0
      (2) half(s(0)) -> 0
      (3) half(s(s(X))) -> s(half(X))

***** We apply the Subterm Criterion Processor on D4 = (P4, R UNION R_?, i, c).

We use the following projection function:

  nu(map#) = 2

We thus have:

  (1) cons(W, P) |>| P

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 UNION R_?, i, c).

We use the following projection function:

  nu(filter#)  = 2
  nu(filter2#) = 4

We thus have:

  (1) cons(U1, V1) |>|  V1
  (2) X2           |>=| X2
  (3) V2           |>=| V2

We may remove the strictly oriented DPs.

Processor output: { D7 = (P6, R UNION R_?, i, c) }, where:

  P6. (1) filter2#(true, I1, P1, X2) => filter#(I1, X2)
      (2) filter2#(false, Z2, U2, V2) => filter#(Z2, V2)

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