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

  Signature: 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 -> b
             map     :: (d -> d) -> e -> e
             nil     :: e
             true    :: c

  Rules: f(g(X, Y), X, U) -> f(U, U, U)
         g(V, W) -> V
         g(P, X1) -> X1
         map(Z1, nil) -> nil
         map(G1, cons(V1, W1)) -> cons(G1(V1), map(G1, W1))
         filter(J1, nil) -> nil
         filter(F2, cons(Y2, U2)) -> filter2(F2(Y2), F2, Y2, U2)
         filter2(true, H2, W2, P2) -> cons(W2, filter(H2, P2))
         filter2(false, F3, Y3, U3) -> filter(F3, U3)

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) f#(g(X, Y), X, U) => f#(U, U, U)
      (2) map#(G1, cons(V1, W1)) => map#(G1, W1)
      (3) filter#(F2, cons(Y2, U2)) => filter2#(F2(Y2), F2, Y2, U2)
      (4) filter2#(true, H2, W2, P2) => filter#(H2, P2)
      (5) filter2#(false, F3, Y3, U3) => filter#(F3, U3)

***** 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: 2 
  3: 4 5 
  4: 3 
  5: 3 

There are 3 SCCs.

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

  P2. (1) f#(g(X, Y), X, U) => f#(U, U, U)

  P3. (1) map#(G1, cons(V1, W1)) => map#(G1, W1)

  P4. (1) filter#(F2, cons(Y2, U2)) => filter2#(F2(Y2), F2, Y2, U2)
      (2) filter2#(true, H2, W2, P2) => filter#(H2, P2)
      (3) filter2#(false, F3, Y3, U3) => filter#(F3, U3)

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

We obtain 0 usable rules (out of 9 rules in the input problem).

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

***** We apply the Subterm Criterion Processor on D3 = (P3, R UNION R_?, i, c).

We use the following projection function:

  nu(map#) = 2

We thus have:

  (1) cons(V1, W1) |>| W1

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 UNION R_?, i, c).

We use the following projection function:

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

We thus have:

  (1) cons(Y2, U2) |>|  U2
  (2) P2           |>=| P2
  (3) U3           |>=| U3

We may remove the strictly oriented DPs.

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

  P5. (1) filter2#(true, H2, W2, P2) => filter#(H2, P2)
      (2) filter2#(false, F3, Y3, U3) => filter#(F3, U3)

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