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

  Signature: 0       :: c
             cons    :: c -> d -> d
             false   :: b
             filter  :: (c -> b) -> d -> d
             filter2 :: b -> (c -> b) -> c -> d -> d
             map     :: (c -> c) -> d -> d
             nil     :: d
             rev     :: d -> d
             rev1    :: c -> d -> c
             rev2    :: c -> d -> d
             s       :: a -> c
             true    :: b

  Rules: rev(nil) -> nil
         rev(cons(Y, X)) -> cons(rev1(Y, X), rev2(Y, X))
         rev1(0, nil) -> 0
         rev1(s(U), nil) -> s(U)
         rev1(W, cons(P, V)) -> rev1(P, V)
         rev2(X1, nil) -> nil
         rev2(U1, cons(V1, Y1)) -> rev(cons(U1, rev2(V1, Y1)))
         map(I1, nil) -> nil
         map(J1, cons(X2, Y2)) -> cons(J1(X2), map(J1, Y2))
         filter(G2, nil) -> nil
         filter(H2, cons(W2, P2)) -> filter2(H2(W2), H2, W2, P2)
         filter2(true, F3, Y3, U3) -> cons(Y3, filter(F3, U3))
         filter2(false, H3, W3, P3) -> filter(H3, P3)

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) rev#(cons(Y, X)) => rev1#(Y, X)
      (2) rev#(cons(Y, X)) => rev2#(Y, X)
      (3) rev1#(W, cons(P, V)) => rev1#(P, V)
      (4) rev2#(U1, cons(V1, Y1)) => rev2#(V1, Y1)
      (5) rev2#(U1, cons(V1, Y1)) => rev#(cons(U1, rev2(V1, Y1)))
      (6) map#(J1, cons(X2, Y2)) => map#(J1, Y2)
      (7) filter#(H2, cons(W2, P2)) => filter2#(H2(W2), H2, W2, P2)
      (8) filter2#(true, F3, Y3, U3) => filter#(F3, U3)
      (9) filter2#(false, H3, W3, P3) => filter#(H3, P3)

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

We compute a graph approximation with the following edges:

  1: 3 
  2: 4 5 
  3: 3 
  4: 4 5 
  5: 1 2 
  6: 6 
  7: 8 9 
  8: 7 
  9: 7 

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) rev1#(W, cons(P, V)) => rev1#(P, V)

  P3. (1) rev#(cons(Y, X)) => rev2#(Y, X)
      (2) rev2#(U1, cons(V1, Y1)) => rev2#(V1, Y1)
      (3) rev2#(U1, cons(V1, Y1)) => rev#(cons(U1, rev2(V1, Y1)))

  P4. (1) map#(J1, cons(X2, Y2)) => map#(J1, Y2)

  P5. (1) filter#(H2, cons(W2, P2)) => filter2#(H2(W2), H2, W2, P2)
      (2) filter2#(true, F3, Y3, U3) => filter#(F3, U3)
      (3) filter2#(false, H3, W3, P3) => filter#(H3, P3)

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

We use the following projection function:

  nu(rev1#) = 2

We thus have:

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

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 7 usable rules (out of 13 rules in the input problem).

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

  R2. (1) rev1(0, nil) -> 0
      (2) rev1(s(U), nil) -> s(U)
      (3) rev1(W, cons(P, V)) -> rev1(P, V)
      (4) rev2(X1, nil) -> nil
      (5) rev2(U1, cons(V1, Y1)) -> rev(cons(U1, rev2(V1, Y1)))
      (6) rev(nil) -> nil
      (7) rev(cons(Y, X)) -> cons(rev1(Y, X), rev2(Y, 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(X2, Y2) |>| Y2

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(W2, P2) |>|  P2
  (2) U3           |>=| U3
  (3) P3           |>=| P3

We may remove the strictly oriented DPs.

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

  P6. (1) filter2#(true, F3, Y3, U3) => filter#(F3, U3)
      (2) filter2#(false, H3, W3, P3) => filter#(H3, P3)

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