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

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

  Rules: f(0) -> f(0)
         0 -> 1
         map(F, nil) -> nil
         map(Z, cons(U, V)) -> cons(Z(U), map(Z, V))
         filter(I, nil) -> nil
         filter(J, cons(X1, Y1)) -> filter2(J(X1), J, X1, Y1)
         filter2(true, G1, V1, W1) -> cons(V1, filter(G1, W1))
         filter2(false, J1, X2, Y2) -> filter(J1, Y2)

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#(0) => 0#
      (2) f#(0) => f#(0)
      (3) map#(Z, cons(U, V)) => map#(Z, V)
      (4) filter#(J, cons(X1, Y1)) => filter2#(J(X1), J, X1, Y1)
      (5) filter2#(true, G1, V1, W1) => filter#(G1, W1)
      (6) filter2#(false, J1, X2, Y2) => filter#(J1, Y2)

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

We compute a graph approximation with the following edges:

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

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#(0) => f#(0)

  P3. (1) map#(Z, cons(U, V)) => map#(Z, V)

  P4. (1) filter#(J, cons(X1, Y1)) => filter2#(J(X1), J, X1, Y1)
      (2) filter2#(true, G1, V1, W1) => filter#(G1, W1)
      (3) filter2#(false, J1, X2, Y2) => filter#(J1, Y2)

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

We obtain 1 usable rules (out of 8 rules in the input problem).

Processor output: { D5 = (P2, R2, i, c) }, where:

  R2. (1) 0 -> 1

***** 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(U, V) |>| V

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(X1, Y1) |>|  Y1
  (2) W1           |>=| W1
  (3) Y2           |>=| Y2

We may remove the strictly oriented DPs.

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

  P5. (1) filter2#(true, G1, V1, W1) => filter#(G1, W1)
      (2) filter2#(false, J1, X2, Y2) => filter#(J1, Y2)

***** No progress could be made on DP problem D5 = (P2, R2, i, c).