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

  Signature: *       :: a -> a -> a
             +       :: a -> a -> a
             0       :: a
             cons    :: c -> d -> d
             fact    :: a -> a
             false   :: b
             filter  :: (c -> b) -> d -> d
             filter2 :: b -> (c -> b) -> c -> d -> d
             map     :: (c -> c) -> d -> d
             nil     :: d
             p       :: a -> a
             s       :: a -> a
             true    :: b

  Rules: p(s(X)) -> X
         fact(0) -> s(0)
         fact(s(Y)) -> *(s(Y), fact(p(s(Y))))
         *(0, U) -> 0
         *(s(V), W) -> +(*(V, W), W)
         +(P, 0) -> P
         +(X1, s(Y1)) -> s(+(X1, Y1))
         map(G1, nil) -> nil
         map(H1, cons(W1, P1)) -> cons(H1(W1), map(H1, P1))
         filter(F2, nil) -> nil
         filter(Z2, cons(U2, V2)) -> filter2(Z2(U2), Z2, U2, V2)
         filter2(true, I2, P2, X3) -> cons(P2, filter(I2, X3))
         filter2(false, Z3, U3, V3) -> filter(Z3, V3)

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)  fact#(s(Y)) => p#(s(Y))
      (2)  fact#(s(Y)) => fact#(p(s(Y)))
      (3)  fact#(s(Y)) => *#(s(Y), fact(p(s(Y))))
      (4)  *#(s(V), W) => *#(V, W)
      (5)  *#(s(V), W) => +#(*(V, W), W)
      (6)  +#(X1, s(Y1)) => +#(X1, Y1)
      (7)  map#(H1, cons(W1, P1)) => map#(H1, P1)
      (8)  filter#(Z2, cons(U2, V2)) => filter2#(Z2(U2), Z2, U2, V2)
      (9)  filter2#(true, I2, P2, X3) => filter#(I2, X3)
      (10) filter2#(false, Z3, U3, V3) => filter#(Z3, V3)

***** 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 
  4:  4 5 
  5:  6 
  6:  6 
  7:  7 
  8:  9 10 
  9:  8 
  10: 8 

There are 5 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) ; D6 = (P6, R UNION R_?, i, c) }, where:

  P2. (1) +#(X1, s(Y1)) => +#(X1, Y1)

  P3. (1) *#(s(V), W) => *#(V, W)

  P4. (1) fact#(s(Y)) => fact#(p(s(Y)))

  P5. (1) map#(H1, cons(W1, P1)) => map#(H1, P1)

  P6. (1) filter#(Z2, cons(U2, V2)) => filter2#(Z2(U2), Z2, U2, V2)
      (2) filter2#(true, I2, P2, X3) => filter#(I2, X3)
      (3) filter2#(false, Z3, U3, V3) => filter#(Z3, V3)

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

We use the following projection function:

  nu(+#) = 2

We thus have:

  (1) s(Y1) |>| Y1

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_?, i, c).

We use the following projection function:

  nu(*#) = 1

We thus have:

  (1) s(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 D4 = (P4, R UNION R_?, i, c).

We obtain 1 usable rules (out of 13 rules in the input problem).

Processor output: { D7 = (P4, R2, i, c) }, where:

  R2. (1) p(s(X)) -> X

***** We apply the Subterm Criterion Processor on D5 = (P5, R UNION R_?, i, c).

We use the following projection function:

  nu(map#) = 2

We thus have:

  (1) cons(W1, P1) |>| P1

All DPs are strictly oriented, and may be removed.  Hence, this DP problem is finite.

Processor output: { }.

***** We apply the Subterm Criterion Processor on D6 = (P6, R UNION R_?, i, c).

We use the following projection function:

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

We thus have:

  (1) cons(U2, V2) |>|  V2
  (2) X3           |>=| X3
  (3) V3           |>=| V3

We may remove the strictly oriented DPs.

Processor output: { D8 = (P7, R UNION R_?, i, c) }, where:

  P7. (1) filter2#(true, I2, P2, X3) => filter#(I2, X3)
      (2) filter2#(false, Z3, U3, V3) => filter#(Z3, V3)

***** No progress could be made on DP problem D7 = (P4, R2, i, c).