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

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

  Rules: h(Y, e(X)) -> h(c(Y), d(Y, X))
         d(U, g(0, 0)) -> e(0)
         d(P, g(V, W)) -> g(e(V), d(P, W))
         d(c(U1), g(g(X1, Y1), 0)) -> g(d(c(U1), g(X1, Y1)), d(U1, g(X1, Y1)))
         g(e(V1), e(W1)) -> e(g(V1, W1))
         map(J1, nil) -> nil
         map(F2, cons(Y2, U2)) -> cons(F2(Y2), map(F2, U2))
         filter(H2, nil) -> nil
         filter(I2, cons(P2, X3)) -> filter2(I2(P2), I2, P2, X3)
         filter2(true, Z3, U3, V3) -> cons(U3, filter(Z3, V3))
         filter2(false, I3, P3, X4) -> filter(I3, X4)

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)  h#(Y, e(X)) => d#(Y, X)
      (2)  h#(Y, e(X)) => h#(c(Y), d(Y, X))
      (3)  d#(P, g(V, W)) => d#(P, W)
      (4)  d#(P, g(V, W)) => g#(e(V), d(P, W))
      (5)  d#(c(U1), g(g(X1, Y1), 0)) => g#(X1, Y1)
      (6)  d#(c(U1), g(g(X1, Y1), 0)) => d#(c(U1), g(X1, Y1))
      (7)  d#(c(U1), g(g(X1, Y1), 0)) => g#(X1, Y1)
      (8)  d#(c(U1), g(g(X1, Y1), 0)) => d#(U1, g(X1, Y1))
      (9)  d#(c(U1), g(g(X1, Y1), 0)) => g#(d(c(U1), g(X1, Y1)), d(U1, g(X1, Y1)))
      (10) g#(e(V1), e(W1)) => g#(V1, W1)
      (11) map#(F2, cons(Y2, U2)) => map#(F2, U2)
      (12) filter#(I2, cons(P2, X3)) => filter2#(I2(P2), I2, P2, X3)
      (13) filter2#(true, Z3, U3, V3) => filter#(Z3, V3)
      (14) filter2#(false, I3, P3, X4) => filter#(I3, X4)

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

We compute a graph approximation with the following edges:

  1:  3 4 5 6 7 8 9 
  2:  1 2 
  3:  3 4 5 6 7 8 9 
  4:  10 
  5:  10 
  6:  3 4 5 6 7 8 9 
  7:  10 
  8:  3 4 5 6 7 8 9 
  9:  10 
  10: 10 
  11: 11 
  12: 13 14 
  13: 12 
  14: 12 

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) g#(e(V1), e(W1)) => g#(V1, W1)

  P3. (1) d#(P, g(V, W)) => d#(P, W)
      (2) d#(c(U1), g(g(X1, Y1), 0)) => d#(c(U1), g(X1, Y1))
      (3) d#(c(U1), g(g(X1, Y1), 0)) => d#(U1, g(X1, Y1))

  P4. (1) h#(Y, e(X)) => h#(c(Y), d(Y, X))

  P5. (1) map#(F2, cons(Y2, U2)) => map#(F2, U2)

  P6. (1) filter#(I2, cons(P2, X3)) => filter2#(I2(P2), I2, P2, X3)
      (2) filter2#(true, Z3, U3, V3) => filter#(Z3, V3)
      (3) filter2#(false, I3, P3, X4) => filter#(I3, X4)

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

We use the following projection function:

  nu(g#) = 1

We thus have:

  (1) e(V1) |>| V1

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(d#) = 1

We thus have:

  (1) P     |>=| P
  (2) c(U1) |>=| c(U1)
  (3) c(U1) |>|  U1

We may remove the strictly oriented DPs.

Processor output: { D7 = (P7, R UNION R_?, i, c) }, where:

  P7. (1) d#(P, g(V, W)) => d#(P, W)
      (2) d#(c(U1), g(g(X1, Y1), 0)) => d#(c(U1), g(X1, Y1))

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

We obtain 4 usable rules (out of 11 rules in the input problem).

Processor output: { D8 = (P4, R2, i, c) }, where:

  R2. (1) d(U, g(0, 0)) -> e(0)
      (2) d(P, g(V, W)) -> g(e(V), d(P, W))
      (3) d(c(U1), g(g(X1, Y1), 0)) -> g(d(c(U1), g(X1, Y1)), d(U1, g(X1, Y1)))
      (4) g(e(V1), e(W1)) -> e(g(V1, W1))

***** 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(Y2, U2) |>| U2

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(P2, X3) |>|  X3
  (2) V3           |>=| V3
  (3) X4           |>=| X4

We may remove the strictly oriented DPs.

Processor output: { D9 = (P8, R UNION R_?, i, c) }, where:

  P8. (1) filter2#(true, Z3, U3, V3) => filter#(Z3, V3)
      (2) filter2#(false, I3, P3, X4) => filter#(I3, X4)

***** We apply the Subterm Criterion Processor on D7 = (P7, R UNION R_?, i, c).

We use the following projection function:

  nu(d#) = 2

We thus have:

  (1) g(V, W)         |>| W
  (2) g(g(X1, Y1), 0) |>| g(X1, Y1)

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

Processor output: { }.

***** No progress could be made on DP problem D8 = (P4, R2, i, c).