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

  Signature: 0       :: a
             cons    :: c -> d -> d
             double  :: a -> a
             false   :: b
             filter  :: (c -> b) -> d -> d
             filter2 :: b -> (c -> b) -> c -> d -> d
             map     :: (c -> c) -> d -> d
             minus   :: a -> a -> a
             nil     :: d
             plus    :: a -> a -> a
             s       :: a -> a
             true    :: b

  Rules: minus(X, 0) -> X
         minus(s(Y), s(U)) -> minus(Y, U)
         double(0) -> 0
         double(s(V)) -> s(s(double(V)))
         plus(0, W) -> W
         plus(s(P), X1) -> s(plus(P, X1))
         plus(s(Y1), U1) -> plus(Y1, s(U1))
         plus(s(V1), W1) -> s(plus(minus(V1, W1), double(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_?, f, c), where:

  P1. (1)  minus#(s(Y), s(U)) => minus#(Y, U)
      (2)  double#(s(V)) => double#(V)
      (3)  plus#(s(P), X1) => plus#(P, X1)
      (4)  plus#(s(Y1), U1) => plus#(Y1, s(U1))
      (5)  plus#(s(V1), W1) => minus#(V1, W1)
      (6)  plus#(s(V1), W1) => double#(W1)
      (7)  plus#(s(V1), W1) => plus#(minus(V1, W1), double(W1))
      (8)  map#(F2, cons(Y2, U2)) => map#(F2, U2)
      (9)  filter#(I2, cons(P2, X3)) => filter2#(I2(P2), I2, P2, X3)
      (10) filter2#(true, Z3, U3, V3) => filter#(Z3, V3)
      (11) filter2#(false, I3, P3, X4) => filter#(I3, X4)

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

We compute a graph approximation with the following edges:

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

There are 5 SCCs.

Processor output: { D2 = (P2, R UNION R_?, f, c) ; D3 = (P3, R UNION R_?, f, c) ; D4 = (P4, R UNION R_?, f, c) ; D5 = (P5, R UNION R_?, f, c) ; D6 = (P6, R UNION R_?, f, c) }, where:

  P2. (1) minus#(s(Y), s(U)) => minus#(Y, U)

  P3. (1) double#(s(V)) => double#(V)

  P4. (1) plus#(s(P), X1) => plus#(P, X1)
      (2) plus#(s(Y1), U1) => plus#(Y1, s(U1))
      (3) plus#(s(V1), W1) => plus#(minus(V1, W1), double(W1))

  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_?, f, c).

We use the following projection function:

  nu(minus#) = 1

We thus have:

  (1) s(Y) |>| Y

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_?, f, c).

We use the following projection function:

  nu(double#) = 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: { }.

***** No progress could be made on DP problem D4 = (P4, R UNION R_?, f, c).