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

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

  Rules: minus(X, 0) -> X
         minus(s(Y), s(U)) -> minus(Y, U)
         f(0) -> s(0)
         f(s(V)) -> minus(s(V), g(f(V)))
         g(0) -> 0
         g(s(W)) -> minus(s(W), f(g(W)))
         map(J, nil) -> nil
         map(F1, cons(Y1, U1)) -> cons(F1(Y1), map(F1, U1))
         filter(H1, nil) -> nil
         filter(I1, cons(P1, X2)) -> filter2(I1(P1), I1, P1, X2)
         filter2(true, Z2, U2, V2) -> cons(U2, filter(Z2, V2))
         filter2(false, I2, P2, X3) -> filter(I2, X3)

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)  f#(s(V)) => f#(V)
      (3)  f#(s(V)) => g#(f(V))
      (4)  f#(s(V)) => minus#(s(V), g(f(V)))
      (5)  g#(s(W)) => g#(W)
      (6)  g#(s(W)) => f#(g(W))
      (7)  g#(s(W)) => minus#(s(W), f(g(W)))
      (8)  map#(F1, cons(Y1, U1)) => map#(F1, U1)
      (9)  filter#(I1, cons(P1, X2)) => filter2#(I1(P1), I1, P1, X2)
      (10) filter2#(true, Z2, U2, V2) => filter#(Z2, V2)
      (11) filter2#(false, I2, P2, X3) => filter#(I2, X3)

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

There are 4 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) }, where:

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

  P3. (1) f#(s(V)) => f#(V)
      (2) f#(s(V)) => g#(f(V))
      (3) g#(s(W)) => g#(W)
      (4) g#(s(W)) => f#(g(W))

  P4. (1) map#(F1, cons(Y1, U1)) => map#(F1, U1)

  P5. (1) filter#(I1, cons(P1, X2)) => filter2#(I1(P1), I1, P1, X2)
      (2) filter2#(true, Z2, U2, V2) => filter#(Z2, V2)
      (3) filter2#(false, I2, P2, X3) => filter#(I2, X3)

***** 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: { }.

***** No progress could be made on DP problem D3 = (P3, R UNION R_?, f, c).