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

  Signature: compose  :: (a -> a) -> (a -> a) -> a -> a
             cons     :: a -> a -> a
             hd       :: a -> a
             init     :: a -> a
             last     :: a -> a
             nil      :: a
             reverse  :: a -> a
             reverse2 :: a -> a -> a
             tl       :: a -> a

  Rules: compose(F, Z, U) -> Z(F(U))
         reverse(V) -> reverse2(V, nil)
         reverse2(nil, W) -> W
         reverse2(cons(X1, Y1), P) -> reverse2(Y1, cons(X1, P))
         hd(cons(U1, V1)) -> U1
         tl(cons(W1, P1)) -> P1
         last -> compose(hd, reverse)
         init -> compose(reverse, compose(tl, reverse))

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)  reverse#(V) => reverse2#(V, nil)
      (2)  reverse2#(cons(X1, Y1), P) => reverse2#(Y1, cons(X1, P))
      (3)  last#(arg1) => hd#(fresh1)
      (4)  last#(arg1) => reverse#(fresh1)
      (5)  last#(arg1) => compose#(hd, reverse, arg1)
      (6)  init#(arg1) => reverse#(fresh1)
      (7)  init#(arg1) => tl#(fresh1)
      (8)  init#(arg1) => reverse#(fresh1)
      (9)  init#(arg1) => compose#(tl, reverse, fresh1)
      (10) init#(arg1) => compose#(reverse, compose(tl, reverse), arg1)

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

We compute a graph approximation with the following edges:

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

There is only one SCC, so all DPs not inside the SCC can be removed.

Processor output: { D2 = (P2, R UNION R_?, i, c) }, where:

  P2. (1) reverse2#(cons(X1, Y1), P) => reverse2#(Y1, cons(X1, P))

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

We use the following projection function:

  nu(reverse2#) = 1

We thus have:

  (1) cons(X1, Y1) |>| Y1

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

Processor output: { }.