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

  Signature: 0    :: a
             cons :: a -> b -> b
             inc  :: b -> b
             map  :: (a -> a) -> b -> b
             nil  :: b
             plus :: a -> a -> a
             s    :: a -> a

  Rules: plus(0, X) -> X
         plus(s(Y), U) -> s(plus(Y, U))
         inc(V) -> map(plus(s(0)), V)
         map(I, nil) -> nil
         map(J, cons(X1, Y1)) -> cons(J(X1), map(J, Y1))

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) plus#(s(Y), U) => plus#(Y, U)
      (2) inc#(V) => plus#(s(0), fresh1)
      (3) inc#(V) => map#(plus(s(0)), V)
      (4) map#(J, cons(X1, Y1)) => map#(J, Y1)

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

We compute a graph approximation with the following edges:

  1: 1 
  2: 1 
  3: 4 
  4: 4 

There are 2 SCCs.

Processor output: { D2 = (P2, R UNION R_?, i, c) ; D3 = (P3, R UNION R_?, i, c) }, where:

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

  P3. (1) map#(J, cons(X1, Y1)) => map#(J, Y1)

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

We use the following projection function:

  nu(plus#) = 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_?, i, c).

We use the following projection function:

  nu(map#) = 2

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: { }.