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

  Signature: 0    :: c
             cons :: c -> b -> b
             map  :: (c -> c) -> b -> b
             nil  :: b
             node :: a -> b -> c
             plus :: c -> c -> c
             s    :: c -> c
             size :: c -> c
             sum  :: b -> c

  Rules: map(F, nil) -> nil
         map(Z, cons(U, V)) -> cons(Z(U), map(Z, V))
         sum(cons(W, P)) -> plus(W, sum(P))
         size(node(X1, Y1)) -> s(sum(map(size, Y1)))
         plus(0, U1) -> 0
         plus(s(V1), W1) -> s(plus(V1, W1))

The system is accessible function passing by a sort ordering with b = c ≻ a.

We start by computing the initial DP problem D1 = (P1, R UNION R_?, i, c), where:

  P1. (1) map#(Z, cons(U, V)) => map#(Z, V)
      (2) sum#(cons(W, P)) => sum#(P)
      (3) sum#(cons(W, P)) => plus#(W, sum(P))
      (4) size#(node(X1, Y1)) => size#(fresh1)
      (5) size#(node(X1, Y1)) => map#(size, Y1)
      (6) size#(node(X1, Y1)) => sum#(map(size, Y1))
      (7) plus#(s(V1), W1) => plus#(V1, W1)

***** 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: 2 3 
  3: 7 
  4: 4 5 6 
  5: 1 
  6: 2 3 
  7: 7 

There are 4 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) }, where:

  P2. (1) map#(Z, cons(U, V)) => map#(Z, V)

  P3. (1) plus#(s(V1), W1) => plus#(V1, W1)

  P4. (1) sum#(cons(W, P)) => sum#(P)

  P5. (1) size#(node(X1, Y1)) => size#(fresh1)

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

We use the following projection function:

  nu(map#) = 2

We thus have:

  (1) cons(U, V) |>| V

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(plus#) = 1

We thus have:

  (1) s(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 D4 = (P4, R UNION R_?, i, c).

We use the following projection function:

  nu(sum#) = 1

We thus have:

  (1) cons(W, P) |>| P

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

Processor output: { }.

***** We apply the Usable Rules Processor on D5 = (P5, R UNION R_?, i, c).

We obtain 0 usable rules (out of 6 rules in the input problem).

Processor output: { D6 = (P5, {}, i, c) }.

***** No progress could be made on DP problem D6 = (P5, {}, i, c).