YES
O(n^4)
TRS:
 {
        f(s(0())) -> s(s(0())),
        f(s(0())) -> *(s(s(0())), f(0())),
           f(0()) -> s(0()),
       f(+(x, y)) -> *(f(x), f(y)),
  f(+(x, s(0()))) -> +(s(s(0())), f(x))
 }
 DUP: We consider a non-duplicating system.
  Trs:
   {
          f(s(0())) -> s(s(0())),
          f(s(0())) -> *(s(s(0())), f(0())),
             f(0()) -> s(0()),
         f(+(x, y)) -> *(f(x), f(y)),
    f(+(x, s(0()))) -> +(s(s(0())), f(x))
   }
  Matrix Interpretation:
   Interpretation class: triangular
       [X7]  [X3]    [1 0 0 0][X7]   [1 0 0 0][X3]   [0]
       [X6]  [X2]    [0 1 1 1][X6]   [0 1 0 0][X2]   [1]
   [+]([X5], [X1]) = [0 0 1 0][X5] + [0 0 0 0][X1] + [0]
       [X4]  [X0]    [0 0 0 0][X4]   [0 0 0 1][X0]   [0]
   
       [X7]  [X3]    [1 0 0 0][X7]   [1 0 0 0][X3]   [0]
       [X6]  [X2]    [0 1 0 0][X6]   [0 0 0 0][X2]   [0]
   [*]([X5], [X1]) = [0 0 0 0][X5] + [0 0 0 0][X1] + [0]
       [X4]  [X0]    [0 0 0 0][X4]   [0 0 0 0][X0]   [0]
   
       [X3]    [1 1 0 1][X3]   [0]
       [X2]    [0 1 1 1][X2]   [1]
   [f]([X1]) = [0 0 0 0][X1] + [0]
       [X0]    [0 0 0 0][X0]   [1]
   
         [1]
         [0]
   [0] = [0]
         [1]
   
       [X3]    [1 0 0 0][X3]   [0]
       [X2]    [0 0 0 1][X2]   [1]
   [s]([X1]) = [0 0 0 0][X1] + [0]
       [X0]    [0 0 0 0][X0]   [1]
   
   
   Qed