YES
O(n^3)
TRS:
 {
                0() -> n__0(),
               f(X) -> n__f(X),
             f(0()) -> cons(0(), n__f(n__s(n__0()))),
          f(s(0())) -> f(p(s(0()))),
          p(s(0())) -> 0(),
               s(X) -> n__s(X),
        activate(X) -> X,
  activate(n__f(X)) -> f(activate(X)),
  activate(n__s(X)) -> s(activate(X)),
   activate(n__0()) -> 0()
 }
 DUP: We consider a non-duplicating system.
  Trs:
   {
                  0() -> n__0(),
                 f(X) -> n__f(X),
               f(0()) -> cons(0(), n__f(n__s(n__0()))),
            f(s(0())) -> f(p(s(0()))),
            p(s(0())) -> 0(),
                 s(X) -> n__s(X),
          activate(X) -> X,
    activate(n__f(X)) -> f(activate(X)),
    activate(n__s(X)) -> s(activate(X)),
     activate(n__0()) -> 0()
   }
  Matrix Interpretation:
   Interpretation class: triangular
              [X2]    [1 1 1][X2]   [1]
   [activate]([X1]) = [0 1 0][X1] + [0]
              [X0]    [0 0 1][X0]   [0]
   
       [X2]    [1 0 0][X2]   [1]
   [s]([X1]) = [0 1 0][X1] + [1]
       [X0]    [0 0 1][X0]   [1]
   
       [X2]    [1 0 0][X2]   [0]
   [p]([X1]) = [0 0 0][X1] + [1]
       [X0]    [0 0 0][X0]   [0]
   
       [X2]    [1 0 1][X2]   [1]
   [f]([X1]) = [0 1 1][X1] + [1]
       [X0]    [0 0 1][X0]   [1]
   
            [0]
   [n__0] = [1]
            [0]
   
          [X2]    [1 0 0][X2]   [0]
   [n__s]([X1]) = [0 1 0][X1] + [1]
          [X0]    [0 0 1][X0]   [1]
   
          [X2]    [1 0 0][X2]   [0]
   [n__f]([X1]) = [0 1 1][X1] + [1]
          [X0]    [0 0 1][X0]   [1]
   
         [1]
   [0] = [1]
         [0]
   
          [X5]  [X2]    [1 0 0][X5]   [1 0 0][X2]   [0]
   [cons]([X4], [X1]) = [0 0 0][X4] + [0 0 0][X1] + [0]
          [X3]  [X0]    [0 0 0][X3]   [0 0 0][X0]   [0]
   
   
   Qed