YES
O(n^3)
TRS:
 {
             a__f(X) -> f(X),
           a__f(0()) -> cons(0(), f(s(0()))),
        a__f(s(0())) -> a__f(a__p(s(0()))),
             a__p(X) -> p(X),
        a__p(s(0())) -> 0(),
  mark(cons(X1, X2)) -> cons(mark(X1), X2),
           mark(0()) -> 0(),
          mark(f(X)) -> a__f(mark(X)),
          mark(s(X)) -> s(mark(X)),
          mark(p(X)) -> a__p(mark(X))
 }
 DUP: We consider a non-duplicating system.
  Trs:
   {
               a__f(X) -> f(X),
             a__f(0()) -> cons(0(), f(s(0()))),
          a__f(s(0())) -> a__f(a__p(s(0()))),
               a__p(X) -> p(X),
          a__p(s(0())) -> 0(),
    mark(cons(X1, X2)) -> cons(mark(X1), X2),
             mark(0()) -> 0(),
            mark(f(X)) -> a__f(mark(X)),
            mark(s(X)) -> s(mark(X)),
            mark(p(X)) -> a__p(mark(X))
   }
  Matrix Interpretation:
   Interpretation class: triangular
       [X2]    [1 0 0][X2]   [0]
   [p]([X1]) = [0 1 1][X1] + [0]
       [X0]    [0 0 0][X0]   [3]
   
          [X2]    [1 3 3][X2]   [0]
   [mark]([X1]) = [0 1 0][X1] + [0]
          [X0]    [0 0 1][X0]   [0]
   
          [X2]    [1 0 0][X2]   [2]
   [a__p]([X1]) = [0 1 1][X1] + [0]
          [X0]    [0 0 0][X0]   [3]
   
          [X2]    [1 0 3][X2]   [3]
   [a__f]([X1]) = [0 1 3][X1] + [0]
          [X0]    [0 0 0][X0]   [2]
   
       [X2]    [1 0 0][X2]   [0]
   [s]([X1]) = [0 1 1][X1] + [0]
       [X0]    [0 0 1][X0]   [3]
   
       [X2]    [1 0 0][X2]   [0]
   [f]([X1]) = [0 1 3][X1] + [0]
       [X0]    [0 0 0][X0]   [2]
   
         [3]
   [0] = [3]
         [2]
   
          [X5]  [X2]    [1 0 2][X5]   [1 0 0][X2]   [1]
   [cons]([X4], [X1]) = [0 1 1][X4] + [0 0 0][X1] + [0]
          [X3]  [X0]    [0 0 0][X3]   [0 0 0][X0]   [1]
   
   
   Qed