YES
O(n^4)
TRS:
 {
     a__f(X1, X2, X3) -> f(X1, X2, X3),
      a__f(a(), X, X) -> a__f(X, a__b(), b()),
               a__b() -> b(),
               a__b() -> a(),
            mark(b()) -> a__b(),
            mark(a()) -> a(),
  mark(f(X1, X2, X3)) -> a__f(X1, mark(X2), X3)
 }
 DUP: We consider a non-duplicating system.
  Trs:
   {
       a__f(X1, X2, X3) -> f(X1, X2, X3),
        a__f(a(), X, X) -> a__f(X, a__b(), b()),
                 a__b() -> b(),
                 a__b() -> a(),
              mark(b()) -> a__b(),
              mark(a()) -> a(),
    mark(f(X1, X2, X3)) -> a__f(X1, mark(X2), X3)
   }
  Matrix Interpretation:
   Interpretation class: triangular
       [X11]  [X7]  [X3]    [1 1 0 0][X11]   [1 0 0 0][X7]   [1 1 0 1][X3]   [0]
       [X10]  [X6]  [X2]    [0 0 0 1][X10]   [0 1 0 0][X6]   [0 1 0 1][X2]   [1]
   [f]([ X9], [X5], [X1]) = [0 0 1 1][X9] + [0 0 1 0][X5] + [0 0 0 1][X1] + [1]
       [ X8]  [X4]  [X0]    [0 0 0 0][X8]   [0 0 0 0][X4]   [0 0 0 0][X0]   [0]
   
          [X3]    [1 1 1 0][X3]   [1]
          [X2]    [0 1 1 0][X2]   [1]
   [mark]([X1]) = [0 0 1 0][X1] + [1]
          [X0]    [0 0 0 0][X0]   [1]
   
         [0]
         [1]
   [a] = [0]
         [1]
   
         [0]
         [0]
   [b] = [1]
         [0]
   
            [1]
            [1]
   [a__b] = [1]
            [1]
   
          [X11]  [X7]  [X3]    [1 1 0 1][X11]   [1 0 0 0][X7]   [1 1 0 1][X3]   [1]
          [X10]  [X6]  [X2]    [0 0 0 1][X10]   [0 1 0 0][X6]   [0 1 0 1][X2]   [1]
   [a__f]([ X9], [X5], [X1]) = [0 0 1 1][X9] + [0 0 1 0][X5] + [0 0 0 1][X1] + [1]
          [ X8]  [X4]  [X0]    [0 0 0 0][X8]   [0 0 0 0][X4]   [0 0 0 0][X0]   [0]
   
   
   Qed