YES
O(n^4)
TRS:
 {p(p(b(a(x0)), x1), p(x2, x3)) -> p(p(b(x2), a(a(b(x1)))), p(x3, x0))}
 DUP: We consider a non-duplicating system.
  Trs:
   {p(p(b(a(x0)), x1), p(x2, x3)) -> p(p(b(x2), a(a(b(x1)))), p(x3, x0))}
  Matrix Interpretation:
   Interpretation class: triangular
       [X3]    [1 1 0 0][X3]   [0]
       [X2]    [0 0 0 0][X2]   [0]
   [a]([X1]) = [0 0 1 0][X1] + [1]
       [X0]    [0 0 0 0][X0]   [0]
   
       [X3]    [1 0 0 0][X3]   [0]
       [X2]    [0 0 0 0][X2]   [0]
   [b]([X1]) = [0 0 1 0][X1] + [0]
       [X0]    [0 0 0 0][X0]   [0]
   
       [X7]  [X3]    [1 0 1 0][X7]   [1 1 0 0][X3]   [0]
       [X6]  [X2]    [0 0 0 0][X6]   [0 0 1 0][X2]   [0]
   [p]([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]
   
   
   Qed