YES

Problem:
 p(a(x0),p(a(b(x1)),x2)) -> p(a(b(a(x2))),p(a(a(x1)),x2))

Proof:
 DP Processor:
  DPs:
   p#(a(x0),p(a(b(x1)),x2)) -> p#(a(a(x1)),x2)
   p#(a(x0),p(a(b(x1)),x2)) -> p#(a(b(a(x2))),p(a(a(x1)),x2))
  TRS:
   p(a(x0),p(a(b(x1)),x2)) -> p(a(b(a(x2))),p(a(a(x1)),x2))
  KBO Processor:
   argument filtering:
    pi(a) = []
    pi(b) = 0
    pi(p) = [1]
    pi(p#) = 1
   weight function:
    w0 = 1
    w(p#) = w(p) = w(b) = w(a) = 1
   precedence:
    p# ~ p ~ b ~ a
   problem:
    DPs:
     p#(a(x0),p(a(b(x1)),x2)) -> p#(a(b(a(x2))),p(a(a(x1)),x2))
    TRS:
     p(a(x0),p(a(b(x1)),x2)) -> p(a(b(a(x2))),p(a(a(x1)),x2))
   Matrix Interpretation Processor: dim=2
    
    interpretation:
     [p#](x0, x1) = [1 0]x1 + [2],
     
                   [1 0]     [0 1]  
     [p](x0, x1) = [2 0]x0 + [0 1]x1,
     
               [0]
     [b](x0) = [1],
     
               [0 1]  
     [a](x0) = [0 0]x0
    orientation:
     p#(a(x0),p(a(b(x1)),x2)) = [0 1]x2 + [3] >= [0 1]x2 + [2] = p#(a(b(a(x2))),p(a(a(x1)),x2))
     
                               [0 1]     [0 1]     [2]    [0 1]     [1]                                
     p(a(x0),p(a(b(x1)),x2)) = [0 2]x0 + [0 1]x2 + [2] >= [0 1]x2 + [2] = p(a(b(a(x2))),p(a(a(x1)),x2))
    problem:
     DPs:
      
     TRS:
      p(a(x0),p(a(b(x1)),x2)) -> p(a(b(a(x2))),p(a(a(x1)),x2))
    Qed