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))
  Matrix Interpretation Processor: dim=2
   
   usable rules:
    p(a(x0),p(a(b(x1)),x2)) -> p(a(b(a(x2))),p(a(a(x1)),x2))
   interpretation:
    [p#](x0, x1) = [1 2]x1 + [1],
    
                  [1 0]     [1 0]     [0]
    [p](x0, x1) = [0 0]x0 + [1 1]x1 + [1],
    
              [0 0]     [0]
    [b](x0) = [0 1]x0 + [3],
    
              [0 1]     [1]
    [a](x0) = [0 0]x0 + [0]
   orientation:
    p#(a(x0),p(a(b(x1)),x2)) = [0 1]x1 + [3 2]x2 + [7] >= [1 2]x2 + [1] = p#(a(a(x1)),x2)
    
    p#(a(x0),p(a(b(x1)),x2)) = [0 1]x1 + [3 2]x2 + [7] >= [3 2]x2 + [4] = p#(a(b(a(x2))),p(a(a(x1)),x2))
    
                              [0 1]     [0 1]     [1 0]     [5]    [1 0]     [5]                                
    p(a(x0),p(a(b(x1)),x2)) = [0 0]x0 + [0 1]x1 + [2 1]x2 + [6] >= [2 1]x2 + [3] = 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