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
   
   interpretation:
    [p#](x0, x1) = [0 1]x1,
    
                  [0 0]     [2 0]     [1]
    [p](x0, x1) = [1 0]x0 + [1 2]x1 + [0],
    
              [3 0]     [1]
    [b](x0) = [0 0]x0 + [1],
    
              [0 1]  
    [a](x0) = [0 0]x0
   orientation:
    p#(a(x0),p(a(b(x1)),x2)) = [1 2]x2 + [1] >= [0 1]x2 = p#(a(a(x1)),x2)
    
    p#(a(x0),p(a(b(x1)),x2)) = [1 2]x2 + [1] >= [1 2]x2 = p#(a(b(a(x2))),p(a(a(x1)),x2))
    
                              [0 0]     [4 0]     [3]    [4 0]     [3]                                
    p(a(x0),p(a(b(x1)),x2)) = [0 1]x0 + [4 4]x2 + [3] >= [4 4]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