YES

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

Proof:
 DP Processor:
  DPs:
   p#(a(x0),p(b(a(x1)),x2)) -> a#(b(a(x1)))
   p#(a(x0),p(b(a(x1)),x2)) -> p#(a(b(a(x1))),x2)
   p#(a(x0),p(b(a(x1)),x2)) -> p#(x1,p(a(b(a(x1))),x2))
   a#(b(a(x0))) -> a#(b(x0))
  TRS:
   p(a(x0),p(b(a(x1)),x2)) -> p(x1,p(a(b(a(x1))),x2))
   a(b(a(x0))) -> b(a(b(x0)))
  Matrix Interpretation Processor: dim=3
   
   usable rules:
    p(a(x0),p(b(a(x1)),x2)) -> p(x1,p(a(b(a(x1))),x2))
    a(b(a(x0))) -> b(a(b(x0)))
   interpretation:
    [a#](x0) = [0 1 0]x0 + [1],
    
    [p#](x0, x1) = [0 1 0]x0 + [0 1 0]x1 + [1],
    
                  [0 0 0]     [0 0 0]  
    [p](x0, x1) = [0 1 0]x0 + [0 1 0]x1
                  [0 0 0]     [0 0 0]  ,
    
              [1 0 0]  
    [b](x0) = [0 0 1]x0
              [0 0 0]  ,
    
              [0 0 1]     [0]
    [a](x0) = [1 0 0]x0 + [1]
              [0 1 1]     [1]
   orientation:
    p#(a(x0),p(b(a(x1)),x2)) = [1 0 0]x0 + [0 1 1]x1 + [0 1 0]x2 + [3] >= [0 1 1]x1 + [2] = a#(b(a(x1)))
    
    p#(a(x0),p(b(a(x1)),x2)) = [1 0 0]x0 + [0 1 1]x1 + [0 1 0]x2 + [3] >= [0 0 1]x1 + [0 1 0]x2 + [2] = p#(a(b(a(x1))),x2)
    
    p#(a(x0),p(b(a(x1)),x2)) = [1 0 0]x0 + [0 1 1]x1 + [0 1 0]x2 + [3] >= [0 1 1]x1 + [0 1 0]x2 + [2] = p#(x1,p(a(b(a(x1))),x2))
    
    a#(b(a(x0))) = [0 1 1]x0 + [2] >= [0 0 1]x0 + [1] = a#(b(x0))
    
                              [0 0 0]     [0 0 0]     [0 0 0]     [0]    [0 0 0]     [0 0 0]     [0]                          
    p(a(x0),p(b(a(x1)),x2)) = [1 0 0]x0 + [0 1 1]x1 + [0 1 0]x2 + [2] >= [0 1 1]x1 + [0 1 0]x2 + [1] = p(x1,p(a(b(a(x1))),x2))
                              [0 0 0]     [0 0 0]     [0 0 0]     [0]    [0 0 0]     [0 0 0]     [0]                          
    
                  [0 0 0]     [0]    [0 0 0]     [0]              
    a(b(a(x0))) = [0 0 1]x0 + [1] >= [0 0 1]x0 + [1] = b(a(b(x0)))
                  [0 1 1]     [2]    [0 0 0]     [0]              
   problem:
    DPs:
     
    TRS:
     p(a(x0),p(b(a(x1)),x2)) -> p(x1,p(a(b(a(x1))),x2))
     a(b(a(x0))) -> b(a(b(x0)))
   Qed