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)))
  TDG 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)))
   graph:
    a#(b(a(x0))) -> a#(b(x0)) -> a#(b(a(x0))) -> a#(b(x0))
    p#(a(x0),p(b(a(x1)),x2)) -> a#(b(a(x1))) ->
    a#(b(a(x0))) -> a#(b(x0))
    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))
    p#(a(x0),p(b(a(x1)),x2)) -> p#(a(b(a(x1))),x2) ->
    p#(a(x0),p(b(a(x1)),x2)) -> p#(a(b(a(x1))),x2)
    p#(a(x0),p(b(a(x1)),x2)) -> p#(a(b(a(x1))),x2) ->
    p#(a(x0),p(b(a(x1)),x2)) -> a#(b(a(x1)))
    p#(a(x0),p(b(a(x1)),x2)) -> p#(x1,p(a(b(a(x1))),x2)) ->
    p#(a(x0),p(b(a(x1)),x2)) -> p#(x1,p(a(b(a(x1))),x2))
    p#(a(x0),p(b(a(x1)),x2)) -> p#(x1,p(a(b(a(x1))),x2)) ->
    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)) ->
    p#(a(x0),p(b(a(x1)),x2)) -> a#(b(a(x1)))
   SCC Processor:
    #sccs: 2
    #rules: 3
    #arcs: 8/16
    DPs:
     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))
    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)))
    Arctic Interpretation Processor:
     dimension: 1
     interpretation:
      [p#](x0, x1) = x1 + 0,
      
      [p](x0, x1) = 1x1 + 4,
      
      [b](x0) = 0,
      
      [a](x0) = 0
     orientation:
      p#(a(x0),p(b(a(x1)),x2)) = 1x2 + 4 >= x2 + 0 = p#(a(b(a(x1))),x2)
      
      p#(a(x0),p(b(a(x1)),x2)) = 1x2 + 4 >= 1x2 + 4 = p#(x1,p(a(b(a(x1))),x2))
      
      p(a(x0),p(b(a(x1)),x2)) = 2x2 + 5 >= 2x2 + 5 = p(x1,p(a(b(a(x1))),x2))
      
      a(b(a(x0))) = 0 >= 0 = b(a(b(x0)))
     problem:
      DPs:
       p#(a(x0),p(b(a(x1)),x2)) -> p#(x1,p(a(b(a(x1))),x2))
      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
      
      interpretation:
       [p#](x0, x1) = [1 0 0]x0 + [0 1 0]x1,
       
                     [0 0 0]     [0 1 0]  
       [p](x0, x1) = [1 0 0]x0 + [0 1 0]x1
                     [0 0 0]     [0 0 0]  ,
       
                 [0 0 1]  
       [b](x0) = [0 1 0]x0
                 [0 0 0]  ,
       
                 [0 1 0]     [1]
       [a](x0) = [0 0 1]x0 + [0]
                 [1 0 1]     [1]
      orientation:
       p#(a(x0),p(b(a(x1)),x2)) = [0 1 0]x0 + [1 0 1]x1 + [0 1 0]x2 + [2] >= [1 0 1]x1 + [0 1 0]x2 + [1] = p#(x1,p(a(b(a(x1))),x2))
       
                                 [0 0 0]     [1 0 1]     [0 1 0]     [1]    [0 0 1]     [0 1 0]     [1]                          
       p(a(x0),p(b(a(x1)),x2)) = [0 1 0]x0 + [1 0 1]x1 + [0 1 0]x2 + [2] >= [1 0 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 1]     [1]    [0 0 1]     [1]              
       a(b(a(x0))) = [0 0 0]x0 + [0] >= [0 0 0]x0 + [0] = b(a(b(x0)))
                     [1 0 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
    
    DPs:
     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)))
    Arctic Interpretation Processor:
     dimension: 1
     interpretation:
      [a#](x0) = x0,
      
      [p](x0, x1) = 1x1,
      
      [b](x0) = x0,
      
      [a](x0) = 3x0
     orientation:
      a#(b(a(x0))) = 3x0 >= x0 = a#(b(x0))
      
      p(a(x0),p(b(a(x1)),x2)) = 2x2 >= 2x2 = p(x1,p(a(b(a(x1))),x2))
      
      a(b(a(x0))) = 6x0 >= 3x0 = b(a(b(x0)))
     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