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)))
    Subterm Criterion Processor:
     simple projection:
      pi(p) = [1,1]
      pi(p#) = [1,1]
     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=4
      
      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:
                 [0 0 1 0]  
                 [0 1 0 0]  
       [b](x0) = [0 0 0 0]x0
                 [0 1 0 0]  ,
       
       [p#](x0, x1) = [1 0 0 0]x0 + [1 0 0 0]x1,
       
                 [0 1 0 0]     [1]
                 [0 0 1 0]     [0]
       [a](x0) = [1 0 1 0]x0 + [1]
                 [0 0 0 1]     [0],
       
                     [1 0 0 0]     [1 0 0 0]     [0]
                     [0 0 0 0]     [0 1 1 0]     [0]
       [p](x0, x1) = [0 0 0 0]x0 + [0 1 1 0]x1 + [0]
                     [0 0 0 0]     [0 0 0 0]     [1]
      orientation:
       p#(a(x0),p(b(a(x1)),x2)) = [0 1 0 0]x0 + [1 0 1 0]x1 + [1 0 0 0]x2 + [2] >= [1 0 1 0]x1 + [1 0 0 0]x2 + [1] = p#(x1,p(a(b(a(x1))),x2))
       
                                 [0 1 0 0]     [1 0 1 0]     [1 0 0 0]     [2]    [1 0 1 0]     [1 0 0 0]     [1]                          
                                 [0 0 0 0]     [0 0 0 0]     [0 2 2 0]     [0]    [0 0 0 0]     [0 2 2 0]     [0]                          
       p(a(x0),p(b(a(x1)),x2)) = [0 0 0 0]x0 + [0 0 0 0]x1 + [0 2 2 0]x2 + [0] >= [0 0 0 0]x1 + [0 2 2 0]x2 + [0] = p(x1,p(a(b(a(x1))),x2))
                                 [0 0 0 0]     [0 0 0 0]     [0 0 0 0]     [1]    [0 0 0 0]     [0 0 0 0]     [1]                          
       
                     [0 0 1 0]     [1]    [0 0 1 0]     [1]              
                     [0 0 0 0]     [0]    [0 0 0 0]     [0]              
       a(b(a(x0))) = [1 0 1 0]x0 + [2] >= [0 0 0 0]x0 + [0] = b(a(b(x0)))
                     [0 0 1 0]     [0]    [0 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)))
    Subterm Criterion Processor:
     simple projection:
      pi(b) = 0
      pi(a#) = 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