YES

Problem:
 a(a(a(b(x1)))) -> b(a(a(a(x1))))
 b(b(x1)) -> a(b(a(b(x1))))

Proof:
 DP Processor:
  DPs:
   a#(a(a(b(x1)))) -> a#(x1)
   a#(a(a(b(x1)))) -> a#(a(x1))
   a#(a(a(b(x1)))) -> a#(a(a(x1)))
   a#(a(a(b(x1)))) -> b#(a(a(a(x1))))
   b#(b(x1)) -> a#(b(x1))
   b#(b(x1)) -> b#(a(b(x1)))
   b#(b(x1)) -> a#(b(a(b(x1))))
  TRS:
   a(a(a(b(x1)))) -> b(a(a(a(x1))))
   b(b(x1)) -> a(b(a(b(x1))))
  Matrix Interpretation Processor: dim=3
   
   interpretation:
    [b#](x0) = [1 0 0]x0,
    
    [a#](x0) = [0 1 0]x0,
    
              [0 2 0]  
    [a](x0) = [0 0 1]x0
              [1 0 0]  ,
    
              [2 1 1]     [1]
    [b](x0) = [0 0 0]x0 + [0]
              [2 0 0]     [0]
   orientation:
    a#(a(a(b(x1)))) = [2 1 1]x1 + [1] >= [0 1 0]x1 = a#(x1)
    
    a#(a(a(b(x1)))) = [2 1 1]x1 + [1] >= [0 0 1]x1 = a#(a(x1))
    
    a#(a(a(b(x1)))) = [2 1 1]x1 + [1] >= [1 0 0]x1 = a#(a(a(x1)))
    
    a#(a(a(b(x1)))) = [2 1 1]x1 + [1] >= [2 0 0]x1 = b#(a(a(a(x1))))
    
    b#(b(x1)) = [2 1 1]x1 + [1] >= [0] = a#(b(x1))
    
    b#(b(x1)) = [2 1 1]x1 + [1] >= [0] = b#(a(b(x1)))
    
    b#(b(x1)) = [2 1 1]x1 + [1] >= [0] = a#(b(a(b(x1))))
    
                     [4 2 2]     [2]    [4 2 2]     [1]                 
    a(a(a(b(x1)))) = [0 0 0]x1 + [0] >= [0 0 0]x1 + [0] = b(a(a(a(x1))))
                     [4 0 0]     [0]    [4 0 0]     [0]                 
    
               [6 2 2]     [3]    [0 0 0]     [0]                 
    b(b(x1)) = [0 0 0]x1 + [0] >= [0 0 0]x1 + [0] = a(b(a(b(x1))))
               [4 2 2]     [2]    [4 1 1]     [2]                 
   problem:
    DPs:
     
    TRS:
     a(a(a(b(x1)))) -> b(a(a(a(x1))))
     b(b(x1)) -> a(b(a(b(x1))))
   Qed