YES

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

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