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) = [0 1 0]x0 + [2],
    
              [0 2 0]  
    [a](x0) = [0 0 1]x0
              [0 1 0]  ,
    
              [0 2 1]     [2]
    [b](x0) = [0 0 0]x0 + [0]
              [1 0 0]     [0]
   orientation:
    a#(a(b(b(x1)))) = [0 2 1]x1 + [4] >= [0 1 0]x1 + [2] = a#(x1)
    
    a#(a(b(b(x1)))) = [0 2 1]x1 + [4] >= [0 0 1]x1 + [2] = a#(a(x1))
    
    a#(a(b(b(x1)))) = [0 2 1]x1 + [4] >= [0 1 0]x1 + [2] = a#(a(a(x1)))
    
    a#(a(b(b(x1)))) = [0 2 1]x1 + [4] >= [0 0 1]x1 + [2] = a#(a(a(a(x1))))
    
    a#(a(b(b(x1)))) = [0 2 1]x1 + [4] >= [0 1 0]x1 + [2] = a#(a(a(a(a(x1)))))
    
                     [0 4 2]     [4]    [0 1 2]     [4]                             
    a(a(b(b(x1)))) = [0 0 0]x1 + [0] >= [0 0 0]x1 + [0] = b(b(b(a(a(a(a(a(x1))))))))
                     [0 2 1]     [2]    [0 2 0]     [2]                             
   problem:
    DPs:
     
    TRS:
     a(a(b(b(x1)))) -> b(b(b(a(a(a(a(a(x1))))))))
   Qed