YES

Problem:
 a(a(b(b(x1)))) -> b(b(b(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)))
  TRS:
   a(a(b(b(x1)))) -> b(b(b(a(a(a(x1))))))
  Matrix Interpretation Processor: dim=5
   
   interpretation:
    [a#](x0) = [1 0 0 0 0]x0,
    
              [0 0 0 1 1]  
              [1 0 1 0 0]  
    [a](x0) = [0 0 0 0 1]x0
              [0 0 0 0 0]  
              [1 0 0 0 0]  ,
    
              [0 0 0 0 0]     [0]
              [1 0 0 1 1]     [1]
    [b](x0) = [1 0 0 0 0]x0 + [0]
              [0 0 1 0 0]     [0]
              [0 1 0 0 0]     [0]
   orientation:
    a#(a(b(b(x1)))) = [2 0 0 1 1]x1 + [1] >= [1 0 0 0 0]x1 = a#(x1)
    
    a#(a(b(b(x1)))) = [2 0 0 1 1]x1 + [1] >= [0 0 0 1 1]x1 = a#(a(x1))
    
    a#(a(b(b(x1)))) = [2 0 0 1 1]x1 + [1] >= [1 0 0 0 0]x1 = a#(a(a(x1)))
    
                     [0 0 0 0 0]     [0]    [0 0 0 0 0]     [0]                       
                     [3 0 0 2 2]     [2]    [1 0 0 2 2]     [2]                       
    a(a(b(b(x1)))) = [0 0 0 0 0]x1 + [0] >= [0 0 0 0 0]x1 + [0] = b(b(b(a(a(a(x1))))))
                     [0 0 0 0 0]     [0]    [0 0 0 0 0]     [0]                       
                     [2 0 0 1 1]     [1]    [2 0 0 1 1]     [1]                       
   problem:
    DPs:
     
    TRS:
     a(a(b(b(x1)))) -> b(b(b(a(a(a(x1))))))
   Qed