YES

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

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