YES

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

Proof:
 DP Processor:
  DPs:
   b#(a(c(b(b(x1))))) -> b#(a(c(x1)))
   b#(a(c(b(b(x1))))) -> b#(b(a(c(x1))))
   b#(a(c(b(b(x1))))) -> b#(b(b(a(c(x1)))))
  TRS:
   b(a(c(b(b(x1))))) -> a(c(b(b(b(a(c(x1)))))))
  Arctic Interpretation Processor:
   dimension: 1
   interpretation:
    [b#](x0) = x0,
    
    [a](x0) = x0,
    
    [c](x0) = x0,
    
    [b](x0) = 8x0
   orientation:
    b#(a(c(b(b(x1))))) = 16x1 >= x1 = b#(a(c(x1)))
    
    b#(a(c(b(b(x1))))) = 16x1 >= 8x1 = b#(b(a(c(x1))))
    
    b#(a(c(b(b(x1))))) = 16x1 >= 16x1 = b#(b(b(a(c(x1)))))
    
    b(a(c(b(b(x1))))) = 24x1 >= 24x1 = a(c(b(b(b(a(c(x1)))))))
   problem:
    DPs:
     b#(a(c(b(b(x1))))) -> b#(b(b(a(c(x1)))))
    TRS:
     b(a(c(b(b(x1))))) -> a(c(b(b(b(a(c(x1)))))))
   Matrix Interpretation Processor: dim=4
    
    interpretation:
     [b#](x0) = [0 0 0 1]x0,
     
               [0 0 0 0]  
               [0 0 0 0]  
     [a](x0) = [0 0 0 0]x0
               [1 0 0 0]  ,
     
               [0 0 1 0]  
               [0 0 0 0]  
     [c](x0) = [0 0 0 0]x0
               [0 0 0 0]  ,
     
               [0 0 1 0]     [0]
               [0 0 0 0]     [1]
     [b](x0) = [1 1 0 0]x0 + [0]
               [0 0 0 1]     [0]
    orientation:
     b#(a(c(b(b(x1))))) = [0 0 1 0]x1 + [1] >= [0 0 1 0]x1 = b#(b(b(a(c(x1)))))
     
                         [0 0 0 0]     [0]    [0]                          
                         [0 0 0 0]     [1]    [0]                          
     b(a(c(b(b(x1))))) = [0 0 0 0]x1 + [0] >= [0] = a(c(b(b(b(a(c(x1)))))))
                         [0 0 1 0]     [1]    [1]                          
    problem:
     DPs:
      
     TRS:
      b(a(c(b(b(x1))))) -> a(c(b(b(b(a(c(x1)))))))
    Qed