YES

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

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