YES

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

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