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)))))))
  Usable Rule 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:
    
   KBO Processor:
    weight function:
     w0 = 1
     w(b#) = w(a) = w(b) = w(c) = 1
    precedence:
     b# > a > b ~ c
    problem:
     DPs:
      
     TRS:
      
    Qed