NO

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

Proof:
 String Reversal Processor:
  b(a(a(x1))) -> a(a(a(c(c(x1)))))
  c(a(x1)) -> a(b(x1))
  Unfolding Processor:
   loop length: 10
   terms:
    b(a(a(a(a(a(a(a(a(x5444)))))))))
    a(a(a(c(c(a(a(a(a(a(a(x5444)))))))))))
    a(a(a(c(a(b(a(a(a(a(a(x5444)))))))))))
    a(a(a(a(b(b(a(a(a(a(a(x5444)))))))))))
    a(a(a(a(b(a(a(a(c(c(a(a(a(x5444)))))))))))))
    a(a(a(a(b(a(a(a(c(a(b(a(a(x5444)))))))))))))
    a(a(a(a(b(a(a(a(a(b(b(a(a(x5444)))))))))))))
    a(a(a(a(b(a(a(a(a(b(a(a(a(c(c(x5444)))))))))))))))
    a(a(a(a(b(a(a(a(a(a(a(a(c(c(a(c(c(x5444)))))))))))))))))
    a(a(a(a(b(a(a(a(a(a(a(a(c(a(b(c(c(x5444)))))))))))))))))
   context: a(a(a(a([]))))
   substitution:
    x5444 -> b(b(c(c(x5444))))
   Qed