NO

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

Proof:
 String Reversal Processor:
  c(b(x1)) -> b(b(a(x1)))
  a(b(x1)) -> b(c(a(x1)))
  Unfolding Processor:
   loop length: 10
   terms:
    c(b(b(b(b(b(x6553))))))
    b(b(a(b(b(b(b(x6553)))))))
    b(b(b(c(a(b(b(b(x6553))))))))
    b(b(b(c(b(c(a(b(b(x6553)))))))))
    b(b(b(c(b(c(b(c(a(b(x6553))))))))))
    b(b(b(c(b(b(b(a(c(a(b(x6553)))))))))))
    b(b(b(c(b(b(b(a(c(b(c(a(x6553))))))))))))
    b(b(b(c(b(b(b(a(b(b(a(c(a(x6553)))))))))))))
    b(b(b(c(b(b(b(b(c(a(b(a(c(a(x6553))))))))))))))
    b(b(b(c(b(b(b(b(c(b(c(a(a(c(a(x6553)))))))))))))))
   context: b(b(b([])))
   substitution:
    x6553 -> b(a(c(a(a(c(a(x6553)))))))
   Qed