NO

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

Proof:
 String Reversal Processor:
  a(x1) -> x1
  a(x1) -> b(x1)
  c(b(b(x1))) -> b(a(c(c(c(a(x1))))))
  Unfolding Processor:
   loop length: 11
   terms:
    c(b(b(b(b(x161959)))))
    b(a(c(c(c(a(b(b(x161959))))))))
    b(a(c(c(c(b(b(b(x161959))))))))
    b(a(c(c(b(a(c(c(c(a(b(x161959)))))))))))
    b(a(c(c(b(b(c(c(c(a(b(x161959)))))))))))
    b(a(c(c(b(b(c(c(c(b(b(x161959)))))))))))
    b(a(c(c(b(b(c(c(b(a(c(c(c(a(x161959))))))))))))))
    b(a(c(c(b(b(c(c(b(b(c(c(c(a(x161959))))))))))))))
    b(a(c(c(b(b(c(b(a(c(c(c(a(c(c(c(a(x161959)))))))))))))))))
    b(a(c(c(b(b(c(b(b(c(c(c(a(c(c(c(a(x161959)))))))))))))))))
    b(a(c(c(b(b(b(a(c(c(c(a(c(c(c(a(c(c(c(a(x161959))))))))))))))))))))
   context: b(a(c([])))
   substitution:
    x161959 -> c(c(c(a(c(c(c(a(c(c(c(a(x161959))))))))))))
   Qed