NO

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

Proof:
 String Reversal Processor:
  a(a(a(x1))) -> a(a(c(b(b(x1)))))
  b(b(x1)) -> b(a(c(x1)))
  c(x1) -> x1
  Unfolding Processor:
   loop length: 14
   terms:
    a(a(a(b(a(a(b(a(x301512))))))))
    a(a(c(b(b(b(a(a(b(a(x301512))))))))))
    a(a(c(b(b(a(c(a(a(b(a(x301512)))))))))))
    a(a(c(b(a(c(a(c(a(a(b(a(x301512))))))))))))
    a(a(c(b(a(a(c(a(a(b(a(x301512)))))))))))
    a(a(c(b(a(a(a(a(b(a(x301512))))))))))
    a(a(c(b(a(a(a(c(b(b(b(a(x301512))))))))))))
    a(a(c(b(a(a(a(b(b(b(a(x301512)))))))))))
    a(a(c(b(a(a(a(b(b(a(c(a(x301512))))))))))))
    a(a(c(b(a(a(a(b(a(c(a(c(a(x301512)))))))))))))
    a(a(c(b(a(a(a(b(a(a(c(a(x301512))))))))))))
    a(a(c(b(a(a(a(b(a(a(a(x301512)))))))))))
    a(a(c(b(a(a(a(b(a(a(c(b(b(x301512)))))))))))))
    a(a(c(b(a(a(a(b(a(a(b(b(x301512))))))))))))
   context: a(a(c(b([]))))
   substitution:
    x301512 -> c(x301512)
   Qed