NO

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

Proof:
 String Reversal Processor:
  a(x1) -> x1
  b(a(x1)) -> a(a(b(c(a(b(x1))))))
  b(x1) -> x1
  c(c(x1)) -> x1
  Unfolding Processor:
   loop length: 8
   terms:
    b(a(a(a(x46599))))
    a(a(b(c(a(b(a(a(x46599))))))))
    a(a(b(c(b(a(a(x46599)))))))
    a(a(b(c(a(a(b(c(a(b(a(x46599)))))))))))
    a(a(b(c(a(b(c(a(b(a(x46599))))))))))
    a(a(b(c(b(c(a(b(a(x46599)))))))))
    a(a(b(c(c(a(b(a(x46599))))))))
    a(a(b(a(b(a(x46599))))))
   context: a(a([]))
   substitution:
    x46599 -> b(c(a(b(x46599))))
   Qed