NO

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

Proof:
 String Reversal Processor:
  a(x1) -> x1
  a(a(a(x1))) -> b(x1)
  c(b(x1)) -> b(c(c(a(x1))))
  c(x1) -> x1
  Unfolding Processor:
   loop length: 10
   terms:
    c(b(b(a(a(x45920)))))
    b(c(c(a(b(a(a(x45920)))))))
    b(c(c(b(a(a(x45920))))))
    b(c(b(c(c(a(a(a(x45920))))))))
    b(c(b(c(c(b(x45920))))))
    b(c(b(c(b(c(c(a(x45920))))))))
    b(c(b(b(c(c(a(c(c(a(x45920))))))))))
    b(c(b(b(c(a(c(c(a(x45920)))))))))
    b(c(b(b(a(c(c(a(x45920))))))))
    b(c(b(b(a(c(a(x45920)))))))
   context: b([])
   substitution:
    x45920 -> x45920
   Qed