NO

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

Proof:
 String Reversal Processor:
  a(a(a(a(x1)))) -> c(c(a(c(a(x1)))))
  c(c(c(x1))) -> a(a(a(x1)))
  Unfolding Processor:
   loop length: 7
   terms:
    a(a(a(a(a(a(a(a(a(a(a(a(a(x2583)))))))))))))
    c(c(a(c(a(a(a(a(a(a(a(a(a(a(x2583))))))))))))))
    c(c(a(c(c(c(a(c(a(a(a(a(a(a(a(x2583)))))))))))))))
    c(c(a(a(a(a(a(c(a(a(a(a(a(a(a(x2583)))))))))))))))
    c(c(a(a(a(a(a(c(c(c(a(c(a(a(a(a(x2583))))))))))))))))
    c(c(a(a(a(a(a(a(a(a(a(c(a(a(a(a(x2583))))))))))))))))
    c(c(a(a(a(a(a(a(a(a(a(c(c(c(a(c(a(x2583)))))))))))))))))
   context: c(c([]))
   substitution:
    x2583 -> c(a(x2583))
   Qed