NO

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

Proof:
 String Reversal Processor:
  a(x1) -> x1
  b(a(x1)) -> a(b(c(a(b(x1)))))
  b(x1) -> a(x1)
  c(c(x1)) -> x1
  Unfolding Processor:
   loop length: 8
   terms:
    b(a(a(x36400)))
    a(b(c(a(b(a(x36400))))))
    a(b(c(b(a(x36400)))))
    a(b(c(a(b(c(a(b(x36400))))))))
    a(b(c(b(c(a(b(x36400)))))))
    a(b(c(a(c(a(b(x36400)))))))
    a(b(c(c(a(b(x36400))))))
    a(b(a(b(x36400))))
   context: a([])
   substitution:
    x36400 -> x36400
   Qed