NO

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

Proof:
 Unfolding Processor:
  loop length: 10
  terms:
   a(b(b(b(x126166))))
   b(b(a(c(c(a(b(b(x126166))))))))
   b(b(a(c(c(b(b(a(c(c(a(b(x126166))))))))))))
   b(b(a(c(b(a(c(c(a(b(x126166))))))))))
   b(b(a(a(c(c(a(b(x126166))))))))
   b(b(a(a(c(c(b(b(a(c(c(a(x126166))))))))))))
   b(b(a(a(c(b(a(c(c(a(x126166))))))))))
   b(b(a(a(a(c(c(a(x126166))))))))
   b(b(a(a(b(c(c(a(x126166))))))))
   b(b(a(b(b(a(c(c(a(c(c(a(x126166))))))))))))
  context: b(b([]))
  substitution:
   x126166 -> c(c(a(c(c(a(x126166))))))
  Qed