NO

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

Proof:
 Unfolding Processor:
  loop length: 7
  terms:
   a(b(c(x9865)))
   b(a(a(c(a(a(c(x9865)))))))
   b(a(a(c(a(c(x9865))))))
   b(a(a(c(c(x9865)))))
   b(a(a(b(x9865))))
   b(a(b(a(a(c(a(a(x9865))))))))
   b(a(b(a(c(a(a(x9865)))))))
  context: b([])
  substitution:
   x9865 -> a(a(x9865))
  Qed