User blog comment:Wythagoras/All my stuff/@comment-10429372-20130715145943/@comment-1605058-20130715163339

But we can simply assume ¥ is never in iota function set, because it leads to circular definition (we can define in English iota function, you actually did it). Actually, iota will never use any function diagonalizing through definability, so such functions very likely outgrow iota.