<html><head><meta http-equiv="Content-Type" content="text/html; charset=us-ascii"></head><body style="word-wrap: break-word; -webkit-nbsp-mode: space; line-break: after-white-space;" class=""><br class=""><div><br class=""><blockquote type="cite" class=""><div class="">On Jul 6, 2022, at 4:35 AM, Simon Peyton Jones <<a href="mailto:simon.peytonjones@gmail.com" class="">simon.peytonjones@gmail.com</a>> wrote:</div><div class=""><div dir="ltr" class=""><div class="gmail_default" style="font-family:tahoma,sans-serif"><br class=""></div><div class="gmail_default" style="font-family:tahoma,sans-serif">I sort of wonder: if we have (Typeable t) should that not give us (Typeable k) where (t :: k)? That would require us to have a function typeRepKind :: TypeRep (a::k) -> TypeRep k, and I don't know how hard that is to get.</div></div></div></blockquote><br class=""></div><div>We have that function. It's called typeRepKind. What we need for this example is a superclass constraint:</div><div><br class=""></div><div>class Typeable k => Typeable (a :: k)</div><div><br class=""></div><div>which would probably require UndecidableSuperClasses, but otherwise might just work.</div><div><br class=""></div><div>Richard</div><br class=""></body></html>