<div dir="ltr">Oh, never mind. Saw Jesper's reply now. Thanks :).<div><br></div><div>/ Ulf</div></div><div class="gmail_extra"><br><br><div class="gmail_quote">On Tue, Jul 15, 2014 at 12:47 PM, Ulf Norell <span dir="ltr"><<a href="mailto:ulf.norell@gmail.com" target="_blank">ulf.norell@gmail.com</a>></span> wrote:<br>
<blockquote class="gmail_quote" style="margin:0 0 0 .8ex;border-left:1px #ccc solid;padding-left:1ex"><div dir="ltr"><div class="gmail_extra"><br><div class="gmail_quote"><div class="">On Tue, Jul 15, 2014 at 11:23 AM, Matthieu Sozeau <span dir="ltr"><<a href="mailto:mattam@mattam.org" target="_blank">mattam@mattam.org</a>></span> wrote:<br>
<blockquote class="gmail_quote" style="margin:0px 0px 0px 0.8ex;border-left-width:1px;border-left-color:rgb(204,204,204);border-left-style:solid;padding-left:1ex"><div><div><br></div></div><div>I think this is not gonna be provable, as even if y = z is not inhabited, tranport p y = z,</div>
<div>which is the equality you get on the second components, might be. In other words, without K you might have transport p y <> y for p : x = x. An example is if p : Bool = Bool is the negation iso, B := \x. x, y := true, z := false. So you need to assume A is an HSet here, or is decidable.</div>
</blockquote><div><br></div></div><div>Ok, that's fair. Fortunately in my case equality on A is decidable (I'm trying to lift decidable equality on A and B to decidable equality on<font face="arial, helvetica, sans-serif"> <span style="font-size:13px">Σ A B)</span></font>. How would I use that to prove the puzzle?</div>
<span class="HOEnZb"><font color="#888888">
<div><br></div><div>/ Ulf</div></font></span></div></div></div>
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