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feat: linear order is isomorphic to lexicographic sum of two intervals #20409
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feat: linear order is isomorphic to lexicographic sum of two intervals #20409
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It would be nice to have the
trans
andrefl
versions of this too.There was a problem hiding this comment.
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Where does this complaining happen? If you're able to make a mwe, can you file an issue with it?
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This complaining is addressed in #20479, I can make this a dependent PR so that this doesn't come up
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Do you mean #20475?
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I think you should be composing with
ofLex
here somehow. Do we haveofLex
as aRelIso
forSum
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There's some pretty subtle def-eq finagling going on here: the reason
ofRelIsoLT
works is becauseSum.Lex (· < ·) (· < ·) (Iio x) (Ici x)
is def-eq to the<
relation onIio x ⊕ₗ Ici x
. I'm not sure this can be made less abusive by just insertingofLex
.In any case things should probably be fine as long as we keep the actual API free of def-eq abuse.
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I think we should introduce a
Sum.toLexRelIso : Sum.Lex (. < .) (. < .) ≃r (fun a b : Lex (Sum _ _) => a < b)
or whatever the true version is, withsimp
lemmas reducing totoLex
, and cast through that hereThere was a problem hiding this comment.
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It'd be better for
simp
to instead puttoLex
on the RHS.There was a problem hiding this comment.
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That suggestion doesn't make any sense to me, the type of this bundled map is
Lex (Sum (Iio x) (Ici x)) ≃o α
, so when applying the map to aSum
you have to calltoLex
first.(I realize I made a type above too...)
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Oh nevermind, I see why what I was saying is nonsense. And in fact I'm missing some other
ofLex
elsewhere.