[Agda] Total: x !<= y ==> y <= x
Daniel Peebles
pumpkingod at gmail.com
Wed Oct 10 19:27:06 CEST 2012
I think I've suggested this before, but there's super hyperlinked
documentation online at
http://www.cse.chalmers.se/~nad/listings/lib/Everything.html#1, so if you
navigate to a module that references something you don't know, you'll be
able to click on symbols you don't recognize (like \u+) and it will take
you to their definitions.
Another option to search is simply to type \u+ into Agda-mode, take the
symbol it gives you, paste that into grep, and grep anyway.
On Wed, Oct 10, 2012 at 11:59 AM, Serge D. Mechveliani <mechvel at botik.ru>wrote:
> People,
> I have a beginner question on usage of DecTotalOrder of Standard
> library.
>
> I have ord : DecTotalOrder,
> _<=_ extracted from this ord,
> ``with x <=? y'' produced no x-!<=-y
> (I think, x-!<=-y : \neg x <= y )
>
> (here I replace \<= with <= ).
>
> And from this x-!<=-y, I need to construct y<=x : y <= x.
> Because I think that this a constructive reifation for a classical
> lemma for (Total <=) of that not (x <= y) ==> y <= x.
> I see in Relation.Binary, Core:
>
> record IsTotalOrder ...
> isPartialOrder : IsPartialOrder _~~_ _<=_
> total : Total _<=_
>
> (here I replace some math symbols)
>
> Total : forall {a l} {A : Set a} -> Rel A l -> Set _
> Total _~_ = forall x y -> (x ~ y) \u+ (y ~ x)
>
> , where \u+ denotes, I think, something close to union of types.
> How to operate with a data from (P \u+ Q) ?
> what are the constructors?
> Probably, one needs to apply total <= x y : (x <= y) \u+ (y <= x)
> and then to use x-!<=-y : \neg x <= y
>
> Where \u+ is defined in the Standard library?
> It is visible a certain `combinator' <\u+>, with the denotation
> [ P , Q ]. I do not know of what does all this mean.
>
> If it was not a math symbol, I could apply grep foo *.agda.
>
> May be, Standard Library provides somewhere a lemma, like
>
> (\neg x <= y) includedInto (y <= x) for _<=_ from TotalOrder ?
>
> Thank you in advance for explanation,
>
> ------
> Sergei
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>
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