The composition function value combination is easy enough to eliminate using fvco, but the equality to the empty set is a type error, because the lhs is a real number and the question of whether the empty set is a real number is deliberately left ambiguous by the real number axioms. So I would like to know what steps got you to this point. There are some function value theorems that assume this as a "convenience" but there should be analogues of them that don't (probably with some other assumption like the function is a set).
On Fri, Mar 6, 2020 at 8:02 AM 'Stanislas Polu' via Metamath < [email protected]> wrote: > Thanks again Mario! > > I made more progress towards the final demonstration of the full IMO > problem. Working on the following lemma: > > ``` > h1::imo72b2lem1.1 |- ( ph -> F : RR --> RR ) > > h2::imo72b2lem1.7 |- ( ph -> E. x e. RR ( F ` x ) =/= 0 ) > > h3::imo72b2lem0.6 |- ( ph -> A. y e. RR ( abs ` ( F ` y ) ) <_ 1 ) > ``` > > I need to prove the following goal which seems pretty obvious but I'm > struggling to find a way to discharge it: > > ``` > d84:: |- ( ( ph /\ x e. RR ) -> ( ( abs o. F ) ` x ) =/= (/) ) > ``` > > Any idea on how to proceed with this? > > Thanks thanks! > > -stan > > On Thu, Mar 5, 2020 at 6:27 PM Mario Carneiro <[email protected]> wrote: > >> There is a theorem specifically for that translation, something like A. x >> e. ( F " A ) P[x] <-> A. y e. A P[( F ` y )]. It's probably called ralima >> but you've caught me on the bus again. >> >> Mario >> >> On Thu, Mar 5, 2020, 8:07 AM 'Stanislas Polu' via Metamath < >> [email protected]> wrote: >> >>> Thanks Mario! >>> >>> I just finished formalizing the following lemma (which is a good chunk >>> of the proof \o/): >>> >>> ``` >>> $d F c x $. $d c ph x $. >>> imo72b2lem.1 $e |- ( ph -> F : RR --> RR ) $. >>> imo72b2lem.2 $e |- ( ph -> G : RR --> RR ) $. >>> imo72b2lem.3 $e |- ( ph -> A e. RR ) $. >>> imo72b2lem.4 $e |- ( ph -> B e. RR ) $. >>> imo72b2lem.5 $e |- ( ph -> ( ( F ` ( A + B ) ) + ( F ` ( A - B ) ) ) >>> = ( 2 x. ( ( F ` A ) x. ( G ` B ) ) ) ) $. >>> imo72b2lem.6 $e |- ( ph -> A. x e. ( abs " ( F " RR ) ) x <_ 1 ) $. >>> imo72b2lem.7 $e |- ( ph -> E. x e. RR ( F ` x ) =/= 0 ) $. >>> >>> imo72b2lem $p |- ( ph -> ( ( abs ` ( F ` A ) ) x. ( abs ` ( G ` B ) >>> ) ) <_ sup ( ( abs " ( F " RR ) ) , RR , < ) ) $= >>> ``` >>> >>> Proof here: >>> https://github.com/spolu/set.mm/commit/454132a35254c17c4e54353b5c2c772eeb2ebb65 >>> >>> One thing I'm quite dissatisfied with is the shape of `imo72b2lem.6`. >>> I'd much rather have the more natural/intuitive expression `|- ( ph -> A. x >>> e. RR ( abs ` ( F `x ) ) <_ 1 )` but I completely failed to >>> prove imo72b2lem.6 from it. Any help on this would be greatly appreciated! >>> >>> -stan >>> >>> On Wed, Mar 4, 2020 at 8:45 PM Mario Carneiro <[email protected]> wrote: >>> >>>> Can't look right now, but you should search for a theorem of the form A >>>> = (/) <-> ( F " A ) = (/) . >>>> >>>> On Wed, Mar 4, 2020, 11:30 AM 'Stanislas Polu' via Metamath < >>>> [email protected]> wrote: >>>> >>>>> I'm now looking to prove that `( abs " ( F " RR ) ) =/= (/)` given `F >>>>> : RR --> RR`. From my exploration of the definition of --> I believe this >>>>> should be enough but I don't see an easy path towards that? Would anybody >>>>> have an example in mind that could give me a little bit of inspiration? >>>>> >>>>> Thanks for the continued support! >>>>> >>>>> -stan >>>>> >>>>> On Wed, Mar 4, 2020 at 6:29 PM Benoit <[email protected]> wrote: >>>>> >>>>>> Stan: you're right about the need to prove this (if using explicit >>>>>> substitution): look for the utility theorems exchanging [. / ]. with >>>>>> other >>>>>> symbols (quantifiers, operations). As said by Jim and Thierry, who are >>>>>> more experienced in proof building, implicit substitution might be easier >>>>>> to use. I think it is instructive to compare the details of both proving >>>>>> styles on a specific example (e.g. ralbidv, suggested by Thierry, would >>>>>> be >>>>>> analogous to exchanging [. / ]. with A.). >>>>>> >>>>>> Still, I think adding what I called rspesbcd could prove useful (if >>>>>> it is not already in set.mm under another label; I cannot search >>>>>> now, but it probably is already somewhere). >>>>>> >>>>>> Benoît >>>>>> >>>>>> -- >>>>>> You received this message because you are subscribed to the Google >>>>>> Groups "Metamath" group. >>>>>> To unsubscribe from this group and stop receiving emails from it, >>>>>> send an email to [email protected]. >>>>>> To view this discussion on the web visit >>>>>> https://groups.google.com/d/msgid/metamath/ebdd296f-6bcd-49e0-88c8-c7fef3628cdd%40googlegroups.com >>>>>> <https://groups.google.com/d/msgid/metamath/ebdd296f-6bcd-49e0-88c8-c7fef3628cdd%40googlegroups.com?utm_medium=email&utm_source=footer> >>>>>> . >>>>>> >>>>> -- >>>>> You received this message because you are subscribed to the Google >>>>> Groups "Metamath" group. >>>>> To unsubscribe from this group and stop receiving emails from it, send >>>>> an email to [email protected]. >>>>> To view this discussion on the web visit >>>>> https://groups.google.com/d/msgid/metamath/CACZd_0zFiVT5n-7%2BYh-YL2mDCLMom6R66gq7gbMT7tgJzTzadQ%40mail.gmail.com >>>>> <https://groups.google.com/d/msgid/metamath/CACZd_0zFiVT5n-7%2BYh-YL2mDCLMom6R66gq7gbMT7tgJzTzadQ%40mail.gmail.com?utm_medium=email&utm_source=footer> >>>>> . >>>>> >>>> -- >>>> You received this message because you are subscribed to the Google >>>> Groups "Metamath" group. >>>> To unsubscribe from this group and stop receiving emails from it, send >>>> an email to [email protected]. >>>> To view this discussion on the web visit >>>> https://groups.google.com/d/msgid/metamath/CAFXXJStosWZ0BEO7zjpeW90VMNT3pf-Kr6nUaqLefJyA%3Dgwa2Q%40mail.gmail.com >>>> <https://groups.google.com/d/msgid/metamath/CAFXXJStosWZ0BEO7zjpeW90VMNT3pf-Kr6nUaqLefJyA%3Dgwa2Q%40mail.gmail.com?utm_medium=email&utm_source=footer> >>>> . >>>> >>> -- >>> You received this message because you are subscribed to the Google >>> Groups "Metamath" group. >>> To unsubscribe from this group and stop receiving emails from it, send >>> an email to [email protected]. >>> To view this discussion on the web visit >>> https://groups.google.com/d/msgid/metamath/CACZd_0zgFkBD0kOaAXNGyB3PL18Qt06uSHV%3DEmwOjh6Rk5j3OQ%40mail.gmail.com >>> <https://groups.google.com/d/msgid/metamath/CACZd_0zgFkBD0kOaAXNGyB3PL18Qt06uSHV%3DEmwOjh6Rk5j3OQ%40mail.gmail.com?utm_medium=email&utm_source=footer> >>> . >>> >> -- >> You received this message because you are subscribed to the Google Groups >> "Metamath" group. >> To unsubscribe from this group and stop receiving emails from it, send an >> email to [email protected]. >> To view this discussion on the web visit >> https://groups.google.com/d/msgid/metamath/CAFXXJSs4iVxdk7Uz1_bW-heUVCJy-CMGENLNtzWd8%2B12rFEVBQ%40mail.gmail.com >> <https://groups.google.com/d/msgid/metamath/CAFXXJSs4iVxdk7Uz1_bW-heUVCJy-CMGENLNtzWd8%2B12rFEVBQ%40mail.gmail.com?utm_medium=email&utm_source=footer> >> . >> > -- > You received this message because you are subscribed to the Google Groups > "Metamath" group. > To unsubscribe from this group and stop receiving emails from it, send an > email to [email protected]. > To view this discussion on the web visit > https://groups.google.com/d/msgid/metamath/CACZd_0x2N3k16mC%2Byc_%3D6HvNQ3OWpAXtQ3fHkMv%3DPqERwFR60Q%40mail.gmail.com > <https://groups.google.com/d/msgid/metamath/CACZd_0x2N3k16mC%2Byc_%3D6HvNQ3OWpAXtQ3fHkMv%3DPqERwFR60Q%40mail.gmail.com?utm_medium=email&utm_source=footer> > . > -- You received this message because you are subscribed to the Google Groups "Metamath" group. 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