(fvco3d is a natural deduction form of fvco3 proved here: https://github.com/metamath/set.mm/commit/454132a35254c17c4e54353b5c2c772eeb2ebb65#diff-12dc1484b058d1ee6cb68a74194d66c7R641421-R641429 )
-stan On Sat, Mar 7, 2020 at 8:54 AM Stanislas Polu <[email protected]> wrote: > Interesting! > > It basically comes from an attempt at satisfying the conditions > for eliman0. Here's the full proof draft: > > ``` > $( <MM> <PROOF_ASST> THEOREM=imo72b2lem1 LOC_AFTER=? > > 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 ) > > !d84:: |- ( ( ph /\ x e. RR ) -> ( ( abs o. F ) ` x ) =/= (/) ) > d83:d84:adantrr |- ( ( ph /\ ( x e. RR /\ ( F ` x ) =/= 0 ) ) -> ( > ( abs o. F ) ` x ) =/= (/) ) > d81:d83:neneqd |- ( ( ph /\ ( x e. RR /\ ( F ` x ) =/= 0 ) ) -> -. ( > ( abs o. F ) ` x ) = (/) ) > > d95:1:ffvelrnda |- ( ( ph /\ x e. RR ) -> ( F ` x ) e. RR ) > d94:d95:adantrr |- ( ( ph /\ ( x e. RR /\ ( F ` x ) =/= 0 ) ) -> ( F > ` x ) e. RR ) > d92:d94:recnd |- ( ( ph /\ ( x e. RR /\ ( F ` x ) =/= 0 ) ) -> ( > F ` x ) e. CC ) > d93::simprr |- ( ( ph /\ ( x e. RR /\ ( F ` x ) =/= 0 ) ) -> ( F > ` x ) =/= 0 ) > d91:d92,d93:absrpcld |- ( ( ph /\ ( x e. RR /\ ( F ` x ) =/= 0 ) ) -> ( > abs ` ( F ` x ) ) e. RR+ ) > d9:d91:rpgt0d |- ( ( ph /\ ( x e. RR /\ ( F ` x ) =/= 0 ) ) -> 0 < > ( abs ` ( F ` x ) ) ) > > d80::simprl |- ( ( ph /\ ( x e. RR /\ ( F ` x ) =/= 0 ) ) -> x e. > RR ) > d82::eliman0 |- ( ( x e. RR /\ -. ( ( abs o. F ) ` x ) = (/) ) -> > ( ( abs o. F ) ` x ) e. ( ( abs o. F ) " RR ) ) > d75:d80,d81,d82:syl2anc |- ( ( ph /\ ( x e. RR /\ ( F ` x ) =/= 0 ) ) -> > ( ( abs o. F ) ` x ) e. ( ( abs o. F ) " RR ) ) > oeqaa::imaco |- ( ( abs o. F ) " RR ) = ( abs " ( F " RR ) ) > d71:d75,oeqaa:syl6eleq |- ( ( ph /\ ( x e. RR /\ ( F ` x ) =/= 0 ) ) -> > ( ( abs o. F ) ` x ) e. ( abs " ( F " RR ) ) ) > d73:1:adantr |- ( ( ph /\ ( x e. RR /\ ( F ` x ) =/= 0 ) ) -> F : > RR --> RR ) > d74::simprl |- ( ( ph /\ ( x e. RR /\ ( F ` x ) =/= 0 ) ) -> x e. > RR ) > d72:d73,d74:fvco3d |- ( ( ph /\ ( x e. RR /\ ( F ` x ) =/= 0 ) ) -> ( ( > abs o. F ) ` x ) = ( abs ` ( F ` x ) ) ) > d7:d72,d71:eqeltrrd |- ( ( ph /\ ( x e. RR /\ ( F ` x ) =/= 0 ) ) -> ( > abs ` ( F ` x ) ) e. ( abs " ( F " RR ) ) ) > > d10::simpr |- ( ( ( ph /\ ( x e. RR /\ ( F ` x ) =/= 0 ) ) /\ > z = ( abs ` ( F ` x ) ) ) -> z = ( abs ` ( F ` x ) ) ) > d8:d10:breq2d |- ( ( ( ph /\ ( x e. RR /\ ( F ` x ) =/= 0 ) ) /\ z > = ( abs ` ( F ` x ) ) ) -> ( 0 < z <-> 0 < ( abs ` ( F ` x ) ) ) ) > > d6:d7,d8,d9:rspcedvd |- ( ( ph /\ ( x e. RR /\ ( F ` x ) =/= 0 ) ) -> E. > z e. ( abs " ( F " RR ) ) 0 < z ) > qed:2,d6:rexlimddv |- ( ph -> E. z e. ( abs " ( F " RR ) ) 0 < z ) > > $) > > -stan > > On Fri, Mar 6, 2020 at 11:58 PM Mario Carneiro <[email protected]> wrote: > >> 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. >> 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/CAFXXJSv0pGnDb%2Bfae5_ZS_mAK_R6MeMQPAL4gsHD-9S5g2XSVQ%40mail.gmail.com >> <https://groups.google.com/d/msgid/metamath/CAFXXJSv0pGnDb%2Bfae5_ZS_mAK_R6MeMQPAL4gsHD-9S5g2XSVQ%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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