That's the first part of the proof. The second part of the proof is to show a contradiction from this, because +oo is in the RHS but not the LHS of that equality.
On Mon, Oct 14, 2024 at 3:24 PM Glauco <[email protected]> wrote: > I think I got it. Thank you very much for the insight. > > h1::temp4.1 |- { x e. RR | ( F ` x ) = 4 } = { a , b } > 2::pnfex |- +oo e. _V > 3::preq1 |- ( a = +oo -> { a , b } = { +oo , b } ) > 4:3:eqeq2d |- ( a = +oo -> ( { x e. RR | ( F ` x ) = 4 } = { a , > b } <-> { x e. RR | ( F ` x ) = 4 } = { +oo , b } ) ) > qed:2,4,1:vtocl |- { x e. RR | ( F ` x ) = 4 } = { +oo , b } > > $= ( cv cfv c4 wceq cr crab cpr cpnf pnfex preq1 eqeq2d vtocl ) > AFBGHIAJKZCFZDF > ZLZIRMTLZICMNSMIUAUBRSMTOPEQ $. > > $d a x > $d a b > $d F a > > > Il giorno lunedì 14 ottobre 2024 alle 15:02:42 UTC+2 Glauco ha scritto: > >> Hi Mario, >> >> I'm sure you're right, and in fact I would use a class A and something >> like ph -> A e. V as an additional hyp. >> >> But I cannot reproduce the contradiction off the top of my head; it would >> be interesting to see it, to gain a deeper understanding. >> >> Can you please show a short proof? (do you proof something like +oo e. >> RR ? ) >> >> Thank you >> Glauco >> >> >> Il giorno lunedì 14 ottobre 2024 alle 14:28:43 UTC+2 [email protected] ha >> scritto: >> >>> It's not safe to make an assumption of the form |- { x e. RR | ( F ` x ) >>> = 4 } = { a , b } because the variables a,b are implicitly universally >>> quantified in the hypothesis, which means you can prove a contradiction >>> from it. (Consider what happens if you use vtocl on the hypothesis to >>> replace a by +oo.) You should either use class variables A,B as I >>> indicated, or add a context ph -> before every assumption (which is >>> *not* assumed to be disjoint from a,b). >>> >>> On Mon, Oct 14, 2024 at 2:12 PM Glauco <[email protected]> wrote: >>> >>>> I take it back, you can get away with { x | ( F ` x ) = 4 } , here's >>>> the proof (but for your larger goal, I doubt it's enough) >>>> >>>> h1::temp3.1 |- F = ( x e. RR |-> ( ( ( k x. ( x ^ 2 ) ) - ( >>>> ( 2 x. k ) x. x ) ) + l ) ) >>>> h2::temp3.2 |- { x | ( F ` x ) = 4 } = { a , b } >>>> 3::fveq2 |- ( x = a -> ( F ` x ) = ( F ` a ) ) >>>> 4:3:eqeq1d |- ( x = a -> ( ( F ` x ) = 4 <-> ( F ` a ) = 4 ) ) >>>> 5::vex |- a e. _V >>>> 6::nfmpt1 |- F/_ x ( x e. RR |-> ( ( ( k x. ( x ^ 2 ) ) - >>>> ( ( 2 x. k ) x. x ) ) + l ) ) >>>> 7:1,6:nfcxfr |- F/_ x F >>>> 8::nfcv |- F/_ x a >>>> 9:7,8:nffv |- F/_ x ( F ` a ) >>>> 10:9:nfeq1 |- F/ x ( F ` a ) = 4 >>>> 11:10,5,4:elabf |- ( a e. { x | ( F ` x ) = 4 } <-> ( F ` a ) = 4 ) >>>> 12::vex |- a e. _V >>>> 13:12:prid1 |- a e. { a , b } >>>> 14:13,2:eleqtrri |- a e. { x | ( F ` x ) = 4 } >>>> qed:14,11:mpbi |- ( F ` a ) = 4 >>>> >>>> $= ( cv cfv c4 wceq cab wcel cpr vex cr c2 co cmul prid1 cexp cmin >>>> caddc nfmpt1 >>>> eleqtrri cmpt nfcxfr nfcv nffv nfeq1 fveq2 eqeq1d elabf mpbi ) >>>> DIZAIZCJZKLZA >>>> >>>> >>>> MZNUPCJZKLZUPUPEIZOUTUPVCDPZUAHUFUSVBAUPAVAKAUPCACAQBIZUQRUBSTSRVETSUQTSUCSF >>>> IUDSZUGGAQVFUEUHAUPUIUJUKVDUQUPLURVAKUQUPCULUMUNUO $. >>>> >>>> $d a x >>>> >>>> Il giorno lunedì 14 ottobre 2024 alle 13:53:35 UTC+2 Glauco ha scritto: >>>> >>>>> Since the >>>>> >>>>> $d F x >>>>> >>>>> constraint would conflict with your F definition, here is an >>>>> alternative version >>>>> >>>>> >>>>> h1::temp3.1 |- F = ( x e. RR |-> ( ( ( k x. ( x ^ 2 ) ) - >>>>> ( ( 2 x. k ) x. x ) ) + l ) ) >>>>> h2::temp3.2 |- { x e. RR | ( F ` x ) = 4 } = { a , b } >>>>> 3::fveq2 |- ( x = a -> ( F ` x ) = ( F ` a ) ) >>>>> 4:3:eqeq1d |- ( x = a -> ( ( F ` x ) = 4 <-> ( F ` a ) = 4 >>>>> ) ) >>>>> 5::nfcv |- F/_ x a >>>>> 6::nfcv |- F/_ x RR >>>>> 7::nfmpt1 |- F/_ x ( x e. RR |-> ( ( ( k x. ( x ^ 2 ) ) >>>>> - ( ( 2 x. k ) x. x ) ) + l ) ) >>>>> 8:1,7:nfcxfr |- F/_ x F >>>>> 9:8,5:nffv |- F/_ x ( F ` a ) >>>>> 10:9:nfeq1 |- F/ x ( F ` a ) = 4 >>>>> 11:5,6,10,4:elrabf |- ( a e. { x e. RR | ( F ` x ) = 4 } <-> ( a e. >>>>> RR /\ ( F ` a ) = 4 ) ) >>>>> 12::vex |- a e. _V >>>>> 13:12:prid1 |- a e. { a , b } >>>>> 14:13,2:eleqtrri |- a e. { x e. RR | ( F ` x ) = 4 } >>>>> 15:14,11:mpbi |- ( a e. RR /\ ( F ` a ) = 4 ) >>>>> qed:15:simpri |- ( F ` a ) = 4 >>>>> >>>>> $= ( cv cr wcel cfv c4 wceq crab wa nfcv c2 co cmul cpr vex prid1 >>>>> eleqtrri cexp >>>>> cmin caddc cmpt nfmpt1 nfcxfr nffv nfeq1 fveq2 eqeq1d elrabf mpbi >>>>> simpri ) D >>>>> >>>>> >>>>> IZJKZURCLZMNZURAIZCLZMNZAJOZKUSVAPURUREIZUAVEURVFDUBUCHUDVDVAAURJAURQZAJQAUT >>>>> >>>>> >>>>> MAURCACAJBIZVBRUESTSRVHTSVBTSUFSFIUGSZUHGAJVIUIUJVGUKULVBURNVCUTMVBURCUMUNUO >>>>> UPUQ $. >>>>> >>>>> $d a x >>>>> >>>>> Il giorno lunedì 14 ottobre 2024 alle 13:42:45 UTC+2 Glauco ha scritto: >>>>> >>>>>> Hi Jorge, >>>>>> >>>>>> with Yamma you get something like this (I doubt you can get away >>>>>> with { x | ( F ` x ) = 4 }, for instance ( F ` +oo ) is not >>>>>> "well-defined" ) >>>>>> >>>>>> ``` >>>>>> $theorem temp3 >>>>>> >>>>>> * comments >>>>>> >>>>>> h1::temp3.1 |- { x e. RR | ( F ` x ) = 4 } = { a , b } >>>>>> 2::fveq2 |- ( x = a -> ( F ` x ) = ( F ` a ) ) >>>>>> 3:2:eqeq1d |- ( x = a -> ( ( F ` x ) = 4 <-> ( F ` a ) = 4 >>>>>> ) ) >>>>>> 4:3:elrab |- ( a e. { x e. RR | ( F ` x ) = 4 } <-> ( a e. >>>>>> RR /\ ( F ` a ) = 4 ) ) >>>>>> 5::vex |- a e. _V >>>>>> 6:5:prid1 |- a e. { a , b } >>>>>> 7:6,1:eleqtrri |- a e. { x e. RR | ( F ` x ) = 4 } >>>>>> 8:7,4:mpbi |- ( a e. RR /\ ( F ` a ) = 4 ) >>>>>> qed:8:simpri |- ( F ` a ) = 4 >>>>>> >>>>>> $= ( cv cr wcel cfv c4 wceq crab wa cpr vex prid1 eleqtrri fveq2 >>>>>> eqeq1d elrab >>>>>> mpbi simpri ) >>>>>> CFZGHZUCBIZJKZUCAFZBIZJKZAGLZHUDUFMUCUCDFZNUJUCUKCOPEQUIUFAUCG >>>>>> UGUCKUHUEJUGUCBRSTUAUB $. >>>>>> >>>>>> $d F x >>>>>> $d a x >>>>>> ``` >>>>>> >>>>>> >>>>>> >>>>>> >>>>>> Il giorno lunedì 14 ottobre 2024 alle 12:40:47 UTC+2 >>>>>> [email protected] ha scritto: >>>>>> >>>>>>> I wanted to check how far I can get in formalizing this problem and >>>>>>> its solution. But I stuck in the very beginning. >>>>>>> >>>>>>> >>>>>>> Firstly, I formalized the initial conditions as Mario suggested: >>>>>>> >>>>>>> >>>>>>> hyp1: |- 0 < k >>>>>>> >>>>>>> hyp2: |- 0 < l >>>>>>> >>>>>>> hyp3: |- F = ( x e. RR |-> ( ( ( k x. ( x ^ 2 ) ) - ( ( 2 x. k ) x. >>>>>>> x ) ) + l ) ) >>>>>>> >>>>>>> hyp4: |- { x | ( F ` x ) = 4 } = { a , b } >>>>>>> >>>>>>> hyp5: |- ( ( ( a - b ) ^ 2 ) + ( ( ( F ` a ) - ( F ` b ) ) ^ 2 ) ) = >>>>>>> ( 6 ^ 2 ) >>>>>>> >>>>>>> hyp6: |- ( ( ( a ^ 2 ) + ( ( F ` a ) ^ 2 ) ) + ( ( b ^ 2 ) + ( ( F ` >>>>>>> b ) ^ 2 ) ) ) = c >>>>>>> >>>>>>> >>>>>>> I used { a , b } instead of { A , B } because then I can easily >>>>>>> prove |- a e. _V, which is used in the proof below. Also, as I >>>>>>> understand, a and b represent x-coordinates of the corresponding points. >>>>>>> >>>>>>> >>>>>>> Next, I wanted to simplify hyp5, by proving that F(a) = F(b) = 4, so >>>>>>> the hyp5 would be |- ( ( a - b ) ^ 2 ) = ( 6 ^ 2 ). But that’s >>>>>>> where I am stuck. I can prove |- [ a / x ] ( F ` x ) = 4 which >>>>>>> looks a right direction to move in: >>>>>>> >>>>>>> >>>>>>> 1| | vex | |- a e. _V >>>>>>> >>>>>>> 2| 1 | prid1 | |- a e. { a , b } >>>>>>> >>>>>>> 3| | hyp4 | |- { x | ( F ` x ) = 4 } = { a , b } >>>>>>> >>>>>>> 4| 3 | eleq2i | |- ( a e. { x | ( F ` x ) = 4 } <-> a e. { a , b >>>>>>> } ) >>>>>>> >>>>>>> 5| 2,4 | mpbir | |- a e. { x | ( F ` x ) = 4 } >>>>>>> >>>>>>> 6| | df-clab | |- ( a e. { x | ( F ` x ) = 4 } <-> [ a / x ] ( F >>>>>>> ` x ) = 4 ) >>>>>>> >>>>>>> 7| 5,6 | mpbi | |- [ a / x ] ( F ` x ) = 4 >>>>>>> >>>>>>> >>>>>>> But I still cannot prove |- ( F ` a ) = 4. Any suggestions what >>>>>>> approaches I can try to prove this? >>>>>>> >>>>>>> On Sunday, July 28, 2024 at 11:54:47 PM UTC+2 [email protected] >>>>>>> wrote: >>>>>>> >>>>>>>> On Sun, Jul 28, 2024 at 11:43 PM Glauco <[email protected]> >>>>>>>> wrote: >>>>>>>> >>>>>>>>> I maybe wrong, but my feeling is that what Jagra calls A , in >>>>>>>>> Mario's translation is actually < A , 4 > (or < A, F(A) > , if you >>>>>>>>> prefer). >>>>>>>> >>>>>>>> >>>>>>>> I meant it to be interpreted as <A, F(A)>, and part of the proof >>>>>>>> would be showing that F(A) = 4 so that the rest of the statement >>>>>>>> simplifies. (But that would seem to be part of the proof, not the >>>>>>>> formalization of the statement, if we want to read it literally.) >>>>>>>> >>>>>>>> -- >>>> 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/161f583c-22bf-4699-b663-92652793f9c3n%40googlegroups.com >>>> <https://groups.google.com/d/msgid/metamath/161f583c-22bf-4699-b663-92652793f9c3n%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/4c4161e9-ccfc-4289-a19d-1bdcb701bbfbn%40googlegroups.com > <https://groups.google.com/d/msgid/metamath/4c4161e9-ccfc-4289-a19d-1bdcb701bbfbn%40googlegroups.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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