Hi Igor,

I haven't tried the proof myself, a little short on time right now, but I
have doubts about why you are trying to prove F(a) = 4 or F(b). It seems to
me this is given, an hypothesis, right?

Regards,
Jorge



On Wed, Oct 16, 2024 at 8:41 AM Igor Ieskov <[email protected]> wrote:

> Hi Glauco,
>
>
> Yes, this answers my question. Thank you very much. Also I found this
> thread <https://groups.google.com/g/metamath/c/1T18-hzCw5E/m/Igpc2gWFBAAJ>
> contains other shortcuts used in the video.
>
>
> I don’t have much experience with mmj2. I used it to repeat proofs from
> David’s video tutorials, but not for creating my own proofs.
>
>
> -
>
> Igor
>
>
> On Tuesday, October 15, 2024 at 11:50:12 PM UTC+2 Glauco wrote:
>
>> Hi Igor,
>>
>> with regard to your question, yamma, for now, is similar to mmj2. Do you
>> have some experience with mmj2?
>>
>> I tend to proof backwards, thus yamma may have a bias toward this
>> approach.
>>
>> Did you see this simple video?
>>
>> https://www.youtube.com/watch?v=N7W0VBsbOcQ
>>
>> As you can see, when it has to prove
>>
>> |- E. x x e. { 1 }
>>
>> if you hit ctrl+space , intellisense shows you the labels that are more
>> likely to move you in the right direction.
>>
>> The one I chose is the 6th one, that is ceqsexv2d   (intellisense details
>> clearly show me that's the one I'm looking for)
>>
>> I would not be able to know it myself, so it saves me some search time.
>>
>> eximii , the first one suggested, is not the right one, but I can easily
>> see why it's, in general, a good candidate for the given statement
>>
>> HTH
>> Glauco
>>
>>
>>
>> Il giorno lunedì 14 ottobre 2024 alle 22:48:36 UTC+2 [email protected]
>> ha scritto:
>>
>>> Thank you Glauco and Mario for helpful answers. I will need some time to
>>> analyze and understand everything you wrote.
>>>
>>>
>>> For now, I have one question. Glauco, you mentioned that you developed
>>> this proof using Yamma. Can you explain at a high level what steps you’ve
>>> done in Yamma to come up with such a proof? I am asking for the proof
>>> of |- ( F ` a ) = 4. No need to explain how to set up Yamma, it can be
>>> found in its readme
>>> <https://github.com/glacode/yamma/blob/master/README.md>. I am more
>>> interested in how much of this proof was constructed by Yamma and how much
>>> was manual work. I didn’t have time yet to try to prove this in Yamma, but
>>> I will surely try. So, what I think would be interesting for me (and
>>> probably for others) is a high level list of actions you’ve done to create
>>> this proof. For example:
>>>
>>>
>>> 1) put all required hypotheses and the goal statement to Yamma.
>>>
>>> 2) I know that for such kind of proofs elrab assertion is usually used,
>>> so type its name in and unify (press CTRL+U). New steps appeared.
>>>
>>> 3) …
>>>
>>>
>>> I will be very surprised if you’ve just provided the hypotheses and the
>>> goal statement and Yamma found everything else!
>>>
>>>
>>> Sorry, if I am requesting too much. You can respond as high level or
>>> detailed depending on how much free time you have. I am asking because I
>>> tried to prove this using mm-lamp. Since I don’t know set.mm assertions
>>> well, I just searched for assertions which I thought could be used in the
>>> proof and tried to combine them somehow. I am interested in how you came up
>>> with this proof (how much automation you used and how much your experience
>>> with set.mm).
>>>
>>>
>>> BTW, it is Igor writing :)
>>>
>>>
>>> -
>>>
>>> Igor
>>>
>>>
>>> On Monday, October 14, 2024 at 5:05:35 PM UTC+2 Glauco wrote:
>>>
>>>> Below is a full contradiction derived.
>>>>
>>>>
>>>> It's amazing to see how ChatGPT can understand set.mm even though I
>>>> guess there's not much content, when compared to other subjects.
>>>>
>>>> Here's a brief "conversation" (not perfect, but close...):
>>>>
>>>> https://chatgpt.com/share/670d3203-1d2c-8010-bb5b-2e4105fc5379
>>>>
>>>>
>>>>
>>>> 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 } ) )
>>>> 5:2,4,1:vtocl         |- { x e. RR | ( F ` x ) = 4 } = { +oo , b }
>>>> 6:2:prid1             |- +oo e. { +oo , b }
>>>> 7:6,5:eleqtrri       |- +oo e. { x e. RR | ( F ` x ) = 4 }
>>>> 8::elrabi            |- ( +oo e. { x e. RR | ( F ` x ) = 4 } -> +oo e.
>>>> RR )
>>>> 9:7,8:ax-mp         |- +oo e. RR
>>>> 10::pnfnre2         |- -. +oo e. RR
>>>> qed:9,10:pm2.24ii  |- F.
>>>>
>>>> $= ( cpnf cr wcel wfal cv cfv c4 wceq crab cpr pnfex prid1 preq1 eqeq2d
>>>> vtocl
>>>>    eleqtrri elrabi ax-mp pnfnre2 pm2.24ii )
>>>> FGHZIFAJBKLMZAGNZHUFFFDJZOZUHFUIPQU
>>>>    HCJZUIOZMUHUJMCFPUKFMULUJUHUKFUIRSETUAUGAFGUBUCUDUE $.
>>>>
>>>>
>>>> $d a x
>>>> $d a b
>>>> $d F a
>>>>
>>>> Il giorno lunedì 14 ottobre 2024 alle 16:25:31 UTC+2 [email protected]
>>>> ha scritto:
>>>>
>>>>> 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>
>>>>>> .
>>>>>>
>>>>> --
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