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.)
>>>>>
>>>>> --
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