Let f::ℝ->ℂ, a computational(deterministic) function and has the properties
(x,a,t∈ℝ, n∈ℤ):
P1: f(x)=f(x+n*t) // period of f is t
P2: f(x)^a=f(a*x) // f has exponential arithmetic property (e.g. Euler's formula)
Lemma: ∀a∈ℝ, f(x)=f(x+a*t) // the n of the above f is real, P2 still holds.
Proof: f(x)= f(x+a*t)
<=> f(x)^(1/a)=f(x+a*t)^(1/a)
<=> f(x/a)= f(x/a+t) // by P2
<=> f(x/a)= f(x/a) // by P1
<=> true

Proposition: ∀a,b∈ℝ, f(a)=f(b)
Proof: f(a)=f(b)
<=> f(a)=f(a+(b-a))
<=> f(a)=f(a+((b-a)/t)*t)
<=> f(a)=f(a) // from Lemma
<=> true

∴ Function that satisfies P1,P2 is equivalent to a constant function.

Ex: Let w(x)= w(x+2πk), and w(x)^y= w(x*y)
w^1= w^2
<=> w^1= w^(1+1)
<=> (w^1)^(2π)= (w^(1+1))^(2π)
<=> w^(2π)= w^(2π+ 2π) // by w(x)^y= w(x*y)
<=> w^(2π)= w^(2π) // by w(x+2πk)= w(x)
<=> true

On 28/09/2022 14:50, wij wrote:
> Let f::ℝ->ℂ, a computational(deterministic) function and has the properties
> (x,a,t∈ℝ, n∈ℤ):
> P1: f(x)=f(x+n*t) // period of f is t
> P2: f(x)^a=f(a*x) // f has exponential arithmetic property (e.g. Euler's formula)
> Lemma: ∀a∈ℝ, f(x)=f(x+a*t) // the n of the above f is real, P2 still holds.
> Proof: f(x)= f(x+a*t)
> <=> f(x)^(1/a)=f(x+a*t)^(1/a)

This doesn't hold generally E.g. 3^2 = (-3)^2 = 9, but 3 != -3.

As a general point - it is normal to write proofs in a 'forwards' direction, starting from true
statements and deducing new statements that follow from them. You are sort of writing your proofs
backwards, which is rather strange!

Regards,
Mike.

> <=> f(x/a)= f(x/a+t) // by P2
> <=> f(x/a)= f(x/a) // by P1
> <=> true
>
> Proposition: ∀a,b∈ℝ, f(a)=f(b)
> Proof: f(a)=f(b)
> <=> f(a)=f(a+(b-a))
> <=> f(a)=f(a+((b-a)/t)*t)
> <=> f(a)=f(a) // from Lemma
> <=> true
>
> ∴ Function that satisfies P1,P2 is equivalent to a constant function.
>
> Ex: Let w(x)= w(x+2πk), and w(x)^y= w(x*y)
> w^1= w^2
> <=> w^1= w^(1+1)
> <=> (w^1)^(2π)= (w^(1+1))^(2π)
> <=> w^(2π)= w^(2π+ 2π) // by w(x)^y= w(x*y)
> <=> w^(2π)= w^(2π) // by w(x+2πk)= w(x)
> <=> true
>

On Wednesday, 28 September 2022 at 23:06:55 UTC+8, Mike Terry wrote:
> On 28/09/2022 14:50, wij wrote:
> > Let f::ℝ->ℂ, a computational(deterministic) function and has the properties
> > (x,a,t∈ℝ, n∈ℤ):
> > P1: f(x)=f(x+n*t) // period of f is t
> > P2: f(x)^a=f(a*x) // f has exponential arithmetic property (e.g. Euler's formula)
> > Lemma: ∀a∈ℝ, f(x)=f(x+a*t) // the n of the above f is real, P2 still holds.
> > Proof: f(x)= f(x+a*t)
> > <=> f(x)^(1/a)=f(x+a*t)^(1/a)
> This doesn't hold generally E.g. 3^2 = (-3)^2 = 9, but 3 != -3.
>
> As a general point - it is normal to write proofs in a 'forwards' direction, starting from true
> statements and deducing new statements that follow from them. You are sort of writing your proofs
> backwards, which is rather strange!
>
> Regards,
> Mike.
> > <=> f(x/a)= f(x/a+t) // by P2
> > <=> f(x/a)= f(x/a) // by P1
> > <=> true
> >
> > Proposition: ∀a,b∈ℝ, f(a)=f(b)
> > Proof: f(a)=f(b)
> > <=> f(a)=f(a+(b-a))
> > <=> f(a)=f(a+((b-a)/t)*t)
> > <=> f(a)=f(a) // from Lemma
> > <=> true
> >
> > ∴ Function that satisfies P1,P2 is equivalent to a constant function.
> >
> > Ex: Let w(x)= w(x+2πk), and w(x)^y= w(x*y)
> > w^1= w^2
> > <=> w^1= w^(1+1)
> > <=> (w^1)^(2π)= (w^(1+1))^(2π)
> > <=> w^(2π)= w^(2π+ 2π) // by w(x)^y= w(x*y)
> > <=> w^(2π)= w^(2π) // by w(x+2πk)= w(x)
> > <=> true
> >

You did not provide any valid refutation.
Find the the place where the deduction is invalid.

Regards,
Me.

On Wednesday, September 28, 2022 at 8:52:46 AM UTC-7, wij wrote:
> On Wednesday, 28 September 2022 at 23:06:55 UTC+8, Mike Terry wrote:
> > On 28/09/2022 14:50, wij wrote:
> > > Let f::ℝ->ℂ, a computational(deterministic) function and has the properties
> > > (x,a,t∈ℝ, n∈ℤ):
> > > P1: f(x)=f(x+n*t) // period of f is t
> > > P2: f(x)^a=f(a*x) // f has exponential arithmetic property (e.g. Euler's formula)
> > > Lemma: ∀a∈ℝ, f(x)=f(x+a*t) // the n of the above f is real, P2 still holds.
> > > Proof: f(x)= f(x+a*t)
> > > <=> f(x)^(1/a)=f(x+a*t)^(1/a)
> > This doesn't hold generally E.g. 3^2 = (-3)^2 = 9, but 3 != -3.
> >
> > As a general point - it is normal to write proofs in a 'forwards' direction, starting from true
> > statements and deducing new statements that follow from them. You are sort of writing your proofs
> > backwards, which is rather strange!
> >
> > Regards,
> > Mike.
> > > <=> f(x/a)= f(x/a+t) // by P2
> > > <=> f(x/a)= f(x/a) // by P1
> > > <=> true
> > >
> > > Proposition: ∀a,b∈ℝ, f(a)=f(b)
> > > Proof: f(a)=f(b)
> > > <=> f(a)=f(a+(b-a))
> > > <=> f(a)=f(a+((b-a)/t)*t)
> > > <=> f(a)=f(a) // from Lemma
> > > <=> true
> > >
> > > ∴ Function that satisfies P1,P2 is equivalent to a constant function.
> > >
> > > Ex: Let w(x)= w(x+2πk), and w(x)^y= w(x*y)
> > > w^1= w^2
> > > <=> w^1= w^(1+1)
> > > <=> (w^1)^(2π)= (w^(1+1))^(2π)
> > > <=> w^(2π)= w^(2π+ 2π) // by w(x)^y= w(x*y)
> > > <=> w^(2π)= w^(2π) // by w(x+2πk)= w(x)
> > > <=> true
> > >
> You did not provide any valid refutation.
> Find the the place where the deduction is invalid.
>
> Regards,
> Me.

So you want to cancel it, or replace it?

Cancel it is easier while replace it though would require extra terms.

Cancellation and replacement are pretty must usual operations,
in writing systems of equations.

Lots of times through and usually numerical methods
and according to operators, has already constrained what
series-by-series terms would have to cancel/replace: also
series-by-series.

Or Fourier and Taylor and differential systems and so on.

On 28/09/2022 16:52, wij wrote:
> On Wednesday, 28 September 2022 at 23:06:55 UTC+8, Mike Terry wrote:
>> On 28/09/2022 14:50, wij wrote:
>>> Let f::ℝ->ℂ, a computational(deterministic) function and has the properties
>>> (x,a,t∈ℝ, n∈ℤ):
>>> P1: f(x)=f(x+n*t) // period of f is t
>>> P2: f(x)^a=f(a*x) // f has exponential arithmetic property (e.g. Euler's formula)
>>> Lemma: ∀a∈ℝ, f(x)=f(x+a*t) // the n of the above f is real, P2 still holds.
>>> Proof: f(x)= f(x+a*t)
>>> <=> f(x)^(1/a)=f(x+a*t)^(1/a)
>> This doesn't hold generally E.g. 3^2 = (-3)^2 = 9, but 3 != -3.

The above deduction is the place where your deduction is invalid, for the reason I explained.

>>
>> As a general point - it is normal to write proofs in a 'forwards' direction, starting from true
>> statements and deducing new statements that follow from them. You are sort of writing your proofs
>> backwards, which is rather strange!
>>
>> Regards,
>> Mike.
>>> <=> f(x/a)= f(x/a+t) // by P2
>>> <=> f(x/a)= f(x/a) // by P1
>>> <=> true
>>>
>>> Proposition: ∀a,b∈ℝ, f(a)=f(b)
>>> Proof: f(a)=f(b)
>>> <=> f(a)=f(a+(b-a))
>>> <=> f(a)=f(a+((b-a)/t)*t)
>>> <=> f(a)=f(a) // from Lemma
>>> <=> true
>>>
>>> ∴ Function that satisfies P1,P2 is equivalent to a constant function.
>>>
>>> Ex: Let w(x)= w(x+2πk), and w(x)^y= w(x*y)
>>> w^1= w^2
>>> <=> w^1= w^(1+1)
>>> <=> (w^1)^(2π)= (w^(1+1))^(2π)
>>> <=> w^(2π)= w^(2π+ 2π) // by w(x)^y= w(x*y)
>>> <=> w^(2π)= w^(2π) // by w(x+2πk)= w(x)
>>> <=> true
>>>
>
> You did not provide any valid refutation.
> Find the the place where the deduction is invalid.

(See above)

>
> Regards,
> Me.
>

On Wednesday, September 28, 2022 at 8:59:19 AM UTC-7, Ross A. Finlayson wrote:
> On Wednesday, September 28, 2022 at 8:52:46 AM UTC-7, wij wrote:
> > On Wednesday, 28 September 2022 at 23:06:55 UTC+8, Mike Terry wrote:
> > > On 28/09/2022 14:50, wij wrote:
> > > > Let f::ℝ->ℂ, a computational(deterministic) function and has the properties
> > > > (x,a,t∈ℝ, n∈ℤ):
> > > > P1: f(x)=f(x+n*t) // period of f is t
> > > > P2: f(x)^a=f(a*x) // f has exponential arithmetic property (e.g. Euler's formula)
> > > > Lemma: ∀a∈ℝ, f(x)=f(x+a*t) // the n of the above f is real, P2 still holds.
> > > > Proof: f(x)= f(x+a*t)
> > > > <=> f(x)^(1/a)=f(x+a*t)^(1/a)
> > > This doesn't hold generally E.g. 3^2 = (-3)^2 = 9, but 3 != -3.
> > >
> > > As a general point - it is normal to write proofs in a 'forwards' direction, starting from true
> > > statements and deducing new statements that follow from them. You are sort of writing your proofs
> > > backwards, which is rather strange!
> > >
> > > Regards,
> > > Mike.
> > > > <=> f(x/a)= f(x/a+t) // by P2
> > > > <=> f(x/a)= f(x/a) // by P1
> > > > <=> true
> > > >
> > > > Proposition: ∀a,b∈ℝ, f(a)=f(b)
> > > > Proof: f(a)=f(b)
> > > > <=> f(a)=f(a+(b-a))
> > > > <=> f(a)=f(a+((b-a)/t)*t)
> > > > <=> f(a)=f(a) // from Lemma
> > > > <=> true
> > > >
> > > > ∴ Function that satisfies P1,P2 is equivalent to a constant function.
> > > >
> > > > Ex: Let w(x)= w(x+2πk), and w(x)^y= w(x*y)
> > > > w^1= w^2
> > > > <=> w^1= w^(1+1)
> > > > <=> (w^1)^(2π)= (w^(1+1))^(2π)
> > > > <=> w^(2π)= w^(2π+ 2π) // by w(x)^y= w(x*y)
> > > > <=> w^(2π)= w^(2π) // by w(x+2πk)= w(x)
> > > > <=> true
> > > >
> > You did not provide any valid refutation.
> > Find the the place where the deduction is invalid.
> >
> > Regards,
> > Me.
> So you want to cancel it, or replace it?
>
> Cancel it is easier while replace it though would require extra terms.
>
> Cancellation and replacement are pretty must usual operations,
> in writing systems of equations.
>
> Lots of times through and usually numerical methods
> and according to operators, has already constrained what
> series-by-series terms would have to cancel/replace: also
> series-by-series.
>
> Or Fourier and Taylor and differential systems and so on.

Don't need a refutation / you didn't say anything, just "about" to, ...,
which would possibly be right, if not wrong. There's definitely conditions..

On Wednesday, 28 September 2022 at 23:59:40 UTC+8, Mike Terry wrote:
> On 28/09/2022 16:52, wij wrote:
> > On Wednesday, 28 September 2022 at 23:06:55 UTC+8, Mike Terry wrote:
> >> On 28/09/2022 14:50, wij wrote:
> >>> Let f::ℝ->ℂ, a computational(deterministic) function and has the properties
> >>> (x,a,t∈ℝ, n∈ℤ):
> >>> P1: f(x)=f(x+n*t) // period of f is t
> >>> P2: f(x)^a=f(a*x) // f has exponential arithmetic property (e.g. Euler's formula)
> >>> Lemma: ∀a∈ℝ, f(x)=f(x+a*t) // the n of the above f is real, P2 still holds.
> >>> Proof: f(x)= f(x+a*t)
> >>> <=> f(x)^(1/a)=f(x+a*t)^(1/a)
> >> This doesn't hold generally E.g. 3^2 = (-3)^2 = 9, but 3 != -3.
> The above deduction is the place where your deduction is invalid, for the reason I explained.

3^2=(-3)^2 ... OK
(-3)^2=9 ... OK

How is the conclusion 3!=3 (or 3=3) derived? What is the relation with the lemma?

On 28/09/2022 17:21, wij wrote:
> On Wednesday, 28 September 2022 at 23:59:40 UTC+8, Mike Terry wrote:
>> On 28/09/2022 16:52, wij wrote:
>>> On Wednesday, 28 September 2022 at 23:06:55 UTC+8, Mike Terry wrote:
>>>> On 28/09/2022 14:50, wij wrote:
>>>>> Let f::ℝ->ℂ, a computational(deterministic) function and has the properties
>>>>> (x,a,t∈ℝ, n∈ℤ):
>>>>> P1: f(x)=f(x+n*t) // period of f is t
>>>>> P2: f(x)^a=f(a*x) // f has exponential arithmetic property (e.g. Euler's formula)
>>>>> Lemma: ∀a∈ℝ, f(x)=f(x+a*t) // the n of the above f is real, P2 still holds.
>>>>> Proof: f(x)= f(x+a*t)
>>>>> <=> f(x)^(1/a)=f(x+a*t)^(1/a)
>>>> This doesn't hold generally E.g. 3^2 = (-3)^2 = 9, but 3 != -3.
>> The above deduction is the place where your deduction is invalid, for the reason I explained.
>
> 3^2=(-3)^2 ... OK
> (-3)^2=9 ... OK
>
> How is the conclusion 3!=3 (or 3=3) derived? What is the relation with the lemma?
>

Your lemma claimed that

f(x)^(1/a) = f(x+a*t)^(1/a) ==> f(x) = f(x+a*t)

This is trying to use a rule which says

a^s = b^s ==> a = b,

but that is not a valid rule. (Exponentiation is not a 1 to 1 function.)

Mike.

On Thursday, 29 September 2022 at 01:22:04 UTC+8, Mike Terry wrote:
> On 28/09/2022 17:21, wij wrote:
> > On Wednesday, 28 September 2022 at 23:59:40 UTC+8, Mike Terry wrote:
> >> On 28/09/2022 16:52, wij wrote:
> >>> On Wednesday, 28 September 2022 at 23:06:55 UTC+8, Mike Terry wrote:
> >>>> On 28/09/2022 14:50, wij wrote:
> >>>>> Let f::ℝ->ℂ, a computational(deterministic) function and has the properties
> >>>>> (x,a,t∈ℝ, n∈ℤ):
> >>>>> P1: f(x)=f(x+n*t) // period of f is t
> >>>>> P2: f(x)^a=f(a*x) // f has exponential arithmetic property (e.g. Euler's formula)
> >>>>> Lemma: ∀a∈ℝ, f(x)=f(x+a*t) // the n of the above f is real, P2 still holds.
> >>>>> Proof: f(x)= f(x+a*t)
> >>>>> <=> f(x)^(1/a)=f(x+a*t)^(1/a)
> >>>> This doesn't hold generally E.g. 3^2 = (-3)^2 = 9, but 3 != -3..
> >> The above deduction is the place where your deduction is invalid, for the reason I explained.
> >
> > 3^2=(-3)^2 ... OK
> > (-3)^2=9 ... OK
> >
> > How is the conclusion 3!=3 (or 3=3) derived? What is the relation with the lemma?
> >
> Your lemma claimed that
>
> f(x)^(1/a) = f(x+a*t)^(1/a) ==> f(x) = f(x+a*t)
>
> This is trying to use a rule which says
>
> a^s = b^s ==> a = b,
>
> but that is not a valid rule. (Exponentiation is not a 1 to 1 function.)
>
>
> Mike.

The point is that exponentiation operation is deterministic (or not?).

If f(x) = f(x+a*t) is true (by definition)
then f(x)^(1/a) = f(x+n*t)^(1/a) must follow.

On Thursday, 29 September 2022 at 23:46:49 UTC+8, wij wrote:
> On Thursday, 29 September 2022 at 01:22:04 UTC+8, Mike Terry wrote:
> > On 28/09/2022 17:21, wij wrote:
> > > On Wednesday, 28 September 2022 at 23:59:40 UTC+8, Mike Terry wrote:
> > >> On 28/09/2022 16:52, wij wrote:
> > >>> On Wednesday, 28 September 2022 at 23:06:55 UTC+8, Mike Terry wrote:
> > >>>> On 28/09/2022 14:50, wij wrote:
> > >>>>> Let f::ℝ->ℂ, a computational(deterministic) function and has the properties
> > >>>>> (x,a,t∈ℝ, n∈ℤ):
> > >>>>> P1: f(x)=f(x+n*t) // period of f is t
> > >>>>> P2: f(x)^a=f(a*x) // f has exponential arithmetic property (e.g.. Euler's formula)
> > >>>>> Lemma: ∀a∈ℝ, f(x)=f(x+a*t) // the n of the above f is real, P2 still holds.
> > >>>>> Proof: f(x)= f(x+a*t)
> > >>>>> <=> f(x)^(1/a)=f(x+a*t)^(1/a)
> > >>>> This doesn't hold generally E.g. 3^2 = (-3)^2 = 9, but 3 != -3.
> > >> The above deduction is the place where your deduction is invalid, for the reason I explained.
> > >
> > > 3^2=(-3)^2 ... OK
> > > (-3)^2=9 ... OK
> > >
> > > How is the conclusion 3!=3 (or 3=3) derived? What is the relation with the lemma?
> > >
> > Your lemma claimed that
> >
> > f(x)^(1/a) = f(x+a*t)^(1/a) ==> f(x) = f(x+a*t)
> >
> > This is trying to use a rule which says
> >
> > a^s = b^s ==> a = b,
> >
> > but that is not a valid rule. (Exponentiation is not a 1 to 1 function.)
> >
> >
> > Mike.
> The point is that exponentiation operation is deterministic (or not?).
>
> If f(x) = f(x+a*t) is true (by definition)
> then f(x)^(1/a) = f(x+n*t)^(1/a) must follow.

Sorry, a mistake.I am changing the statement in you way of thinking.

On 29/09/2022 16:46, wij wrote:
> On Thursday, 29 September 2022 at 01:22:04 UTC+8, Mike Terry wrote:
>> On 28/09/2022 17:21, wij wrote:
>>> On Wednesday, 28 September 2022 at 23:59:40 UTC+8, Mike Terry wrote:
>>>> On 28/09/2022 16:52, wij wrote:
>>>>> On Wednesday, 28 September 2022 at 23:06:55 UTC+8, Mike Terry wrote:
>>>>>> On 28/09/2022 14:50, wij wrote:
>>>>>>> Let f::ℝ->ℂ, a computational(deterministic) function and has the properties
>>>>>>> (x,a,t∈ℝ, n∈ℤ):
>>>>>>> P1: f(x)=f(x+n*t) // period of f is t
>>>>>>> P2: f(x)^a=f(a*x) // f has exponential arithmetic property (e.g. Euler's formula)
>>>>>>> Lemma: ∀a∈ℝ, f(x)=f(x+a*t) // the n of the above f is real, P2 still holds.
>>>>>>> Proof: f(x)= f(x+a*t)
>>>>>>> <=> f(x)^(1/a)=f(x+a*t)^(1/a)
>>>>>> This doesn't hold generally E.g. 3^2 = (-3)^2 = 9, but 3 != -3.
>>>> The above deduction is the place where your deduction is invalid, for the reason I explained.
>>>
>>> 3^2=(-3)^2 ... OK
>>> (-3)^2=9 ... OK
>>>
>>> How is the conclusion 3!=3 (or 3=3) derived? What is the relation with the lemma?
>>>
>> Your lemma claimed that
>>
>> f(x)^(1/a) = f(x+a*t)^(1/a) ==> f(x) = f(x+a*t)
>>
>> This is trying to use a rule which says
>>
>> a^s = b^s ==> a = b,
>>
>> but that is not a valid rule. (Exponentiation is not a 1 to 1 function.)
>>
>>
>> Mike.
>
> The point is that exponentiation operation is deterministic (or not?).
>
> If f(x) = f(x+a*t) is true (by definition)
> then f(x)^(1/a) = f(x+n*t)^(1/a) must follow.
>

Yes, that's using a = b ==> a^s = b^s, which is valid. The problem is with the reverse inference:
a^s = b^s ==> a = b [NOT VALID]

Mike.