|
|
 | | From: | |-|erc | | Subject: | Re: How many flips of DIAG are on the infintie list of infinite con flippers ? | | Date: | Fri, 21 Jan 2005 13:04:36 +1000 |
|
|
 | "The Ghost In The Machine" wrote in
> >> it is just incoherent. I can always come up with a new
> >> anything, period, if there are more than a finite number
> >> of that kind of thing. You can't prove that they've all
> >> been thought of already.
> >
> >
> > Sure I can. Think of a natural number not on this list.
> >
> > defun nats (nat(0))
> >
> > defun nat(n) (cons n (nat(plus ( n 1 ))))
> >
> > nats
> > <1 2 3 4 5 6 7 8..>
> >
>
> All natural numbers are on that list, by definition. Did
> you have a point here?
can you come up with a new natural, or did I cover every one?
>
> >
> >
> >
> >>
> >>
> >> > AntiDiag =
> >> > |<------ How Many flips ? ------->|
> >>
> >> Your calling these "flips" is stupid.
> >> They are just letters. This is just a string.
>
> They are also flips. A coin flip can be modeled as
> letter strings, binary digit sequences (0101010101...),
> raw bits (which are hard to represent in ASCII directly;
> one usually uses letters or binary digit sequences),
> photon/non-photon, red/green, pointer at 100% / pointer at 0%
> on a hypothetical dial, current pulse/absence, +5V/-5V, etc.
>
> Of course it would help if |-|erc gave us a complete
> specification for these flips, which implies a function
> spec -- and here's where it gets interesting.
>
> If we define F(i,j) as a function defining |-|erc's flips
> (domain: N x N, range, boolean), one can define an antidiagonal
> function AntiDiag(F,i)
> (domain: {functions with domain NxN and range boolean} x N,
> range, boolean).
>
> It is clear that AntiDiag cannot be on F's list, because
> of a simple issue with typing (they aren't compatible).
> One can attempt to fix this by asking the more intelligent
> question
>
> for what i is F(i,*) = AntiDiag(F,*) [*]
>
> or asking whether
>
> (Ei)(Aj)(F(i,j) = AntiDiag(F,j))
>
> is true. For most F this will obviously be false. In
> fact, for all F this is provably false, since one can
> derive:
>
> 1. (Ei)(Aj)(F(i,j) = AntiDiag(F,j))
> 2. (Aj)(F(k,j) = AntiDiag(F,j)) [1, EI]
> 3. (F(k,k) = AntiDiag(F,k)) [2, UI]
>
> which is clearly false because of the construction
> method of AntiDiag(F,j), which is defined as
>
> AntiDiag(F,j) = !(F(j,j))
>
> Of course this is not to be confused with the
> proposition
>
> (Aj)(Ei)(F(i,j) = AntiDiag(F,j))
>
> which for most F is true. |-|erc, you've had this problem before.
>
> [rest snipped]
>
> [*] the notation is borrowed from the Illiac IV.
> Basically, if f : N x N -> Boolean, then
> g_i = f(i,*) is a function mapping N to Boolean,
> such that g_i(j) = f(i,j).
>
More importantly
(Aj)(Ei)(F(i,k) = AntiDiag(F,k) 0<=k<=j
but the assumption is F covers every possible finite sequence as
part of its infinite sequences.
This does not crop up anywhere in Cantor's proof.
There is no maximum to j.
There is no maximum to the amount of natural numbers
-> There are oo amount of natural numbers.
There is no maximum to the amount of flips of AD on F.
-> There are oo amount of flips of AD on F.
Herc
|
|
| | From: | The Ghost In The Machine | | Subject: | Re: How many flips of DIAG are on the infintie list of infinite con flippers ? | | Date: | Fri, 21 Jan 2005 05:01:44 GMT |
|
|
| In sci.logic, |-|erc
wrote
on Fri, 21 Jan 2005 13:04:36 +1000
<35b9piF4kkhqiU1@individual.net>:
> "The Ghost In The Machine" wrote in
>
>> >> it is just incoherent. I can always come up with a new
>> >> anything, period, if there are more than a finite number
>> >> of that kind of thing. You can't prove that they've all
>> >> been thought of already.
>> >
>> >
>> > Sure I can. Think of a natural number not on this list.
>> >
>> > defun nats (nat(0))
>> >
>> > defun nat(n) (cons n (nat(plus ( n 1 ))))
>> >
>> > nats
>> > <1 2 3 4 5 6 7 8..>
>> >
>>
>> All natural numbers are on that list, by definition. Did
>> you have a point here?
>
> can you come up with a new natural, or did I cover every one?
You covered every one. Did you have a point here?
[1] If P(1) and P(i) => P(i+1), then P(n) for all n in N.
[2] If S is a subset of N, 1 is an element of S, and
(As)(s in S => (s+1) in S), then one can state S = N.
>
>
>>
>> >
>> >
>> >
>> >>
>> >>
>> >> > AntiDiag =
>> >> > |<------ How Many flips ? ------->|
>> >>
>> >> Your calling these "flips" is stupid.
>> >> They are just letters. This is just a string.
>>
>> They are also flips. A coin flip can be modeled as
>> letter strings, binary digit sequences (0101010101...),
>> raw bits (which are hard to represent in ASCII directly;
>> one usually uses letters or binary digit sequences),
>> photon/non-photon, red/green, pointer at 100% / pointer at 0%
>> on a hypothetical dial, current pulse/absence, +5V/-5V, etc.
>>
>> Of course it would help if |-|erc gave us a complete
>> specification for these flips, which implies a function
>> spec -- and here's where it gets interesting.
>>
>> If we define F(i,j) as a function defining |-|erc's flips
>> (domain: N x N, range, boolean), one can define an antidiagonal
>> function AntiDiag(F,i)
>> (domain: {functions with domain NxN and range boolean} x N,
>> range, boolean).
>>
>> It is clear that AntiDiag cannot be on F's list, because
>> of a simple issue with typing (they aren't compatible).
>> One can attempt to fix this by asking the more intelligent
>> question
>>
>> for what i is F(i,*) = AntiDiag(F,*) [*]
>>
>> or asking whether
>>
>> (Ei)(Aj)(F(i,j) = AntiDiag(F,j))
>>
>> is true. For most F this will obviously be false. In
>> fact, for all F this is provably false, since one can
>> derive:
>>
>> 1. (Ei)(Aj)(F(i,j) = AntiDiag(F,j))
>> 2. (Aj)(F(k,j) = AntiDiag(F,j)) [1, EI]
>> 3. (F(k,k) = AntiDiag(F,k)) [2, UI]
>>
>> which is clearly false because of the construction
>> method of AntiDiag(F,j), which is defined as
>>
>> AntiDiag(F,j) = !(F(j,j))
>>
>> Of course this is not to be confused with the
>> proposition
>>
>> (Aj)(Ei)(F(i,j) = AntiDiag(F,j))
>>
>> which for most F is true. |-|erc, you've had this problem before.
>>
>> [rest snipped]
>>
>> [*] the notation is borrowed from the Illiac IV.
>> Basically, if f : N x N -> Boolean, then
>> g_i = f(i,*) is a function mapping N to Boolean,
>> such that g_i(j) = f(i,j).
>>
>
> More importantly
> (Aj)(Ei)(F(i,k) = AntiDiag(F,k) 0<=k<=j
I'd phrase that as
(Aj)(Ei)(Ek)(F(i,k) = AntiDiag(F,k), 1 <= k <= j)
mostly because I'm 1-based and you forgot a quantifier *and* a ')'.
Or did you mean
(Aj)(Ei)(Ak)(F(i,k) = AntiDiag(F,k), 1 <= k <= j) ?
Not that it matters.
Depending on F, both are most likely true, but all you're
doing is playing shuffle. AntiDiag(F,i) != F(i,i),
therefore AntiDiag(F,*) is not on the list.
Or did you want to claim that pi is a rational number?
All finite prefixes of pi are, after all, in Q (in fact,
they're in TX_10). That quickly gets ridiculous.
>
> but the assumption is F covers every possible finite sequence as
> part of its infinite sequences.
>
> This does not crop up anywhere in Cantor's proof.
It doesn't crop up in Cantor's first proof, either.
>
> There is no maximum to j.
>
> There is no maximum to the amount of natural numbers
> -> There are oo amount of natural numbers.
There exist real numbers not in any denumerable list of real numbers
-> there are C > Aleph_0 real numbers.
Spot The Difference.
>
> There is no maximum to the amount of flips of AD on F.
> -> There are oo amount of flips of AD on F.
Not quite that simple, and even if one were able to prove the
first phrase
"the finite subsequence of length 1 of AD is in F"
and the implication
"the finite subsequence of length n of AD is in F" implies
"the finite subsequence of length n+1 of AD is in F"
somehow, and thereby prove that
"all finite subsequences of AD are in F"
by induction (which one can easily do for certain F),
one still has the little problem that one is dealing with
an *infinite* sequence (namely, AD), and all elements in
F are not equal to this sequence (since they differ by at
least one digit).
1/3 is not an element of the set {.3, .33, .333, ...},
even though all finite subsequences of 1/3 are.
pi is not a rational number.
You are not presenting here a real proof. :-P
>
> Herc
>
--
#191, ewill3@earthlink.net
It's still legal to go .sigless.
|
|
| | From: | |-|erc | | Subject: | Re: How many flips of DIAG are on the infintie list of infinite con flippers ? | | Date: | Fri, 21 Jan 2005 15:17:28 +1000 |
|
|
| "The Ghost In The Machine" wrote
> >
> >> >> it is just incoherent. I can always come up with a new
> >> >> anything, period, if there are more than a finite number
> >> >> of that kind of thing. You can't prove that they've all
> >> >> been thought of already.
> >> >
> >> >
> >> > Sure I can. Think of a natural number not on this list.
> >> >
> >> > defun nats (nat(0))
> >> >
> >> > defun nat(n) (cons n (nat(plus ( n 1 ))))
> >> >
> >> > nats
> >> > <1 2 3 4 5 6 7 8..>
> >> >
> >>
> >> All natural numbers are on that list, by definition. Did
> >> you have a point here?
> >
> > can you come up with a new natural, or did I cover every one?
>
> You covered every one. Did you have a point here?
yes, you don't read the post you are replying to.
>
> [1] If P(1) and P(i) => P(i+1), then P(n) for all n in N.
> [2] If S is a subset of N, 1 is an element of S, and
> (As)(s in S => (s+1) in S), then one can state S = N.
>
> >
> >
> >>
> >> >
> >> >
> >> >
> >> >>
> >> >>
> >> >> > AntiDiag =
> >> >> > |<------ How Many flips ? ------->|
> >> >>
> >> >> Your calling these "flips" is stupid.
> >> >> They are just letters. This is just a string.
> >>
> >> They are also flips. A coin flip can be modeled as
> >> letter strings, binary digit sequences (0101010101...),
> >> raw bits (which are hard to represent in ASCII directly;
> >> one usually uses letters or binary digit sequences),
> >> photon/non-photon, red/green, pointer at 100% / pointer at 0%
> >> on a hypothetical dial, current pulse/absence, +5V/-5V, etc.
> >>
> >> Of course it would help if |-|erc gave us a complete
> >> specification for these flips, which implies a function
> >> spec -- and here's where it gets interesting.
> >>
> >> If we define F(i,j) as a function defining |-|erc's flips
> >> (domain: N x N, range, boolean), one can define an antidiagonal
> >> function AntiDiag(F,i)
> >> (domain: {functions with domain NxN and range boolean} x N,
> >> range, boolean).
> >>
> >> It is clear that AntiDiag cannot be on F's list, because
> >> of a simple issue with typing (they aren't compatible).
> >> One can attempt to fix this by asking the more intelligent
> >> question
> >>
> >> for what i is F(i,*) = AntiDiag(F,*) [*]
> >>
> >> or asking whether
> >>
> >> (Ei)(Aj)(F(i,j) = AntiDiag(F,j))
> >>
> >> is true. For most F this will obviously be false. In
> >> fact, for all F this is provably false, since one can
> >> derive:
> >>
> >> 1. (Ei)(Aj)(F(i,j) = AntiDiag(F,j))
> >> 2. (Aj)(F(k,j) = AntiDiag(F,j)) [1, EI]
> >> 3. (F(k,k) = AntiDiag(F,k)) [2, UI]
> >>
> >> which is clearly false because of the construction
> >> method of AntiDiag(F,j), which is defined as
> >>
> >> AntiDiag(F,j) = !(F(j,j))
> >>
> >> Of course this is not to be confused with the
> >> proposition
> >>
> >> (Aj)(Ei)(F(i,j) = AntiDiag(F,j))
> >>
> >> which for most F is true. |-|erc, you've had this problem before.
> >>
> >> [rest snipped]
> >>
> >> [*] the notation is borrowed from the Illiac IV.
> >> Basically, if f : N x N -> Boolean, then
> >> g_i = f(i,*) is a function mapping N to Boolean,
> >> such that g_i(j) = f(i,j).
> >>
> >
> > More importantly
> > (Aj)(Ei)(F(i,k) = AntiDiag(F,k) 0<=k<=j
>
> I'd phrase that as
>
> (Aj)(Ei)(Ek)(F(i,k) = AntiDiag(F,k), 1 <= k <= j)
no. that's identical to your near meaningless 1st attempt.
YOU DO KNOW HOW QUANTIFIERS WORK DON'T YOU?
DO YOU?
DO YOU?
DO YOU?
>
> mostly because I'm 1-based and you forgot a quantifier *and* a ')'.
>
> Or did you mean
>
> (Aj)(Ei)(Ak)(F(i,k) = AntiDiag(F,k), 1 <= k <= j) ?
>
> Not that it matters.
There you go. Completely different assertions actually.
You do know how quantifiers work don't you?
Do you know the difference between Ax a+b and Ex a+b ??
HUH?
>
> Depending on F, both are most likely true, but all you're
> doing is playing shuffle. AntiDiag(F,i) != F(i,i),
> therefore AntiDiag(F,*) is not on the list.
>
> Or did you want to claim that pi is a rational number?
> All finite prefixes of pi are, after all, in Q (in fact,
> they're in TX_10). That quickly gets ridiculous.
>
> >
> > but the assumption is F covers every possible finite sequence as
> > part of its infinite sequences.
> >
> > This does not crop up anywhere in Cantor's proof.
>
> It doesn't crop up in Cantor's first proof, either.
>
> >
> > There is no maximum to j.
> >
> > There is no maximum to the amount of natural numbers
> > -> There are oo amount of natural numbers.
>
> There exist real numbers not in any denumerable list of real numbers
> -> there are C > Aleph_0 real numbers.
>
> Spot The Difference.
Are you challenging my claim on oo?
I didn't see you answer the question on the claim.
>
> >
> > There is no maximum to the amount of flips of AD on F.
> > -> There are oo amount of flips of AD on F.
>
> Not quite that simple, and even if one were able to prove the
> first phrase
HAHAHAH 'fraid so you mass of stupid sheep.
>
> "the finite subsequence of length 1 of AD is in F"
>
> and the implication
>
> "the finite subsequence of length n of AD is in F" implies
> "the finite subsequence of length n+1 of AD is in F"
>
> somehow, and thereby prove that
>
> "all finite subsequences of AD are in F"
>
> by induction (which one can easily do for certain F),
> one still has the little problem that one is dealing with
> an *infinite* sequence (namely, AD), and all elements in
> F are not equal to this sequence (since they differ by at
> least one digit).
X in on Y to Z many digits
Get used to it, its called formulation of knowledge.
>
> 1/3 is not an element of the set {.3, .33, .333, ...},
> even though all finite subsequences of 1/3 are.
>
> pi is not a rational number.
>
> You are not presenting here a real proof. :-P
>
I know, but your counter to the assertions is only for a certain class of sets.
Herc
|
|
| | From: | The Ghost In The Machine | | Subject: | Re: How many flips of DIAG are on the infintie list of infinite con flippers ? | | Date: | Fri, 21 Jan 2005 23:01:41 GMT |
|
|
| In sci.logic, |-|erc
wrote
on Fri, 21 Jan 2005 15:17:28 +1000
<35bhimF4kbbseU1@individual.net>:
> "The Ghost In The Machine" wrote
>> >
>> >> >> it is just incoherent. I can always come up with a new
>> >> >> anything, period, if there are more than a finite number
>> >> >> of that kind of thing. You can't prove that they've all
>> >> >> been thought of already.
>> >> >
>> >> >
>> >> > Sure I can. Think of a natural number not on this list.
>> >> >
>> >> > defun nats (nat(0))
>> >> >
>> >> > defun nat(n) (cons n (nat(plus ( n 1 ))))
>> >> >
>> >> > nats
>> >> > <1 2 3 4 5 6 7 8..>
>> >> >
>> >>
>> >> All natural numbers are on that list, by definition. Did
>> >> you have a point here?
>> >
>> > can you come up with a new natural, or did I cover every one?
>>
>> You covered every one. Did you have a point here?
>
> yes, you don't read the post you are replying to.
>
>
>
>>
>> [1] If P(1) and P(i) => P(i+1), then P(n) for all n in N.
>> [2] If S is a subset of N, 1 is an element of S, and
>> (As)(s in S => (s+1) in S), then one can state S = N.
>>
>> >
>> >
>> >>
>> >> >
>> >> >
>> >> >
>> >> >>
>> >> >>
>> >> >> > AntiDiag =
>> >> >> > |<------ How Many flips ? ------->|
>> >> >>
>> >> >> Your calling these "flips" is stupid.
>> >> >> They are just letters. This is just a string.
>> >>
>> >> They are also flips. A coin flip can be modeled as
>> >> letter strings, binary digit sequences (0101010101...),
>> >> raw bits (which are hard to represent in ASCII directly;
>> >> one usually uses letters or binary digit sequences),
>> >> photon/non-photon, red/green, pointer at 100% / pointer at 0%
>> >> on a hypothetical dial, current pulse/absence, +5V/-5V, etc.
>> >>
>> >> Of course it would help if |-|erc gave us a complete
>> >> specification for these flips, which implies a function
>> >> spec -- and here's where it gets interesting.
>> >>
>> >> If we define F(i,j) as a function defining |-|erc's flips
>> >> (domain: N x N, range, boolean), one can define an antidiagonal
>> >> function AntiDiag(F,i)
>> >> (domain: {functions with domain NxN and range boolean} x N,
>> >> range, boolean).
>> >>
>> >> It is clear that AntiDiag cannot be on F's list, because
>> >> of a simple issue with typing (they aren't compatible).
>> >> One can attempt to fix this by asking the more intelligent
>> >> question
>> >>
>> >> for what i is F(i,*) = AntiDiag(F,*) [*]
>> >>
>> >> or asking whether
>> >>
>> >> (Ei)(Aj)(F(i,j) = AntiDiag(F,j))
>> >>
>> >> is true. For most F this will obviously be false. In
>> >> fact, for all F this is provably false, since one can
>> >> derive:
>> >>
>> >> 1. (Ei)(Aj)(F(i,j) = AntiDiag(F,j))
>> >> 2. (Aj)(F(k,j) = AntiDiag(F,j)) [1, EI]
>> >> 3. (F(k,k) = AntiDiag(F,k)) [2, UI]
>> >>
>> >> which is clearly false because of the construction
>> >> method of AntiDiag(F,j), which is defined as
>> >>
>> >> AntiDiag(F,j) = !(F(j,j))
>> >>
>> >> Of course this is not to be confused with the
>> >> proposition
>> >>
>> >> (Aj)(Ei)(F(i,j) = AntiDiag(F,j))
>> >>
>> >> which for most F is true. |-|erc, you've had this problem before.
>> >>
>> >> [rest snipped]
>> >>
>> >> [*] the notation is borrowed from the Illiac IV.
>> >> Basically, if f : N x N -> Boolean, then
>> >> g_i = f(i,*) is a function mapping N to Boolean,
>> >> such that g_i(j) = f(i,j).
>> >>
>> >
>> > More importantly
>> > (Aj)(Ei)(F(i,k) = AntiDiag(F,k) 0<=k<=j
>>
>> I'd phrase that as
>>
>> (Aj)(Ei)(Ek)(F(i,k) = AntiDiag(F,k), 1 <= k <= j)
>
> no. that's identical to your near meaningless 1st attempt.
>
> YOU DO KNOW HOW QUANTIFIERS WORK DON'T YOU?
>
> DO YOU?
> DO YOU?
> DO YOU?
>
>
>
>>
>> mostly because I'm 1-based and you forgot a quantifier *and* a ')'.
>>
>> Or did you mean
>>
>> (Aj)(Ei)(Ak)(F(i,k) = AntiDiag(F,k), 1 <= k <= j) ?
>>
>> Not that it matters.
>
> There you go. Completely different assertions actually.
>
> You do know how quantifiers work don't you?
>
> Do you know the difference between Ax a+b and Ex a+b ??
>
> HUH?
[rest snipped]
Do you know the difference between
(Ai)(Ej)(F(i,j) = AntiDiag(j))
and
(Ei)(Aj)(F(i,j) = AntiDiag(j))
?
If so, explain it to me. I'm curious.
Hopefully you can Copi this request, Irving M. (HINT HINT)
--
#191, ewill3@earthlink.net
It's still legal to go .sigless.
|
|
| | From: | |-|erc | | Subject: | Re: How many flips of DIAG are on the infintie list of infinite con flippers ? | | Date: | Sat, 22 Jan 2005 11:47:25 +1000 |
|
|
| "The Ghost In The Machine" wrote in > >>
> >> >> AntiDiag(F,j) = !(F(j,j)) [1]
> >> >>
> >> >> Of course this is not to be confused with the
> >> >> proposition
> >> >>
> >> >> (Aj)(Ei)(F(i,j) = AntiDiag(F,j)) [2]
> >> >>
> >> >> which for most F is true.
> >> >
> >> > More importantly
> >> > (Aj)(Ei)(F(i,k) = AntiDiag(F,k) 0<=k<=j [3]
> >>
> >> I'd phrase that as
> >>
> >> (Aj)(Ei)(Ek)(F(i,k) = AntiDiag(F,k), 1 <= k <= j) [4]
> >
> > no. that's identical to your near meaningless 1st attempt.
> >> Or did you mean
> >>
> >> (Aj)(Ei)(Ak)(F(i,k) = AntiDiag(F,k), 1 <= k <= j) ? [5]
> >>
> >> Not that it matters.
> >
> > There you go. Completely different assertions actually.
> Do you know the difference between
>
> (Ai)(Ej)(F(i,j) = AntiDiag(j)) [6]
>
> and
>
> (Ei)(Aj)(F(i,j) = AntiDiag(j)) [7]
>
> ?
>
> If so, explain it to me. I'm curious.
6 - all reals in F have an identical digit to antidiag (same position, same value)
7 - antidiag is a member of F
That was a courtesy, remember its YOU who are making useless formula in this thread!
what is the difference between 2 and 4 ?
Herc
|
|
| | From: | The Ghost In The Machine | | Subject: | Re: How many flips of DIAG are on the infintie list of infinite con flippers ? | | Date: | Sat, 22 Jan 2005 03:00:09 GMT |
|
|
| In sci.logic, |-|erc
wrote
on Sat, 22 Jan 2005 11:47:25 +1000
<35dpkoF4lstkvU1@individual.net>:
> "The Ghost In The Machine" wrote in > >>
>
>> >> >> AntiDiag(F,j) = !(F(j,j)) [1]
>> >> >>
>> >> >> Of course this is not to be confused with the
>> >> >> proposition
>> >> >>
>> >> >> (Aj)(Ei)(F(i,j) = AntiDiag(F,j)) [2]
>> >> >>
>> >> >> which for most F is true.
>
>> >> >
>> >> > More importantly
>> >> > (Aj)(Ei)(F(i,k) = AntiDiag(F,k) 0<=k<=j [3]
>> >>
>> >> I'd phrase that as
>> >>
>> >> (Aj)(Ei)(Ek)(F(i,k) = AntiDiag(F,k), 1 <= k <= j) [4]
>> >
>> > no. that's identical to your near meaningless 1st attempt.
>
>> >> Or did you mean
>> >>
>> >> (Aj)(Ei)(Ak)(F(i,k) = AntiDiag(F,k), 1 <= k <= j) ? [5]
>> >>
>> >> Not that it matters.
>> >
>> > There you go. Completely different assertions actually.
>
>
>> Do you know the difference between
>>
>> (Ai)(Ej)(F(i,j) = AntiDiag(j)) [6]
>>
>> and
>>
>> (Ei)(Aj)(F(i,j) = AntiDiag(j)) [7]
>>
>> ?
>>
>> If so, explain it to me. I'm curious.
>
>
>
> 6 - all reals in F have an identical digit to antidiag
> (same position, same value)
>
> 7 - antidiag is a member of F
There's hope for you yet.
>
>
> That was a courtesy, remember its YOU who are making useless
> formula in this thread!
> what is the difference between 2 and 4 ?
[2]: antidiag has a digit in the real list in the same (finite) place
[4]: antidiag has a digit in the real list in the same finite place
as opposed to
[5]: for every antidiag prefix, the prefix is in the real list somewhere
You're mostly right, [2] and [4] aren't much different. I'm not sure
how to prove it, though.
Of course, is F(i) = AntiDiag for some i? Which i? There is
a problem since F(i,i) != AntiDiag(i). So pick a i2 where
F(i2,i) = AntiDiag(i). But then F(i2,i2) != AntiDiag(i2).
AntiDiag() is not a computable function -- though it is well-defined.
>
> Herc
>
--
#191, ewill3@earthlink.net
It's still legal to go .sigless.
|
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| | From: | |-|erc | | Subject: | Re: How many flips of DIAG are on the infintie list of infinite con flippers ? | | Date: | Sat, 22 Jan 2005 13:17:55 +1000 |
|
|
| "The Ghost In The Machine" wrote in > >
> >> >> >> AntiDiag(F,j) = !(F(j,j)) [1]
> >> >> >>
> >> >> >> Of course this is not to be confused with the
> >> >> >> proposition
> >> >> >>
> >> >> >> (Aj)(Ei)(F(i,j) = AntiDiag(F,j)) [2]
> >> >> >>
> >> >> >> which for most F is true.
> >
> >> >> >
> >> >> > More importantly
> >> >> > (Aj)(Ei)(F(i,k) = AntiDiag(F,k) 0<=k<=j [3]
> >> >>
> >> >> I'd phrase that as
> >> >>
> >> >> (Aj)(Ei)(Ek)(F(i,k) = AntiDiag(F,k), 1 <= k <= j) [4]
> >> >
> >> > no. that's identical to your near meaningless 1st attempt.
> >
> >> >> Or did you mean
> >> >>
> >> >> (Aj)(Ei)(Ak)(F(i,k) = AntiDiag(F,k), 1 <= k <= j) ? [5]
> >> >>
> >> >> Not that it matters.
> >> >
> >> > There you go. Completely different assertions actually.
> >
> >
> >> Do you know the difference between
> >>
> >> (Ai)(Ej)(F(i,j) = AntiDiag(j)) [6]
> >>
> >> and
> >>
> >> (Ei)(Aj)(F(i,j) = AntiDiag(j)) [7]
> >>
> >> ?
> >>
> >> If so, explain it to me. I'm curious.
> >
> >
> >
> > 6 - all reals in F have an identical digit to antidiag
> > (same position, same value)
> >
> > 7 - antidiag is a member of F
>
> There's hope for you yet.
>
> >
> >
> > That was a courtesy, remember its YOU who are making useless
> > formula in this thread!
> > what is the difference between 2 and 4 ?
>
> [2]: antidiag has a digit in the real list in the same (finite) place
> [4]: antidiag has a digit in the real list in the same finite place
>
> as opposed to
>
> [5]: for every antidiag prefix, the prefix is in the real list somewhere
>
> You're mostly right, [2] and [4] aren't much different. I'm not sure
> how to prove it, though.
Ek
Ej, Ek, 1<=k<=j
are the same thing for k, if k is otherwise free from j.
k is a natural. <=> k is between 1 and some natural.
the entire prefix matching requires *all* k to have the property, from k = 1 to k = j.
I used to put it F(i, 1), F(i, 2) ... F(i, j) but Tim Little showed me that one.
>
> Of course, is F(i) = AntiDiag for some i? Which i? There is
> a problem since F(i,i) != AntiDiag(i). So pick a i2 where
> F(i2,i) = AntiDiag(i). But then F(i2,i2) != AntiDiag(i2).
>
> AntiDiag() is not a computable function -- though it is well-defined.
>
except all its digits occur in sequence on the list anyway.
either a finite amount of antidiags digits occur in order, or an infinite amount.
finite per row is not middle ground.
Herc
|
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| | From: | The Ghost In The Machine | | Subject: | Re: How many flips of DIAG are on the infintie list of infinite con flippers ? | | Date: | Sat, 22 Jan 2005 05:01:40 GMT |
|
|
| In sci.logic, |-|erc
wrote
on Sat, 22 Jan 2005 13:17:55 +1000
<35duueF4j3bgqU1@individual.net>:
> "The Ghost In The Machine" wrote in > >
>> >> >> >> AntiDiag(F,j) = !(F(j,j)) [1]
>> >> >> >>
>> >> >> >> Of course this is not to be confused with the
>> >> >> >> proposition
>> >> >> >>
>> >> >> >> (Aj)(Ei)(F(i,j) = AntiDiag(F,j)) [2]
>> >> >> >>
>> >> >> >> which for most F is true.
>> >
>> >> >> >
>> >> >> > More importantly
>> >> >> > (Aj)(Ei)(F(i,k) = AntiDiag(F,k) 0<=k<=j [3]
>> >> >>
>> >> >> I'd phrase that as
>> >> >>
>> >> >> (Aj)(Ei)(Ek)(F(i,k) = AntiDiag(F,k), 1 <= k <= j) [4]
>> >> >
>> >> > no. that's identical to your near meaningless 1st attempt.
>> >
>> >> >> Or did you mean
>> >> >>
>> >> >> (Aj)(Ei)(Ak)(F(i,k) = AntiDiag(F,k), 1 <= k <= j) ? [5]
>> >> >>
>> >> >> Not that it matters.
>> >> >
>> >> > There you go. Completely different assertions actually.
>> >
>> >
>> >> Do you know the difference between
>> >>
>> >> (Ai)(Ej)(F(i,j) = AntiDiag(j)) [6]
>> >>
>> >> and
>> >>
>> >> (Ei)(Aj)(F(i,j) = AntiDiag(j)) [7]
>> >>
>> >> ?
>> >>
>> >> If so, explain it to me. I'm curious.
>> >
>> >
>> >
>> > 6 - all reals in F have an identical digit to antidiag
>> > (same position, same value)
>> >
>> > 7 - antidiag is a member of F
>>
>> There's hope for you yet.
>>
>> >
>> >
>> > That was a courtesy, remember its YOU who are making useless
>> > formula in this thread!
>> > what is the difference between 2 and 4 ?
>>
>> [2]: antidiag has a digit in the real list in the same (finite) place
>> [4]: antidiag has a digit in the real list in the same finite place
>>
>> as opposed to
>>
>> [5]: for every antidiag prefix, the prefix is in the real list somewhere
>>
>> You're mostly right, [2] and [4] aren't much different. I'm not sure
>> how to prove it, though.
>
> Ek
> Ej, Ek, 1<=k<=j
>
> are the same thing for k, if k is otherwise free from j.
> k is a natural. <=> k is between 1 and some natural.
>
> the entire prefix matching requires *all* k to have the property,
> from k = 1 to k = j.
And they do. Just not j+1.
>
> I used to put it F(i, 1), F(i, 2) ... F(i, j) but Tim Little
> showed me that one.
>
>
>
>>
>> Of course, is F(i) = AntiDiag for some i? Which i? There is
>> a problem since F(i,i) != AntiDiag(i). So pick a i2 where
>> F(i2,i) = AntiDiag(i). But then F(i2,i2) != AntiDiag(i2).
>>
>> AntiDiag() is not a computable function -- though it is well-defined.
>>
>
> except all its digits occur in sequence on the list anyway.
But not the actual sequence. Yes, it's a strange result.
>
> either a finite amount of antidiags digits occur in order,
> or an infinite amount. finite per row is not middle ground.
>
> Herc
>
--
#191, ewill3@earthlink.net
It's still legal to go .sigless.
|
|
| | From: | |-|erc | | Subject: | Re: How many flips of DIAG are on the infintie list of infinite con flippers ? | | Date: | Sat, 22 Jan 2005 15:27:26 +1000 |
|
|
| "The Ghost In The Machine" wrote
> >> AntiDiag() is not a computable function -- though it is well-defined.
> >>
> >
> > except all its digits occur in sequence on the list anyway.
>
> But not the actual sequence. Yes, it's a strange result.
that's your result. The claim is every possible comibination to oo length
is on one list, the claim stands. When you flip_digits you should CHECK
that you actually have a new sequence.
Can you see the difference between these 2 techniques?
Technique 1
LIST
<123>
<456>
<789>
MAKE A NEW SEQ
<260>
YES <260> is new.
Technique 2
LIST
<123..>
<456..>
<789..>
MAKE A NEW SEQ
<123..>
NO <123..> is not new.
The difference is Technique 1 actually formed a new_sequence_of_digits.
Herc
>
> >
> > either a finite amount of antidiags digits occur in order,
> > or an infinite amount. finite per row is not middle ground.
> >
> > Herc
> >
>
>
> --
> #191, ewill3@earthlink.net
> It's still legal to go .sigless.
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