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Topic: RE: Matheology 400: Quantifier Confusion
Replies: 188   Last Post: Dec 13, 2013 5:48 AM

 Messages: [ Previous | Next ]
 ross.finlayson@gmail.com Posts: 2,720 Registered: 2/15/09
Re: Matheology 400: Quantifier Confusion
Posted: Dec 5, 2013 11:12 PM

On Wednesday, December 4, 2013 1:13:59 AM UTC-8, Zeit Geist wrote:
> WM recently wrote:
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> Quantifier confusion is not logically valid.
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> You prove for every n in |N: d_1, d_2, ..., d_n differs from the first n entries of the list.
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> You conclude for all n: d_1, d_2, ... differs from all terms of the list.
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> With the same "logic" you could prove:
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> For every n in |N: d_1, d_2, ..., d_n is in the remaining part of the (rationals-complete) list
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> and conclude for all n: d_1, d_2, ... is in the remaining part of the list.
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> Same quantifier confusion.
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> Or
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> For every n in |N: d_1, d_2, ..., d_n is rational
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> conclusion for all n: d_1, d_2, ... is rational.
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> Tell me where this logical structure differs in my three examples.
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> You answer that the contents has to be considered first?
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> Then you do not use logic but interpretation and arbitrariness.
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> End WM Except.
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> I will be more than happy to tell you exactly where logical structure differs.
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> However, only for the first two. The third is Equivalent to the second.
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> I've written it in very simple terms, so that anyone can easily follow.
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> This post is not meant to be condescending, except to those who deserve it.
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> The best best way show that two Conjectures are of the same Logical Structure is to show that the proofs of the two Conjectures process in the same manner. So that what I will do, prove the Conjectures. Or will I?
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> Alright then. Let's Prove some Conjectures!
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> Remember, before we write a Proof, we must first formalize the statement. This allows to understand clearly what the Natural Language statement actually means.
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> Case I:
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> (assume) for every n in |N: d_1, d_2, ..., d_n differs from the first n entries of the list.
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> conclude for all n: d_1, d_2, ... differs from all terms of the list.
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> Writing the Premise formally, we obtain
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> A( n, m e N ) ( m <= n --> FIS( d_n ) ~= l_m ).
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> For Any Natural Numbers n and m, if m <= n ( if m is one of the first n entries of the list ) then the FIS of length n of d is Not equal the m-th line.
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> Now, the conclusion is a little trickier.
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> I interpret "for all n: d_1, d_2, ..." To mean "the full decimal sequence of d" or just simply d.
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> Actually, it wasn't that tricky. It was just written weird.
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> So, the conclusion is simply that d differs from all terms of the list. Or, formally,
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> A( n e N )( d ~= l_n ).
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> This is a Universal Statement, and we proceed Directly.
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> Assume the Premise.
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> We then choose an arbitrary k e N, substitute k for both n and m in the assumed premise. Obtaining
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> A( k e N )( k <= k --> FIS( d_k ) ~= l_k ).
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> Indeed k <= k, so we arrive at the conclusion that FIS( d_k ) ~= l_k. This is a Sufficient Condition to conclude that d ~= l_k.
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> For the slow kids: If a line is Not equal to a FIS of some Number, then it couldn't possible be equal to the full decimal expansion of that number.
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> Now, since k was chosen arbitrarily, we find that For All, Each, and Every line of the list d is Not equal to that line.
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> qed
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> Case II:
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> (assume) For every n in |N: d_1, d_2, ..., d_n is in the remaining part of the (rationals-complete) list
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> conclude for all n: d_1, d_2, ... is in the remaining part of the list.
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> Here the premise say that for any natural number, there is line in the (rations-complete) list equal to the FIS of length n of d. Formally,
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> A( n e N )( E( m e N )( FIS( d_n ) = l_m ) ).
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> The conclusion states some line equals the diagonal.
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> E( n e N )( d = l_n ).
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> Now, do you still wish to say that the two set of statements possess the same Logic Structure?
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> The conclusion of the First Set of Statements is Universal, whereas that of the Second is Existential.
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> Basically, it boils down to that if you say, "It is Not equal to Any." is actually saying "All are Not equal to It.".
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> The Second Conclusion says "There is Some ( at least one ) equal to It.". You disguise it by starting the Natural Language sentence with "For All".
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> Now, onto a "Proof" of the Second Conclusion.
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> Since this is an Existential Statement, we have two options.
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> 1. Construct a line, l_n, such that the d = l_n is satisfied.
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> 2. Use a proof by Contradiction. That is, we assume that For Every line, d is Not equal to that line and derive a Contradiction.
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> So, Dear Professor, could you please explain how either one of these "Proofs" would continue to the desired Result?
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> Now, please tell me whose Methods the Quantifier Confusion resides in?
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> ZG
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> NEXT!

What is this about? Quantifier exchange leading to quantifier confusion is
from where quantifier exchange is reasonable or not.

In set theory, it's the transfer principle, for example, that allows one to
change quantifiers to the object, to quantifiers of sets of the object (here
all sets). For example, ordinals are this way.

Quantifier exchange is exact and simple. For example, in analysis, it's the
order of components. Algebra, similarly, simply maintains associativity of
events in operation.

Quantifier confusion is also "for any x, exists y", and "for any y, exists x".
To combine them requires defining their complement, that x <=> y from x => y
and y=>x. Where that is usual or normal, for example, that is exact and
simple.

Case 1: anti-diagonal, Case 2: Finite rationals on the tree are the only thing
that build irrationals (and some real numbers are rational). diagonalize the
tree.

EF or the sweep function as it is, here the sweep or range function, is good
for the expansions or the tree. It's zero to one, as the natural numbers go
from 0 to infinity, the real numbers go from zero to one. In the antidiagonal
the antidiagonal's always on the end then shown to be in the range as it's the
limit of the functions that build the sweep function, it goes from zero to one
and furthermore is symmetrical about between zero and 1 in the range, then
domain. As the sweep in the tree, here from what ZG actually says:

E( n e N )( d = l_n )

Here the identity I though is zero, instead of unit.

"... we assume that For Every line, d is Not equal to that line and derive a

this is basically making the line a diagonal, basically establishing otherwise
the space that it would be. The sweep as a sequence starts as all zeroes,
would see d = I_n.

Regards, Ross Finlayson

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