Deeper than primes - Continuation

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It is nice to know that some people think that the set of real numbers does not exist (even in the abstract sense).:rolleyes:

I do not think that would be nice at all in any conceivable way.

But opinions aside, was this an attempt to disguise that you ran too fast, fell flat on your face and did a 'scorpion'? (metaphorically speaking of course).
 
It is always true (it is tautology) that contradiction logically means "always false" (or in a simpler way: It is a tautology that contradiction is itself)

It is always false (it is contradiction) that tautology logically means "not always true" (or in a simpler way: It is a contradiction that tautology is not itself).

So, it is a tautology that X is itself in case that X means contradiction, or it is a contradiction that X is not itself in case that X means tautology.

By tautology, contradiction ("always false") is itself ("always false").

By contradiction, tautology ("always true") is not itself (not "always true").

By using tautology we define contradiction.

By using contradiction we can't define tautology.

So any framework that defines tautology or contradiction, is based on tautology, and in the case of a framework of sets, the concept of set is a tautology, such that set is the tautological existence of such framework and it enables to deduce further expressions within that framework, where one of them is, for example, a set (a tautological existence) with no members at all (where no members at all can be logically a contradiction about members of a set (where set is a tautological existence) with no members at all).
 
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Let's improve the last part of the previous post:

So any framework that defines tautology or contradiction, is based on tautology, and in the case of a framework of sets, the concept of set is a tautology, such that set is the tautological existence of such framework and it enables to deduce further expressions within that framework, where one of them is, for example, a set (a tautological existence) with no members at all (where "no members at all" can be logically a contradiction about members of a set (where set is a tautological existence) with no members at all).
 
We are still waiting for you, Doron, to show that ∃x is a well-formed formula. You current ramblings are dependent upon it.

Should be simple, no? You just need to show how ∃x can be constructed using the rules for well-formed formulae. How hard can that be?
 
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We are still waiting for you, Doron, to show that ∃x is a well-formed formula.

If Ψ is a formula and x is "a variable serving to represent a hypothetical object from the domain of discourse", then ∃x Ψ is a formula.

x is "a variable serving to represent a hypothetical object from the domain of discourse" is the same as there exists x as "a variable serving to represent a hypothetical object from the domain of discourse" that is the same as ∃x (as used in the blue expression in the following part: ", then ∃x Ψ is a formula.").

So what is written in blue (in the first sentence of this post) is the same as "If Ψ is a formula and ∃x, then ∃x Ψ is a formula."

In other words, what is written in blue (in the first sentence of this post) has no meaning if ∃x is not wff.
 
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what is written in blue has no meaning if ∃x is not wff.


Great! Then you should have no trouble at all showing how ∃x can be constructed from the rules for well-formed formulae.

We patiently await your construction.
 
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Great! Then you should have no trouble at all showing how ∃x can be constructed from the rules for well-formed formulae.

We patiently await your construction.

"If Ψ is a formula and ∃x, then ∃x Ψ is a formula." is wff because also ∃x is wff, exactly as shown in http://www.internationalskeptics.com/forums/showpost.php?p=10052935&postcount=4006.

Also please look at http://www.internationalskeptics.com/forums/showpost.php?p=10050365&postcount=3983.

Changing "set" by "a variable serving to represent a hypothetical object from the domain of discourse" has no impact on the validity of my argument.
 
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EDIT:



"If Ψ is a formula and ∃x, then ∃x Ψ is a formula." is wff because also ∃x is wff, exactly as shown in http://www.internationalskeptics.com/forums/showpost.php?p=10052935&postcount=4006.

Also please look at http://www.internationalskeptics.com/forums/showpost.php?p=10050365&postcount=3983.

Changing "set" by "a variable serving to represent a hypothetical object from the domain of discourse" has no impact on the validity of my argument.

In the referenced posts there is no demonstration nor construction as to *why* it follows that ∃x is a wff...

Are you someone that, when on holiday, starts talking louder and louder at people in your own language because "if they fail to understand you, they must be deaf"?
 
...is wff because also ∃x is wff...

That would be the part you have not yet shown. You've simply asserted it as true.

There are rules for constructing well-formed formulae. To show that ∃x is such a formula, all you need do is show the steps for its construction using those rules.

How hard can that be?
 
Indeed How hard can that be to follow http://www.internationalskeptics.com/forums/showpost.php?p=10052962&postcount=4008 in order to realize that ∃x is wff?

Indeed, that post in that link is not hard to follow.

In fact, it clearly demonstrates a complete lack of demonstration on how to demonstrate that ∃x is wff.

Even more so, it says nothing to make you realize ∃x is wff but rather only says '...also ∃x is wff'.

And we all know that *that* is not an answer. Another scorpion, Doron?
 
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The simple fact that changing "set" (http://www.internationalskeptics.com/forums/showpost.php?p=10050365&postcount=3983) by "a variable serving to represent a hypothetical object from the domain of discourse" (http://www.internationalskeptics.com/forums/showpost.php?p=10052935&postcount=4006) has no impact on the validity of my argument, can't be deduced by jsfisher.

As a result he can't follow after what already has been given in http://www.internationalskeptics.com/forums/showpost.php?p=10050365&postcount=3983 about his wff, which is:

jsfisher said:
If Ψ is a formula and x is a set, then ∃x Ψ is a formula.
 
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...no impact on the validity of my argument, can't be deduced by jsfisher.

So far all you have done is alleged. No argument whatsoever, just claim.

Since you seem to be having trouble locating them on your own, here, try this link. Those are the basic rules. Now all you need to do is show how ∃x fits within the rules to be a well-formed formula.
 
If Ψ is a formula and x is "a variable serving to represent a hypothetical object from the domain of discourse", then ∃x Ψ is a formula.

If the domain of discourse is some theory and this theory can't be deduced unless there exists some hypothetical object that is used by it, then this hypothetical object has tautological existence in the domain of discourse, notated as ∃x, where ∃x is the wff "x always exits" in the domain of discourse.

So, quantifier elimination (http://en.wikipedia.org/wiki/Quantifier_elimination) can't be used in this case, otherwise there is no hypothetical object to talk about in the domain of discourse.

ZFC is a domain of discourse, where x is "a variable serving to represent a hypothetical object from it", known as set.

So within ZFC ∃x is the wff "set's existence is always true", which is not the same as the wff "member's existence is not always true", within ZFC, exactly because within ZFC there exists (∃) at least one set with no members at all.
 
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If Ψ is a formula and x is "a variable serving to represent a hypothetical object from the domain of discourse", then ∃x Ψ is a formula.

If the domain of discourse is some theory and this theory can't be deduced unless there exists some hypothetical object that is used by it, then this hypothetical object has tautological existence in the domain of discourse, notated as ∃x, where ∃x is the wff "x always exits" in the domain of discourse.

So, quantifier elimination (http://en.wikipedia.org/wiki/Quantifier_elimination) can't be used in this case, otherwise there is no hypothetical object to talk about in the domain of discourse.

ZFC is a domain of discourse, where x is "a variable serving to represent a hypothetical object from it", known as set.

So within ZFC ∃x is the wff "set's existence is always true", which is not the same as the wff "member's existence is not always true", within ZFC, exactly because within ZFC there exists (∃) at least one set with no members at all.

Doron, maybe I can help you to understand what you are so completely and utterly missing:

At which point do you validate ∃ with x as being a formula. The ∃ appears magically without any preamble.

You are talking after the fact, not even hypothetical, because you do not even construct a hypothesis (yes, that is where hypothetical comes from).

Your reasoning is really of the ilk 'if I can see only three legs of a table then it follows all tables have three legs'; tobacco and rocking-chair wisdom that can be peddled to the gullible, but fails miserably here.
 
Yes, you have posted that before. Now, how about showing how to construct ∃x from the rules for constructing well-formed formulae?

The rest of us all think you can't because it isn't.
Again ∃x means "x existence is always true" (tautological existence) within a given domain of discourse, where in the case of ZFC, x is the concept of set, which its existence is not an hypothesis (it is a platonic objective elements that is discovered (not invented) within ZFC).

Now please show how the rules for constructing wff deal with platonic objective elements that are discovered (not invented) within a given domain of discourse like ZFC (it has to be clear that without the platonic objective element known as set ZFC can't be used as a mathematical theory).

If you can't show it then there is a fundamental problem in the rules for constructing wff.

So please show it.
 
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Again ∃x means "x existence is always true"

That would be something you made up. And you making things up, as you are wont to do, has no influence on what it actually means.

But all that is just a dodge on your part. You claimed ∃x was a well-formed formula. If so, as you have claimed, then ∃x could be constructed within the rules for well-formed formulae.

It can't; you were wrong, as you are wont to be.
 
That would be something you made up.
Platonic realm is discovered.

"made up" means "invented".

The concept of set (notated by x) within ZFC is platonic ("x existence is always true").

So, without ∃x as wff within ZFC, ZFC can't be used as a mathematical theory.

If you can't show that ∃x is wff by the rules for constructing wff, these rules simply do not cover the Platonic realm, and they have to be corrected in order to also cover it (which is not the current state of these rules).
 
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