Proof that, (1/0 = 1/0) is false.

[chronon]

The damned point is that (1/0 =1/0) is false.

Of course (1/0) is not a real number, who said it was?

(Owen Holden) is not a real number

Therefore (Owen Holden=Owen Holden) is false?

 

[Owen Holden]

You mean that it is trivially true? Well then, what do you make of Owen's proof that ~(1/0 = 1/0). Is it, then, also false? Or is it meaningless altogether, i.e. is 1/0 = 1/0 trivially true, or meaningless since 1) it is an equation using undefined terms and 2) it seems to be both true and false if treated like a meaningful equation?

1/0 = 1/0 is trivially true, is contradictrory for me.

Non-existent things are not self-identical.

x=y =df AF(Fx <-> Fy), only applies when x and y are existent objects.

(the x: Fx & ~(Fx)), does not exist!

Therefore, (the x: Fx & ~(Fx))=(the x: Fx & ~(Fx)), is false.

 

[Hurkyl]

1/0 = 1/0 is trivially true, is contradictrory for me.

And it's neither for me, because "1/0 = 1/0" is not a valid string of symbols in the language of the arithmetic of real numbers.

 

[AKG]

Edit: Wait, if you let G(1/0) be meaningless, you can't derive anything from it- it isn't a proposition and doesn't get in the door. That's what I don't understand- why limit the things you let in the door? Seriously, that's not meant as a rhetorical question.

If we let any talk about things like 1/0 in the door, then we can derive all sorts of contradictions, like G(1/0) & ~(G(1/0)). In general, especially in mathematics, we don't treat undefined things as false, but don't treat them altogether, i.e. as far as we're concerned, they're meaningless. There is nothing to be gained by talking about undefined things, or things that are not well-defined, moreover, refraining from talking about them insulates us from whatever contradictions arise from talking about them.

 

[arildno]

The actual argument is vacous in the extreme, it relies on the propostion that z*1/z = 1 when z is not equal to zero, it then uses this propostion to conclude that if z = 0 then z*1/z = 1 (!) and 0 is not equal to 0!

If this is what he "proves", then his "proof" is invalid, because it uses an undefined term.
Only terms rigourously defined "exists" (in a mathematical/logical sense) , they certainly cannot pop up of nowhere in some "proof".

 

[AKG]

Wrong. I said ~G(1/0) is ambiguous, ~(G(1/0)) is not ambiguous.

Why do you claim ~(G(1/0)) <-> (1/0=1/0) ?

G(1/0) is false for all G, therefore ~(G(1/0)) is true.
(1/0=1/0) is false, therefore, the equivalence fails.

Suppose we have:

G(x) <-> x=x
F(x) <-> ~(G(x)) <-> ~(x=x)

I can prove ~(F(1/0)) by replacing G with F in the proof, so I can prove:
~(~(G(1/0))) <-> G(1/0) <-> 1/0 = 1/0

Of course, we can prove ~(G(1/0)) <-> ~(1/0 = 1/0)

We get (1/0 = 1/0) and ~(1/0 = 1/0), contradiction. I believe the problem arises when you say z=0 in the second or third step, and then continue on with the deduction as though it were meaningful. z=0 is not a valid substitution, because on the left side you get G(1/0) which, according to you, has a false truth value, but I would refrain from giving it, or any statement which talks about something with no referrent, a truth value. The main reason being that, as I believe I have shown above, we can derive any contradiction we want.

 

[AKG]

If this is what he "proves", then his "proof" is invalid, because it uses an undefined term.
Only terms rigourously defined "exists" (in a mathematical/logical sense) , they certainly cannot pop up of nowhere in some "proof".

Well then, care to point out which step of the argument is invalid, and why?

 

[arildno]

Whenever he uses 1/0 as a meaningful term. It is not.

 

[AKG]

Owen

D1. 1/z =df (the x: 1=x*z & ~(z=0))

If we try to plug z=0 into this, there is no x that satisfies this, so is it not right to say that you can't plug 0 into this in the first place? Otherwise, we get 1/0 =df (nothing)? Certainly, mathematicians treat it this way, that we don't plug 0 into the denominator at all. We don't say that 1/0 =df nothing, we simply disallow any talk of 1/0 in the first place, it's meaningless in mathematics. So, when we say that we can't substitute z=0 in the first place, then the step of your argument (step 2 or 3, can't remember) which does that very substitution would be considered inadmissible.

 

[jcsd]

1/0 isn't necessarily menaigless in mathematics but it is in the real numbers and other rings. So the staemnt 1/0 = 1/0 doesn't mena anything in the context of ring theory, but 1/0 =1/0 is trivial were it has meaning.

 

[moonlight]

1/0 or 0/0 or some other forms are meaningless without specification of their origins (functions).These,generally written as here, are not numbers and sign "=" becomes meaningless for comparations.

 

[honestrosewater]

If we let any talk about things like 1/0 in the door, then we can derive all sorts of contradictions, like G(1/0) & ~(G(1/0)). In general, especially in mathematics, we don't treat undefined things as false, but don't treat them altogether, i.e. as far as we're concerned, they're meaningless. There is nothing to be gained by talking about undefined things, or things that are not well-defined, moreover, refraining from talking about them insulates us from whatever contradictions arise from talking about them.

Yes, I understand that. Is my question about changing the rules off-topic?

 

[AKG]

Yes, I understand that. Is my question about changing the rules off-topic?

Not exactly sure what you're asking. What rule am I changing or adding?

 

[honestrosewater]

Not exactly sure what you're asking. What rule am I changing or adding?

You aren't changing the rules. I'm asking why you don't consider changing the rules. Say a system has symbols, strings, and formulas, with their intuitive definitions, where formulas have truth-values. You have a consistent system where, for example, "1/0=1/0" is a string but not a formula. I understand why "1/0=1/0" is not a formula in this system. I'm asking why you don't consider switching to an alternative, consistent system with different rules where, for example, "1/0=1/0" is a formula. It sounds like you and others think the choice between two such systems would just be a matter of taste. But do you know that the latter system would not be more useful, powerful, or interesting than the former? Does someone already know that the ability to talk about non-existent objects adds nothing desirable to a system? I thought the ability to talk about non-existent objects has already been shown to be desirable in other systems.

Maybe it would make more sense if you put yourself in the position of the rule-maker. Say you've made up some symbols and rules for constructing strings from those symbols. Say you're in the process of making up rules for constructing formulas from those strings, and you already have some idea of what you want your formulas to be. Your strings fall into 3 broad categories: strings that are completely unintelligible (and definitely not formulas), strings you definitely want to be formulas, and strings that aren't completely unintelligible and are very similar to the strings you want to be formulas. For example, "The present King of France is bald" and "1/0=1/0" would fall into that last category. Why would you choose rules that exclude the last category of strings from being formulas without considering choosing rules that would allow the last cateogry to be formulas?

 

[Owen Holden]

Suppose we have:

G(x) <-> x=x
F(x) <-> ~(G(x)) <-> ~(x=x)

I can prove ~(F(1/0)) by replacing G with F in the proof, so I can prove:
~(~(G(1/0))) <-> G(1/0) <-> 1/0 = 1/0

Of course, we can prove ~(G(1/0)) <-> ~(1/0 = 1/0)

We get (1/0 = 1/0) and ~(1/0 = 1/0), contradiction.

Not so. [G(1/0) <-> (1/0=1/0)] & ~(G(1/0)) -> ~(1/0=1/0), is true.
[~(~(G(1/0))) <-> (1/0=1/0)] & ~(G(1/0)) -> (1/0=1/0) is false.

How do you derive that (1/0=1/0) is true??

You cannot derive both ~(1/0=1/0) and (1/0=1/0).

I believe the problem arises when you say z=0 in the second or third step, and then continue on with the deduction as though it were meaningful. z=0 is not a valid substitution, because on the left side you get G(1/0) which, according to you, has a false truth value, but I would refrain from giving it, or any statement which talks about something with no referrent, a truth value. The main reason being that, as I believe I have shown above, we can derive any contradiction we want.

You have not shown any contradiction here at all.

 

[Hurkyl]

How do you derive that (1/0=1/0) is true??

Just as he said: apply your method to F.

 

[Owen Holden]

Just as he said: apply your method to F.

Suppose we have:

G(x) <-> x=x
F(x) <-> ~(G(x)) <-> ~(x=x)

I can prove ~(F(1/0)) by replacing G with F in the proof, so I can prove:
~(~(G(1/0))) <-> G(1/0) <-> 1/0 = 1/0

No you cannot. If you replace G(x) with ~(G(x))

"D2. G(the x: Hx) =df Ey(Ax(x=y <-> Hx) & Gy)"

~(G(the xx)) <-> ~Ey(Ax(x=y <-> Hx) & Gy).

Which is true and not false. i.e. we do not get ~(~(G(1/0)) as a theorem.

(1/0=1/0) is false!

Of course, we can prove ~(G(1/0)) <-> ~(1/0 = 1/0)

We get (1/0 = 1/0) and ~(1/0 = 1/0), contradiction.

Both you and AKG are wrong.

 

[AKG]

honestrosewater

Why would I consider changing the rules? You suggest that it is no better or worse to do so, so why bother?

I'm asking why you don't consider switching to an alternative, consistent system with different rules where, for example, "1/0=1/0" is a formula.

And what would its truth value be? In fact, what does the string "1/0" refer to?

Your strings fall into 3 broad categories: strings that are completely unintelligible (and definitely not formulas), strings you definitely want to be formulas, and strings that aren't completely unintelligible and are very similar to the strings you want to be formulas. For example, "The present King of France is bald" and "1/0=1/0" would fall into that last category. Why would you choose rules that exclude the last category of strings from being formulas without considering choosing rules that would allow the last cateogry to be formulas?

Why? Because we gain nothing from speaking about 1/0, it is meaningless. I also think there is a difference between imaginary and hypothetical things, and things which would be contradictory, by definition. Consider the definition Owen gave for 1/z. If z = 0, then the 1/z is, by definition some x that satisfies, among other conditions, ~(z = 0). Therefore, it must satisfy the conditions z=0 and ~(z=0). It is possible and conceivable for there to be a King of France, it is impossible and unconceivable for any number x to satisfy the conditions z=0 and ~(z=0).

We can speak of the King of France existing in a hypothetical or imaginary sense, we can't speak of 1/0 existing (as a real number) in any sense since it is inherently contradictory. "The King of France" can exist, it simply depends on the domain. If it is the domain of hypothetical beings, he's there, and if it is in the domain of physical, "actual" beings, he's not. In certain domains, perhaps 1/0 does exist (I believe someone on this thread, perhaps jcsd, hinted at that), but with respect to reals, it is a meaningless string, it does not exist in the domain of reals.

 

[AKG]

Owen

We can prove that \forall \, G\, (\neg (G(1/0)). So, why does this not hold for G(x) \Leftrightarrow \neg (x = x)? If it does hold, then doesn't this prove \neg (\neg (1/0 = 1/0)), i.e. 1/0 = 1/0?

Also, given your definition D1, wouldn't you say that it doesn't make sense for z=0, since we get:

1/0 \equiv x : (x \times 0 = 1 \wedge \neg (0 = 0))

and of course, no such x can satisfy \neg (0 = 0). Don't you agree that we simply cannot substitute z=0 into the definition of 1/z? If so, we certainly can't do it in your proof given in your first post.

 

[honestrosewater]

Why would I consider changing the rules? You suggest that it is no better or worse to do so, so why bother?

If anything, I'm suggesting it's better to not exclude strings unnecessarily. People make use of their ability to talk and reason about non-existent things in their natural language. So why would you deny yourself that ability when constructing a formal system? I understand wanting to exclude the self-contradictory and such; I don't understand wanting to exclude the non-existent and making other such assumptions about existence and emptiness. To be clear, I'm not disagreeing with you about what follows from some set of rules. But when you say you think so and so should be treated in such and such a way, you're clearly not following rules anymore; You're making them up or, at least, refining them or making them explicit.

And what would its truth value be?

I don't know- it would depend on the rules of course. I haven't mastered any set of rules yet, so I can't easily see what would or wouldn't work. I'm waiting on some books that don't make as many assumptions as others, so I'll hopefully have an answer soon.

In fact, what does the string "1/0" refer to?

I don't know- it's just a string of symbols, and I don't know how to interpret them. I suppose it refers to the multiplicative inverse of the additive identity of some field- I guess the reals, but I don't recall anyone saying- and that may be neither correct nor the answer you're looking for.

Why? Because we gain nothing from speaking about 1/0, it is meaningless.

Great, that's what I was asking, but I don't know what makes something meaningless- contradiction or failure to be a formula, theorem, to exist, refer, denote, have a non-empty extension, whatever.

I also think there is a difference between imaginary and hypothetical things, and things which would be contradictory, by definition. Consider the definition Owen gave for 1/z. If z = 0, then the 1/z is, by definition some x that satisfies, among other conditions, ~(z = 0). Therefore, it must satisfy the conditions z=0 and ~(z=0). It is possible and conceivable for there to be a King of France, it is impossible and unconceivable for any number x to satisfy the conditions z=0 and ~(z=0).

We can speak of the King of France existing in a hypothetical or imaginary sense, we can't speak of 1/0 existing (as a real number) in any sense since it is inherently contradictory. "The King of France" can exist, it simply depends on the domain. If it is the domain of hypothetical beings, he's there, and if it is in the domain of physical, "actual" beings, he's not. In certain domains, perhaps 1/0 does exist (I believe someone on this thread, perhaps jcsd, hinted at that), but with respect to reals, it is a meaningless string, it does not exist in the domain of reals.

Okay. So would you prefer a system that can analyze hypothetical things, or would you rather exclude them from the start? Or do you not care either way? I thought you wanted to exclude them; That's why I asked this in the first place.

 

[Owen Holden]

Owen

We can prove that \forall \, G\, (\neg (G(1/0)). So, why does this not hold for G(x) \Leftrightarrow \neg (x = x)? If it does hold, then doesn't this prove \neg (\neg (1/0 = 1/0)), i.e. 1/0 = 1/0?

No it does not prove ~~(1/0 =1/0), because it does not hold.

D1. 1/z =df (the x: x*z=1 & ~(z=0))

1/0 =df (the x: x*0=1 & ~(0=0)).

(1/0) is defined by the description (the x: x*0=1 & ~(0=0)).

Both (x*0=1) and ~(0=0) are contradictory, because Ax(x*0=0) and (0=0) and ~(0=1) are theorems.

Therefore, (1/0), (the x: x*0=1 & ~(0=0)), does not exist.
(1/0) is not undefined here.

As I said in the original post; we can drop the condition ~(z=0) if we assume that Ax(x*0=0).

"If we can assert that Ax(x*0=0), then D1. is simplified.
D1a. 1/z =df (the x: 1=x*z), and the proof still works."

Therefore, (1/0), (the x: x*0=1), does not exist.

D2. G(the x:Fx) =df Ey(Ax(x=y <-> Fx) & Gy)

That is, 1/0 = 1/0 <-> Ey(Ax(x=y <-> x*0=1) & y=y)
Which was proven false! i.e. 1/0 = 1/0 is false.

~(1/0 = 1/0) <-> ~Ey(Ax(x=y <-> x*0=1) & y=y).
which is obviously true. Because, Ey(Ax(x=y <-> x*0=1) & y=y) is false.
i.e. ~(1/0 =1/0) is true.


Note: It is incorrect to say, ~(1/0 =1/0) <-> Ey(Ax(x=y <-> x*0=1) & ~(y=y)). (which seems to be what you want to do).

This is an abuse of description theory, and I think Strawson wants to do the same thing.

[Also, given your definition D1, wouldn't you say that it doesn't make sense for z=0, since we get:

1/0 \equiv x : (x \times 0 = 1 \wedge \neg (0 = 0))

Rather, we should say: 1/0 = (the x: contradiction).
And it is a theorem of descriptions that (the x: contradiction) does not exist!
Therefore, (1/0) does not exist, even though it is defined.

and of course, no such x can satisfy \neg (0 = 0). Don't you agree that we simply cannot substitute z=0 into the definition of 1/z? If so, we certainly can't do it in your proof given in your first post.

Of course it makes sense to substitute 0 for z, because z includes 0.

D1. 1/z =df (the x: x*z=1 & ~(z=0)), and, D1a. 1/z =df (the x: 1=x*z), are both definitions which have sense for all values of z, including 0.

Az(1/z = (the x: x*z=1)) is true, by the definition D1a.

 

[arildno]

"Of course it makes sense to substitute 0 for z, because z includes 0."
Worthless crap.

 

[dextercioby]

Not really worthless.It was worth the effort of writing a post,pressing "submit",...

Daniel.

 

[arildno]

Not really worthless.It was worth the effort of writing a post,pressing "submit",...

Daniel.

Still crap, though.

 

[AKG]

honestrosewater

We can and do talk about hypothetical things all the time. We can easily speak of them existing in a different sense, or in a different domain, and we can deal with them no differently from how we deal with physical things, numbers, etc. I have no reason to exclude talk of such things, but I think we agree that self-contradictory things should be excluded.

 

[AKG]

No it does not prove ~~(1/0 =1/0), because it does not hold.

Why not? In post 2 of this thread, I show ~(G(1/0)) without specifying what G(x) is, in other words, this holds for arbitrary G. So, let G(x) <-> ~(x=x).

If my post 2 is correct, then:

~(G(1/0)) for any G.

If G(x) <-> ~(x=x) is a reasonable definition for G, then:

~(~(1/0 = 1/0))

Assuming you agree with your own "proof" that ~(1/0 = 1/0), let P = ~(1/0 = 1/0), then we have P (your proof) and ~P (my proof), a contradiction. So which is it: is my post 2 incorrect, or is, for some reason, G(x) <-> ~(x=x) an unreasonable definition for G? I think it's neither. It's no surprise that, when talking about something which is has a contradictory definition, like 1/0, that we derive other contradictions.

 

[Owen Holden]

Quote:
Originally Posted by Owen Holden
No it does not prove ~~(1/0 =1/0), because it does not hold.

Why not? In post 2 of this thread, I show ~(G(1/0)) without specifying what G(x) is, in other words, this holds for arbitrary G. So, let G(x) <-> ~(x=x).

If my post 2 is correct, then:

~(G(1/0)) for any G.

Yes, 8. ~(G(1/0)), of your post 2 is correct.

If G(x) <-> ~(x=x) is a reasonable definition for G, then:

~(~(1/0 = 1/0))

Incorrect.

Assuming you agree with your own "proof" that ~(1/0 = 1/0), let P = ~(1/0 = 1/0), then we have P (your proof) and ~P (my proof), a contradiction. So which is it: is my post 2 incorrect, or is, for some reason, G(x) <-> ~(x=x) an unreasonable definition for G? I think it's neither.

Your proof (~P) is faulty.

D1. 1/z =df (the x: 1=x*z & ~(z=0))

D2. G(the x: Fx) =df Ey(Ax(x=y <-> Fx) & Gy)

You have not shown that ~(~(G(1/0))) is true.
Your substitution is incorrect.

If we substitute (~G) for G in ~(G(1/0)) we do not get ~(~(G(1/0))),
rather, we do get ~((~G)(1/0)) which is also true.

~(~(G(1/0))) <-> ~((~G)(1/0)) is contradictory.

1/0 has the property G is false for all G.
1/0 has the property non-G, (~G), is false for all G.

~(Gx) <-> (~G)x, iff, x exists!

There is no property that 1/0 has! Because it does not exist.

G(1/0) is false for all G, that is, F(1/0) is false and (~F)(1/0) is false and
(F -> H)(1/0) is false etc. etc.

All predications of the form (G) are contradictory for the non-referring description 1/0, (the x: 1= x*0 & ~(0=0)).

1. (1/0 = 1/0) <-> Ey(Ax(x=y <-> (x*0=1 & ~(0=0))) & y=y), by D1.
2. ~(1/0 = 1/0) <-> ~Ey(Ax(x=y <-> (x*0=1 & ~(0=0))) & y=y),
by ~p <-> ~(p).


=|= means 'is not equal to'.

3. (1/0 =|= 1/0) <-> Ey(Ax(x=y <-> (x*0=1 & ~(0=0))) & y =|= y),
By my post #1 and by your post#2, ~(1/0 =|= 1/0) is a theorem.
i.e. (1/0 =|= 1/0) is false.

~(~(1/0 = 1/0)) <-> ~(1/0 =|= 1/0) is contradictory.

4. ~(1/0 =|= 1/0) <-> ~Ey(Ax(x=y <-> (x*0=1 & ~(0=0))) & y =|= y).
i.e. ~(1/0 =|= 1/0) is true.

It's no surprise that, when talking about something which is has a contradictory definition, like 1/0, that we derive other contradictions.

I disagree, you cannot derive a contradiction from the theory of descriptions.

 

[honestrosewater]

If we substitute (~G) for G in ~(G(1/0)) we do not get ~(~(G(1/0))),
rather, we do get ~((~G)(1/0)) which is also true.

I still don't know what rule allows you to make this distinction. I thought we established that FOL makes no such distinction.

I disagree, you cannot derive a contradiction from the theory of descriptions.

How do you respond to the following?

• Counterexample to Russell’s Theory: ‘Augustus Caesar worshipped
Jupiter’, which is true, is analyzed as ‘Augustus Caesar worshipped
the most powerful Roman god’, which in turn, becomes analyzed as:
There exists a unique most powerful Roman god and Caesar
worshipped it’:
Ex[Mx & Ay(My -> y= x) & Wcx]
This is false, contrary to historical fact.
-http://ist-socrates.berkeley.edu/~fitelson/125/lecture_zalta.pdf

 

[AKG]

Owen

You're making it much more difficult than it needs to be. Define G by G(x) <-> ~(x = x). By my post #2, we get ~(G(1/0)). By definition of G, we get:

~(G(1/0))
~(~(1/0 = 1/0))

I'm not substituting ~G for G. If that's all I were doing, then yes, we'd get ~((~G)(1/0)), but that's not the substitution I'm making, I'm substituting P for G, where P(x) <-> ~(G(x)). I know that you can argue that ~(~(G(1/0))) is not equivalent to ~((~G)(1/0)) which is why I'm not talking about ~((~G)(1/0)). You've already answered the question that I asked about the soundness of my post #2. So, all I need you to accept that the following definition is acceptable:

G(x) <-> ~(x = x)

I believe it is. THAT IS THE ONLY POSSIBLE POINT OF CONTENTION. If you have no problem with that definition for G, then you cannot consistently assert that there is some problem with ~(G(1/0)), i.e. ~(~(1/0 = 1/0)). Upon accepting this, you will have to admit that we get a contradiction.

 

[Owen Holden]

Owen

You're making it much more difficult than it needs to be. Define G by G(x) <-> ~(x = x). By my post #2, we get ~(G(1/0)). By definition of G, we get:

~(G(1/0))
~(~(1/0 = 1/0))

I'm not substituting ~G for G. If that's all I were doing, then yes, we'd get ~((~G)(1/0)), but that's not the substitution I'm making, I'm substituting P for G, where P(x) <-> ~(G(x)). I know that you can argue that ~(~(G(1/0))) is not equivalent to ~((~G)(1/0)) which is why I'm not talking about ~((~G)(1/0)). You've already answered the question that I asked about the soundness of my post #2. So, all I need you to accept that the following definition is acceptable:

G(x) <-> ~(x = x)

"You're making it much more difficult than it needs to be."
Then why don't you prove that (1/0 = 1/0) is true !?

No, it is not acceptable when x is a described object.
If G(x) means x=x then ~(G(x)) means ~(x=x).
That is F(x) <-> ~(G(x)), is not a substitution instance of G(x),
in G(x) <-> x=x, for described objects.

~(G(x)) is not an instance of F(x), where x is a description!

Clearly, you do not understand descriptions.
[G](ix:Fx) =df Ey(Ax(x=y <-> Fx) & Gy) applies to all [G]
That is, it does not apply to f(G(ix:Fx)).

When x is a described object, which is what we are talking about,
then, G(1/0) <-> Ey(Ax(x=y <-> x*0=1) & Gy), by definition D2.

D2. G(ix:Fx) =df Ey(Ax(x=y <-> Fx) & Gy).


(1/0 =1/0) means (the x: x*0=1)=(the x: x*0=1).

(the x: x*0=1)=(the x: x*0=1) means Ey(Ax(x=y <-> x*0=1) & y=y), by D2.

~(1/0 = 1/0) means ~((the x: x*0 =1) = (the x: x*0 = 1)).

~((the x: x*0 =1) = (the x: x*0 = 1)) means ~Ey(Ax(x=y <-> x*0=1) & y=y).

We cannot derive your claim that, ~(~[(the x: x*0) = (the x: x*0=1)]) is true!

Please demonstrate that, (the x: x*0=1) = (the x: x*0=1) is true.
Can You??

I don't think so.

 

[AKG]

No, it is not acceptable when x is a described object.
If G(x) means x=x then ~(G(x)) means ~(x=x).
That is F(x) <-> ~(G(x)), is not a substitution instance of G(x),
in G(x) <-> x=x, for described objects.

I don't follow this. We can define G by:

G(x) <-> x = x

But not

G(x) <-> ~(x = x)

Why in the world not? Are the following acceptable?

G(x) <-> x is odd
G(x) <-> x is even
G(x) <-> ~(x is odd)

It really seems to me that you don't understand what I'm saying. Could you provide a link that formally states the rules for what G can be, and what it can't be?

We have established that, for all G, ~(G(1/0)), regardless of the choice of G, right? So why can't I just choose G by defining it G(x) <-> ~(x = x)? I understand that if F(x) <-> (x = x), then G(x) <-> ~(F(x)), but why does that matter? I'm not talking about F. Just tell me simply why we can't choose G such that G(x) <-> ~(x=x).

Please demonstrate that, (the x: x*0=1) = (the x: x*0=1) is true.
Can You??

I can't see why G(x) <-> ~(x = x) is not acceptable, but if it is, then assuming z=0 is an acceptable substitution, I can show it. Otherwise, it still holds that you can't substitute z=0, i.e. "(the x: x*0=1) = (the x: x*0=1)" is essentially meaningless.

 

[Hurkyl]

Since nobody seems to be convincing anyone of anything, thread locked.