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

[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.

:rofl:

 

[arildno]

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

Daniel.

:rofl:

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.