A proof of G in F

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PeteOlcott
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Joined: Mon Jul 25, 2016 6:55 pm

A proof of G in F

Post by PeteOlcott »

The scope of this post is ∃G ∈ F (G ↔ ¬(F ⊢ G))
There exists a G in F such that G is logically equivalent to its own
unprovability in F.

The idea here is to examine the philosophical foundation of the
mathematical notion of incompleteness making sure that it is coherent.

A proof of G in F that proves that G cannot be proved in F is simply
self-contradictory, thus no such G exists in F.

The conventional definition of incompleteness:
Incomplete(T) ↔ ∃φ ((T ⊬ φ) ∧ (T ⊬ ¬φ))

Thus proving that when the above G is neither provable nor refutable
in F it is because G is self-contradictory in F thus not because F is incomplete.
Skepdick
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Joined: Fri Jun 14, 2019 11:16 am

Re: A proof of G in F

Post by Skepdick »

PeteOlcott wrote: Mon Mar 27, 2023 4:09 pm The scope of this post is ∃G ∈ F (G ↔ ¬(F ⊢ G))
There exists a G in F such that G is logically equivalent to its own
unprovability in F.

The idea here is to examine the philosophical foundation of the
mathematical notion of incompleteness making sure that it is coherent.

A proof of G in F that proves that G cannot be proved in F is simply
self-contradictory, thus no such G exists in F.

The conventional definition of incompleteness:
Incomplete(T) ↔ ∃φ ((T ⊬ φ) ∧ (T ⊬ ¬φ))

Thus proving that when the above G is neither provable nor refutable
in F it is because G is self-contradictory in F thus not because F is incomplete.
Did comp.theory get tired of your bullshit?

Here's the decision problem all over again. Only one of these expressions is true:

1. ∃G ∈ F (G ↔ ¬(F ⊢ G))
2. ¬∃G ∈ F (G ↔ ¬(F ⊢ G))

Give me the decision procedure for the above.
PeteOlcott
Posts: 1514
Joined: Mon Jul 25, 2016 6:55 pm

Re: A proof of G in F

Post by PeteOlcott »

Skepdick wrote: Mon Mar 27, 2023 4:22 pm Here's the decision problem all over again. Only one of these expressions is true:

1. ∃G ∈ F (G ↔ ¬(F ⊢ G))
2. ¬∃G ∈ F (G ↔ ¬(F ⊢ G))

Give me the decision procedure for the above.
If there exists a proof of G in F that G is unprovable in F then this forms a contradiction and proves that there is no such G in F.

If a barber claims that he shaves all those that do not shave themselves then ZFC knows that this barber must be lying because he is saying that he both shaves himself and never shaves himself.
Skepdick
Posts: 14364
Joined: Fri Jun 14, 2019 11:16 am

Re: A proof of G in F

Post by Skepdick »

PeteOlcott wrote: Mon Mar 27, 2023 5:32 pm
Skepdick wrote: Mon Mar 27, 2023 4:22 pm Here's the decision problem all over again. Only one of these expressions is true:

1. ∃G ∈ F (G ↔ ¬(F ⊢ G))
2. ¬∃G ∈ F (G ↔ ¬(F ⊢ G))

Give me the decision procedure for the above.
If there exists a proof of G in F that G is unprovable in F then this forms a contradiction and proves that there is no such G in F.
Sure. So you've proven G ⊢ ¬G

That's called proof of negation and it's a valid proof in Intuitionistic logic.
It's different to proof by contradiction which is ¬G ⊢G. Which is not generally valid in Intuitionistic logic.

Which logic are you using? Classical or intuitionistic?

https://math.andrej.com/2010/03/29/proo ... radiction/
PeteOlcott
Posts: 1514
Joined: Mon Jul 25, 2016 6:55 pm

Re: A proof of G in F

Post by PeteOlcott »

Skepdick wrote: Mon Mar 27, 2023 6:22 pm
PeteOlcott wrote: Mon Mar 27, 2023 5:32 pm
Skepdick wrote: Mon Mar 27, 2023 4:22 pm Here's the decision problem all over again. Only one of these expressions is true:

1. ∃G ∈ F (G ↔ ¬(F ⊢ G))
2. ¬∃G ∈ F (G ↔ ¬(F ⊢ G))

Give me the decision procedure for the above.
If there exists a proof of G in F that G is unprovable in F then this forms a contradiction and proves that there is no such G in F.
Sure. So you've proven G ⊢ ¬G

That's called proof of negation and it's a valid proof in Intuitionistic logic.
It's different to proof by contradiction which is ¬G ⊢G. Which is not generally valid in Intuitionistic logic.

Which logic are you using? Classical or intuitionistic?

https://math.andrej.com/2010/03/29/proo ... radiction/
I define my own system of correct reasoning:
(by extending the notion of a syllogism)

Semantic Necessity operator: ⊨□
(a) Some expressions of language L are stipulated to have the semantic
property of Boolean true.

(b) Some expressions of language L are a semantically necessary consequence of others.

T is a subset of (a)
True(L,X) means X ∈ (a) or T ⊨□ X
False(L,X) means T ⊨□ ~X

This abolishes Tarski Undefinability and Gödel Incompleteness.
Skepdick
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Joined: Fri Jun 14, 2019 11:16 am

Re: A proof of G in F

Post by Skepdick »

PeteOlcott wrote: Mon Mar 27, 2023 6:34 pm
Skepdick wrote: Mon Mar 27, 2023 6:22 pm
PeteOlcott wrote: Mon Mar 27, 2023 5:32 pm

If there exists a proof of G in F that G is unprovable in F then this forms a contradiction and proves that there is no such G in F.
Sure. So you've proven G ⊢ ¬G

That's called proof of negation and it's a valid proof in Intuitionistic logic.
It's different to proof by contradiction which is ¬G ⊢G. Which is not generally valid in Intuitionistic logic.

Which logic are you using? Classical or intuitionistic?

https://math.andrej.com/2010/03/29/proo ... radiction/
I define my own system of correct reasoning:
(by extending the notion of a syllogism)

Semantic Necessity operator: ⊨□
(a) Some expressions of language L are stipulated to have the semantic
property of Boolean true.

(b) Some expressions of language L are a semantically necessary consequence of others.

T is a subset of (a)
True(L,X) means X ∈ (a) or T ⊨□ X
False(L,X) means T ⊨□ ~X

This abolishes Tarski Undefinability and Gödel Incompleteness.
What the fuck does this have to do with anything I said?
PeteOlcott
Posts: 1514
Joined: Mon Jul 25, 2016 6:55 pm

Re: A proof of G in F

Post by PeteOlcott »

"Which logic are you using? Classical or intuitionistic?"
Neither I am using my own system of correct reasoning.
Skepdick
Posts: 14364
Joined: Fri Jun 14, 2019 11:16 am

Re: A proof of G in F

Post by Skepdick »

PeteOlcott wrote: Mon Mar 27, 2023 6:38 pm "Which logic are you using? Classical or intuitionistic?"
Neither I am using my own system of correct reasoning.
So lets call this system O for Olcott.

What's Correct(O) provable in?
PeteOlcott
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Joined: Mon Jul 25, 2016 6:55 pm

Re: A proof of G in F

Post by PeteOlcott »

Skepdick wrote: Mon Mar 27, 2023 6:41 pm
PeteOlcott wrote: Mon Mar 27, 2023 6:38 pm "Which logic are you using? Classical or intuitionistic?"
Neither I am using my own system of correct reasoning.
So lets call this system O for Olcott.

What's Correct(O) provable in?
Anything (such as the principle of explosion) that diverges from the foundation of correct reasoning that I have established is incorrect reasoning.

Foundation of correct reasoning
Semantic Necessity operator: ⊨□

(a) Some expressions of language L are stipulated to have the semantic property of Boolean true.
(b) Some expressions of language L are a semantically necessary consequence of others.

T is a subset of (a)
True(L,X) means X ∈ (a) or T ⊨□ X
False(L,X) means T ⊨□ ~X

The above Foundation of correct reasoning is simply how the body of analytic truth really works.
promethean75
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Re: A proof of G in F

Post by promethean75 »

My god skep he's right. I checked his work.

I never thought I'd see it in my lifetime. But he's done it. He's bloody done it.
Skepdick
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Joined: Fri Jun 14, 2019 11:16 am

Re: A proof of G in F

Post by Skepdick »

PeteOlcott wrote: Mon Mar 27, 2023 6:57 pm
Skepdick wrote: Mon Mar 27, 2023 6:41 pm
PeteOlcott wrote: Mon Mar 27, 2023 6:38 pm "Which logic are you using? Classical or intuitionistic?"
Neither I am using my own system of correct reasoning.
So lets call this system O for Olcott.

What's Correct(O) provable in?
Anything (such as the principle of explosion) that diverges from the foundation of correct reasoning that I have established is incorrect reasoning.

Foundation of correct reasoning
Semantic Necessity operator: ⊨□

(a) Some expressions of language L are stipulated to have the semantic property of Boolean true.
(b) Some expressions of language L are a semantically necessary consequence of others.

T is a subset of (a)
True(L,X) means X ∈ (a) or T ⊨□ X
False(L,X) means T ⊨□ ~X

The above Foundation of correct reasoning is simply how the body of analytic truth really works.
Is there any reason why you are avoiding my question?

You said that you are using your system of "correct" reasoning. So to use your own operator: ⊨□ Correct(O)

In what system is the "Correctness" property of your system provable in?
PeteOlcott
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Re: A proof of G in F

Post by PeteOlcott »

"In what system is the "Correctness" property of your system provable in?"
The entire body of analytic truth specified in any formal or natural language that exists or ever will exist.

How do you know that baby kittens are a type of animal and not any type of ten story office building?

(a) Some expressions of language L are stipulated to have the semantic property of Boolean true.

(b) Some expressions of language L are a semantically necessary consequence of others.
Skepdick
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Joined: Fri Jun 14, 2019 11:16 am

Re: A proof of G in F

Post by Skepdick »

PeteOlcott wrote: Mon Mar 27, 2023 8:55 pm "In what system is the "Correctness" property of your system provable in?"
The entire body of analytic truth specified in any formal or natural language that exists or ever will exist.
That's not a proof system.
PeteOlcott wrote: Mon Mar 27, 2023 8:55 pm How do you know that baby kittens are a type of animal and not any type of ten story office building?

(a) Some expressions of language L are stipulated to have the semantic property of Boolean true.

(b) Some expressions of language L are a semantically necessary consequence of others.
Idiot has mixed up kittens and Booleans.

It's like he doesn't understand the difference between the abstract and the concrete.
PeteOlcott
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Joined: Mon Jul 25, 2016 6:55 pm

Re: A proof of G in F

Post by PeteOlcott »

"It's like he doesn't understand the difference between the abstract and the concrete."

Every expression of any language that can be completely verified as totally true entirely on the basis of its meaning is an element of the set of analytic truth.

Since the finite string "baby kittens" are a defined set of attributes and the finite string "ten story office buildings" are a defined set of attributes and these two sets are disjoint we know analytically that "baby kittens" are not any type of "ten story office building".
Skepdick
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Joined: Fri Jun 14, 2019 11:16 am

Re: A proof of G in F

Post by Skepdick »

PeteOlcott wrote: Mon Mar 27, 2023 9:16 pm Every expression of any language that can be completely verified as totally true entirely on the basis of its meaning is an element of the set of analytic truth.
You are confusing the semantics of formal languages with what we call "meaning" in natural languages. They aren't even remotely related.

But don't take it from me: https://youtu.be/E4KhK3kktcM?t=2490
Very likely that language doesn't have semantics in the technical sense. It has meaning, but meaning is just some broad loose thing - we don't know what that is.
Here, I have a meme for you.
meme.jpg
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