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Wouldn't mind a second opinion on this. Wouldn't mind a second opinion on this.

06-11-2020 , 01:39 AM
Quote:
Originally Posted by ecriture d'adulte
My system does NOT have the statement Con(PA) being true anywhere in it. Godel's 2nd specifically tells us that it cannot.
Edit: I see it now. It's a misunderstanding: "Your system" = The system created by the statements provided in the specific post, not "Your system L."

"(P and not(Q)) and Q" is a logical contradiction, yes? This has nothing in particular to do with Godel. This is just a basic truth table. The statement contains both Q and not(Q), and so gives a logical contradiction. This is the construction that I am considering the in bolded statement in question.

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I don't know why you are switching to "logical contradiction" instead of just inconsistent.
It highlights the somewhat obvious distinction that you are refusing to make.

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If PA is consistent, do you think ~Con(PA) by itself is a logical contradiction?
No. I'm saying "if Con(PA) and not(Con(PA)), then contradiction."

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That's really the only way to save what you said, but it really tortures the meaning of contradiction into just a synonym for false.

The error your making is confusing basic consistency with soundness or intuitively true or a higher level concept like w consistency.
A contradiction is a statement. False is a truth value. I don't really see how they're synonyms for each other if they aren't even the same type of object. I'll lay out all of my definitions:

Math:
* A logical contradiction is a statement that is false regardless of the truth values of the propositions contained within that statement. For example, P and not(P) is a contradiction.
* A set of statements S is consistent if there are no derivable statements P (from S) such that both P and not(P) are derivable.
* An argument is sound if the assumptions are true and the deductive logic is valid.
* An axiom is a statement that is accepted to be true without proof.

Edit: To be absolutely clear....

--- False is a truth value of a statement: "The statement P is false."
--- A contradiction is a statement: "The statement 'P and not(P)' is a contradiction because 'P and not(P)' is false for all possible truth values of P."
--- Consistent is a property of a collection of statements: "The collection of statements S is consistent because you cannot derive a contradiction."

Philosophy:
* A statement is a contingent truth if there exist possible universes in which the statement is false.
* A statement is a necessary truth if there do not exist possible universes in which the statement is false.

You may or may not accept the philosophy definitions. But do you accept the mathematical definitions? Because if you do:

Quote:
Originally Posted by wiki
Now, assuming PA is really consistent, it follows that PA + ¬Con(PA) is also consistent
This is correct.

Quote:
Originally Posted by me
I think PA + not(Con(PA)) [leads to] a logical contradiction if "PA is consistent" is true [in addition].
This is correct as well, with the language cleaned up to be tighter.

Quote:
Originally Posted by you
The easiest example would be the set of axioms 1&2 taken together called L:

L:
1. All the axioms of Peano Arithmetic
2. The axioms of Peano Arithmetic are inconsistent.
We are taking these statements to be true. And if that's the case, then if you further assume that it's true that the axioms of Peano Arithmetic are consistent ("really consistent"), then that statement combined with PA + not(Con(PA)) gives a logical contradiction.

But this does not imply that PA + not(Con(PA)) is inconsistent. And I don't think I ever claimed that anywhere.

Last edited by Aaron W.; 06-11-2020 at 02:01 AM.
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06-11-2020 , 02:16 AM
Your cleaned up quote is still not correct. PA + not(Con(PA)) does not lead to a logical contradiction. You already admitted it was consistent.

Let L be: PA + not(Con(PA))

You ask L: "Can PA prove 1=0?"
L says "Yes"
To find a contradiction:
You ask PA: "Can PA prove 1=0?"
PA says "I don't know"

So there is no contradiction and that's why L is consistent.

And while this seems nitty, it's actually hugely important, bc with the completeness theorem we have to have a model for the L integers which was my original claim. They cannot be rejected on the basis of consistency alone.

Last edited by ecriture d'adulte; 06-11-2020 at 02:31 AM.
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06-11-2020 , 11:54 AM
Quote:
Originally Posted by ecriture d'adulte
Your cleaned up quote is still not correct. PA + not(Con(PA)) does not lead to a logical contradiction. You already admitted it was consistent.
If you assume

1) PA
2) not(Con(PA))
3) Con(PA)

Then you have a contradiction.

Quote:
To find a contradiction:
You ask PA: "Can PA prove 1=0?"
PA says "I don't know"
Bolded for emphasis. You are only asking what can be proven.

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They cannot be rejected on the basis of consistency alone.
I don't think I ever rejected anything you've said based on consistency alone. If I did, then that was an error.

This is where we have to talk about the meaning of "possible worlds." The ability to assert a set of consistent axioms does not immediately imply that there is a possible world that coheres to those axioms.
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06-11-2020 , 12:29 PM
You said agreed with wiki when it

Quote:
Originally Posted by wiki
Now, assuming PA is really consistent, it follows that PA + ¬Con(PA) is also consistent
Do you still agree with that? Note that you are correct on (1), (2) and (3) being inconsistent/a contradiction. The problem is (1), (2) and (3) is not the same as the wiki quote above. You are confusing yourself by conflating them.

You also claim
Quote:
Originally Posted by you
I think PA + not(Con(PA)) [leads to] a logical contradiction if "PA is consistent" is true [in addition].
How do you square that with the wiki definition below of consistent,given you agree that PA + ¬Con(PA) is consistent?
Quote:
In classical deductive logic, a consistent theory is one that does not entail a contradiction.
Maybe the easiest way to see it is to simply ask for the contradiction PA + not(Con(PA)) "leads to". You can't claim any wiggle room with "well I'm not talking about only what can be proven" because leading to a contradiction is a syntacial condition that you correctly defined earlier about proving P and ~P.

Last edited by ecriture d'adulte; 06-11-2020 at 12:58 PM.
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06-11-2020 , 01:54 PM
Quote:
Originally Posted by lagtight
Maybe this mathy stuff should be moved to SMP?
I think of RGT as SMP+RGT. RGT was originally created because some of the SMP regs got tired of talking about religion. That's fine, but doesn't mean we can't also talk about SMP topics here.
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06-11-2020 , 02:17 PM
I think it probably makes sense to just recombine SMP and RGT now. Traffic was much higher when they were split. Now each forum only seems to be getting a handful of posts a day.
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06-11-2020 , 02:56 PM
I think of all the off topic forums as my future fiefdoms
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06-11-2020 , 08:09 PM
Quote:
Originally Posted by well named
I think of all the off topic forums as my future fiefdoms
Yes!

I still say if you substitute PA with Probable Assh*le. All this stuff makes much more sense.

You ask PA: "Can PA prove 1=0?"
PA says "I don't know"

Just sayin'.
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06-11-2020 , 09:53 PM
Quote:
Originally Posted by Original Position
I think of RGT as SMP+RGT. RGT was originally created because some of the SMP regs got tired of talking about religion. That's fine, but doesn't mean we can't also talk about SMP topics here.
Kewl!
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06-11-2020 , 11:14 PM
Quote:
Originally Posted by Original Position
I think of RGT as SMP+RGT. RGT was originally created because some of the SMP regs got tired of talking about religion. That's fine, but doesn't mean we can't also talk about SMP topics here.
I think this is essentially my take also; in a sense this makes RGT a more broad free-flowing forum and adds to its panache.

And I disagree about combining the forums. A separation is useful and meaningful and has a practical application, even if there is some cross pollination. And slow burning extinction for either or both forums is not a concern. Peruse some Marcus Aurelius or Confucius if you are confused. This to shall pass, as all things do.

Last edited by Zeno; 06-11-2020 at 11:19 PM.
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06-11-2020 , 11:16 PM
Your distinctiveness will be added to our own.
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06-11-2020 , 11:21 PM
Quote:
Originally Posted by ecriture d'adulte
In classical deductive logic, a consistent theory is one that does not entail a contradiction.
Here's what I said:

Quote:
Originally Posted by me
* A set of statements S is consistent if there are no derivable statements P (from S) such that both P and not(P) are derivable.
I don't see any difference.

Quote:
Maybe the easiest way to see it is to simply ask for the contradiction PA + not(Con(PA)) "leads to".
If you also assert Con(PA), it leads to the contradiction Con(PA) and not(Con(PA)). I don't know how much more plain I can make that.
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06-12-2020 , 12:49 AM
Quote:
Originally Posted by Aaron W.
Here's what I said:



I don't see any difference.
So you think PA + not Con(PA) is consistent. But you also think that it leads to a logical contradiction? If a set of axioms is consistent it means you can never derive a logical contradiction from them.

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If you also assert Con(PA), it leads to the contradiction Con(PA) and not(Con(PA)). I don't know how much more plain I can make that.

Sigh....I am talking about PA+ not Con(PA). You keep acting like that is the same thing as PA+ not Con(PA)+ Con(PA). It isn't. One is consistent and one is inconsistent. You are putting in a ton of work just to not learn Godel's theorems. It's far easier just to learn them.
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06-12-2020 , 01:23 AM
I honestly can't stop laughing.

I can't wait to see what (P)robable (A)sshole arguments are waiting for me in the morning.
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06-12-2020 , 04:54 AM
I need a break.

I think I'm gonna self-ban for a week or so.

I have to relocate in a few days.

See y'all later. Stay well!
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06-12-2020 , 10:55 AM
Quote:
Originally Posted by ecriture d'adulte
Maybe the easiest way to see it is to simply ask for the contradiction PA + not(Con(PA)) "leads to". You can't claim any wiggle room with "well I'm not talking about only what can be proven" because leading to a contradiction is a syntactical condition that you correctly defined earlier about proving P and ~P.
I'll just go ahead and complete this, but using an axiom system that actually does "lead to a contradiction".

Define F: F= PA + 6 is not even
(This is the axioms of PA plus the axiom that 6 is not an even number)

F clearly proves 6 is not even
PA proves 6 is even
So F proves 6 is even and 6 is not even.
That satisfies the definition of inconsistent or "leads to a contradiction" in your terminology.
If PA is actually consistent, you can not possibly do what I did above for F with PA + not(Con(PA)). I know this because Godel proved it a 100+ years ago.
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06-12-2020 , 11:49 PM
lagtight
self-banned

Damn, you dudes are a harsh bunch.

Don't even think it will happen to me.
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06-13-2020 , 05:05 AM
Quote:
Originally Posted by RoundGuy
lagtight
self-banned

Damn, you dudes are a harsh bunch.

Don't even think it will happen to me.
Are you planning to ever make a useful contribution to this thread? What you're doing is pretty rude really, but worse, it's just useless. Try and be useful....
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06-13-2020 , 06:32 PM
Quote:
Originally Posted by ecriture d'adulte
So you think PA + not Con(PA) is consistent. But you also think that it leads to a logical contradiction?
It's clear that you have no interest in reading what I write. You were confused earlier as to why I talked about "logical contradiction" instead of "consistency" and I explained the difference in language. I've laid out the specific list of thins that are being assumed to be true that lead to the logical contradiction. It's not at all clear that you've even taken the time to understand it.

What I've said is simple, clear, and obvious. You are straining to misread it for reasons that are beyond my comprehension.
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06-13-2020 , 08:15 PM
Nah. I get what you are saying. It's the same way tons of students reacts to Godel's 2nd when first presented it. It's just wrong. The set of axioms PA + not Con(PA) does NOT lead to a contradiction if PA is consistent. It seems like it should, because it asserts a set of axioms only to assert those set of axioms lead to a contradiction. But surprisingly they don't. I've explained why and how this works to you multiple times. You can start an SMP or even better reddit thread if want to confirm. But nobody will tell you anything different from what I've already said and it doesn't seem like you care about actually learning the logic 101 stuff. We never really got to the main part, which is how PA+ not Con(PA) forms a different model of the integers. But you'll never get that if you don't get that PA + not Con(PA) does NOT lead to a contradiction
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06-14-2020 , 01:27 AM
Quote:
Originally Posted by Mightyboosh
Are you planning to ever make a useful contribution to this thread? What you're doing is pretty rude really, but worse, it's just useless. Try and be useful....
I figure it's my job here to try and jerk you morons back into the real world with some wit, humor, a little intelligence, a little stupidity, and a ton of complete bullshit.

Original can ban me if needed.
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06-16-2020 , 01:39 AM
Quote:
Originally Posted by ecriture d'adulte
Nah. I get what you are saying.
If you did, you wouldn't be giving the same wrong analysis of my position over and over again. You're analyzing something different.
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06-16-2020 , 08:59 AM
Something can't come from nothing. God always was.
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06-16-2020 , 10:36 AM
Quote:
Originally Posted by Aaron W.
If you did, you wouldn't be giving the same wrong analysis of my position over and over again. You're analyzing something different.
Don't you find it remarkable that the above has been your standard answer for several years?

Maybe the problem has been you (your pathetic arrogance and your inability to clearly state your position) all along?
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06-16-2020 , 10:52 AM
Quote:
Originally Posted by Aaron W.
If you did, you wouldn't be giving the same wrong analysis of my position over and over again. You're analyzing something different.
Too bad for you I have your own analysis of your position.

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I think PA + not(Con(PA)) [leads to] a logical contradiction if "PA is consistent" is true.
The only surprising thing is that you are fighting so much. It's clear you know 0 logic , because nobody that knows any would say the above. Given this is a religion sub on a poker forum, not knowing any logic is completely expected/standard. Not being able to stop, think and learn is problematic even on just a poker forum.
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