Inconsistent Theories Metatheoretically Prove Trivialism

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Mosesquine
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Re: Inconsistent Theories Metatheoretically Prove Trivialism

Post by Mosesquine »

paulemok wrote: April 26th, 2018, 9:04 pm
Mosesquine wrote: April 26th, 2018, 5:16 amI think that the trick is an assumption that 'p' and 'not-p' are true at the same time. Your hypothetic theory T contains the propositions as follows:

(1) p
(2) It is not the case that p

You are irresponsibly assuming that (1) and (2) above are true at the same time, but logical contradictions like (1) and (2) above are not allowed in traditional logic.
There is no assumption that "(1) is true in T." There is no assumption that "(2) is true in T." The propositions "(1) is true in T" and "(2) is true in T" are logical consequences.
Mosesquine wrote: April 26th, 2018, 5:16 amYour whole argument is a failure, unless you prove that contradictions can be established.
The very argument we are discussing proves that contradictions can be established.

Consider the following truth table:

p ... not-p ... p & not-p
T ... F ........ F
F ... T ........ F

It seems that no true contradiction can be established. If it can be, then suggest a relevant truth table.
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paulemok
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Re: Inconsistent Theories Metatheoretically Prove Trivialism

Post by paulemok »

Mosesquine wrote: April 27th, 2018, 12:26 am It seems that no true contradiction can be established. If it can be, then suggest a relevant truth table.
One problem with the truth table you provided is that it presumes p has exactly one truth value. The table covers the case where p is true and the case where p is false, but it does not cover the case where p is true and false.

No truth table has been employed in the proof in my original post, so no truth table is relevant. One sound proof alone is sufficient to make my case, and I have already provided one sound proof. In order to prove me wrong, a problem must be shown to exist with the proof I have already provided.
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The Beast
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Re: Inconsistent Theories Metatheoretically Prove Trivialism

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And the dream said: “ I will think you happy in this opinion, if only you add one thing”
“Do you think there is anything in our not your or your not our mortal and perishable affairs which could grant a state of this sort?”
In my waking moment it was clear: To p or not to p.
P: is a good start. Carry on.
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Mosesquine
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Re: Inconsistent Theories Metatheoretically Prove Trivialism

Post by Mosesquine »

paulemok wrote: April 27th, 2018, 7:14 am
Mosesquine wrote: April 27th, 2018, 12:26 am It seems that no true contradiction can be established. If it can be, then suggest a relevant truth table.
One problem with the truth table you provided is that it presumes p has exactly one truth value. The table covers the case where p is true and the case where p is false, but it does not cover the case where p is true and false.

No truth table has been employed in the proof in my original post, so no truth table is relevant. One sound proof alone is sufficient to make my case, and I have already provided one sound proof. In order to prove me wrong, a problem must be shown to exist with the proof I have already provided.

All negated true statements are false statements. Your proof amounts to which there is no false statements. Since truth value is either true or false, your proof is failed.
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Re: Inconsistent Theories Metatheoretically Prove Trivialism

Post by paulemok »

Mosesquine wrote: April 28th, 2018, 2:32 am Your proof amounts to which there is no false statements.
That's true and false. My proof establishes that if T is an inconsistent theory and p is a proposition, then trivialism is true.

Example 3 describes an inconsistent theory that exists.

Example 3. B is a theory that has the following two postulates.

Postulate 1. 1 equals 1.
Postulate 2. It is not true that "1 equals 1."

Since Postulate 2 is the negation of Postulate 1, some contradiction exists in B. So, by definition of inconsistent theory, B is inconsistent. Despite the fact that B is inconsistent, B exists. This concludes the example.


As Example 3 makes evident, B exists. By existential introduction, some inconsistent theory exists. So, the proposition "some inconsistent theory exists" exists. Thus, by existential introduction, some proposition exists. By conjunction introduction, some inconsistent theory exists and some proposition exists. Thus, the hypothesis of my proof is satisfiable. So, as my proof implies, trivialism is true. By definition of trivialism, all propositions are true. Since all propositions are true, the proposition "all propositions are true and false" is true. Therefore, by simplification, all propositions are true and false.
Mosesquine wrote: April 28th, 2018, 2:32 am Since truth value is either true or false, your proof is failed.
Since all propositions are true, the proposition "my proof is sound and unsound" is true. Therefore, by simplification, my proof is sound and unsound.
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Mosesquine
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Re: Inconsistent Theories Metatheoretically Prove Trivialism

Post by Mosesquine »

paulemok wrote: April 28th, 2018, 9:55 pm
Mosesquine wrote: April 28th, 2018, 2:32 am Your proof amounts to which there is no false statements.
That's true and false. My proof establishes that if T is an inconsistent theory and p is a proposition, then trivialism is true.

Example 3 describes an inconsistent theory that exists.

Example 3. B is a theory that has the following two postulates.

Postulate 1. 1 equals 1.
Postulate 2. It is not true that "1 equals 1."

Since Postulate 2 is the negation of Postulate 1, some contradiction exists in B. So, by definition of inconsistent theory, B is inconsistent. Despite the fact that B is inconsistent, B exists. This concludes the example.


As Example 3 makes evident, B exists. By existential introduction, some inconsistent theory exists. So, the proposition "some inconsistent theory exists" exists. Thus, by existential introduction, some proposition exists. By conjunction introduction, some inconsistent theory exists and some proposition exists. Thus, the hypothesis of my proof is satisfiable. So, as my proof implies, trivialism is true. By definition of trivialism, all propositions are true. Since all propositions are true, the proposition "all propositions are true and false" is true. Therefore, by simplification, all propositions are true and false.
Mosesquine wrote: April 28th, 2018, 2:32 am Since truth value is either true or false, your proof is failed.
Since all propositions are true, the proposition "my proof is sound and unsound" is true. Therefore, by simplification, my proof is sound and unsound.


Negation symbol (~) is a truth functional connective. If 'p' is true, then '~p' (not-p) is, truth functionally false, and if 'p' is false, then '~p' (not-p) is, truth functionally true. This is an elementary discipline of logic. Your proof of inconsistent theories and trivialism is no more than rejecting truth functional connectives and the existence of false statements.
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Re: Inconsistent Theories Metatheoretically Prove Trivialism

Post by Atreyu »

Perhaps someone here could define 'trivialism' for me?
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Re: Inconsistent Theories Metatheoretically Prove Trivialism

Post by Eduk »

You can define it yourself Atreyu.
Unknown means unknown.
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paulemok
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Re: Inconsistent Theories Metatheoretically Prove Trivialism

Post by paulemok »

Mosesquine wrote: April 29th, 2018, 3:34 am Your proof of inconsistent theories and trivialism is no more than rejecting truth functional connectives and the existence of false statements.
My proof implies truth functional connectives exist and do not exist. It implies false statements exist and do not exist.

I feel that one of the most questionable aspects of my proof is the existence of inconsistent theory. Naive set theory, for example, is inconsistent, by definition of inconsistent theory, since, as explained on page 432 of the textbook Language, Proof and Logic (1999, 2000, 2002, 2003, 2007, 2008) by Jon Barwise and John Etchemendy, some contradiction, Russell's Paradox, exists in it.

For the sake of a proof by contradiction, assume naive set theory is true. As claimed in the previous paragraph, some contradiction exists in naive set theory. So, since naive set theory is true, some contradiction exists. Discharge the assumption. Completing the proof by contradiction method, naive set theory is false. Nonetheless, it seems to be accepted that naive set theory exists. As the following proof shows, any such acceptance is wrong. For the sake of a proof by contradiction, assume naive set theory exists. Then, as stated in the previous paragraph, some contradiction exists in naive set theory. Thus, since
paulemok wrote: April 25th, 2018, 3:57 pm The proposition "some contradiction exists in a theory" is logically equivalent to the proposition "the theory is right and wrong."
naive set theory is right and wrong. By definition of wrong, naive set theory is right and not right. But that's a contradiction. Discharge the assumption. Therefore, by completing the proof by contradiction method, naive set theory does not exist. So, we, we as in everybody, are entitled to regard naive set theory as formally nonexistent. Nonetheless, there is strong evidence for its existence. As exhibited in this post and in chapter 15 of the aforementioned textbook by Barwise and Etchemendy (1999, 2000, 2002, 2003, 2007, 2008), we can morally consider naive set theory as existent. Thus, since we are justified either way, we can morally consider naive set theory as existent and nonexistent simultaneously. Therefore, through ex contradictione quodlibet, we can morally regard trivialism as true.

We may epistemically, with respect to all I know, go as far as to deny not only the truth of the proposition named the Axiom of Unrestricted Comprehension, but its very existence.
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Re: Inconsistent Theories Metatheoretically Prove Trivialism

Post by Mosesquine »

paulemok wrote: April 29th, 2018, 10:04 pm
Mosesquine wrote: April 29th, 2018, 3:34 am Your proof of inconsistent theories and trivialism is no more than rejecting truth functional connectives and the existence of false statements.
My proof implies truth functional connectives exist and do not exist. It implies false statements exist and do not exist.

I feel that one of the most questionable aspects of my proof is the existence of inconsistent theory. Naive set theory, for example, is inconsistent, by definition of inconsistent theory, since, as explained on page 432 of the textbook Language, Proof and Logic (1999, 2000, 2002, 2003, 2007, 2008) by Jon Barwise and John Etchemendy, some contradiction, Russell's Paradox, exists in it.

For the sake of a proof by contradiction, assume naive set theory is true. As claimed in the previous paragraph, some contradiction exists in naive set theory. So, since naive set theory is true, some contradiction exists. Discharge the assumption. Completing the proof by contradiction method, naive set theory is false. Nonetheless, it seems to be accepted that naive set theory exists. As the following proof shows, any such acceptance is wrong. For the sake of a proof by contradiction, assume naive set theory exists. Then, as stated in the previous paragraph, some contradiction exists in naive set theory. Thus, since
paulemok wrote: April 25th, 2018, 3:57 pm The proposition "some contradiction exists in a theory" is logically equivalent to the proposition "the theory is right and wrong."
naive set theory is right and wrong. By definition of wrong, naive set theory is right and not right. But that's a contradiction. Discharge the assumption. Therefore, by completing the proof by contradiction method, naive set theory does not exist. So, we, we as in everybody, are entitled to regard naive set theory as formally nonexistent. Nonetheless, there is strong evidence for its existence. As exhibited in this post and in chapter 15 of the aforementioned textbook by Barwise and Etchemendy (1999, 2000, 2002, 2003, 2007, 2008), we can morally consider naive set theory as existent. Thus, since we are justified either way, we can morally consider naive set theory as existent and nonexistent simultaneously. Therefore, through ex contradictione quodlibet, we can morally regard trivialism as true.

We may epistemically, with respect to all I know, go as far as to deny not only the truth of the proposition named the Axiom of Unrestricted Comprehension, but its very existence.

It follows that, according to your proof, inconsistent theories exist and do not exist. It further follows that your proof is right and is not right. You are stupid and are not stupid, and you are a creature with kidneys and are not a creature with kidneys. Huh???
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Re: Inconsistent Theories Metatheoretically Prove Trivialism

Post by Londoner »

paulemok wrote: April 28th, 2018, 9:55 pm
Example 3 describes an inconsistent theory that exists.

Example 3. B is a theory that has the following two postulates.

Postulate 1. 1 equals 1.
Postulate 2. It is not true that "1 equals 1."

Since Postulate 2 is the negation of Postulate 1, some contradiction exists in B. So, by definition of inconsistent theory, B is inconsistent. Despite the fact that B is inconsistent, B exists. This concludes the example.
I do not agree 'B exists'. I do not see any 'theory' or understand why you use that word. A theory has to be a theory about something but you do not give a theory, you simply list two propositions, or rather a single proposition with alternative truth values. That is not a contradiction; it is the nature of propositions that they can be either true or false.

In order to create a contradiction then their truth or falsity cannot be just be the interchangeable values in the formal system of logic, they must be descriptive of something in the world. True or false in the sense of being a fact. We would understand 'what fact' according to what theory B was about. But you do not give the theory.
Since all propositions are true, the proposition "my proof is sound and unsound" is true. Therefore, by simplification, my proof is sound and unsound.
'True' as a description of propositions is simply a formal value. A proposition can be either true or false in the same way that a number can be positive or negative. It has nothing to do with soundness.

If "my proof is sound and unsound" was in a proposition within piece of logic we would replace it with a symbol, because such propositions only have the value T or F. They do not mean anything. Within logic, we do not need to make propositions stand for anything, any more than 1 + 1 + 2 requires us to know 'One what?'

So the paradox is created not because the propositions contradict each other, or because "my proof is sound and unsound" is self-contradictory, but because you are mixing the purely formal system of logic with factual/empirical notions of 'soundness' and 'theories'. To put it simply, In logic, we are free to simply assume the truth or falsity of any proposition. But if i say a theory is sound then I have to be able to explain why.
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Re: Inconsistent Theories Metatheoretically Prove Trivialism

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Mosesquine wrote: April 30th, 2018, 5:27 am Huh???
Yes, that’s correct.
Londoner wrote: April 30th, 2018, 7:46 am I do not agree 'B exists'. I do not see any 'theory' or understand why you use that word. A theory has to be a theory about something but you do not give a theory, you simply list two propositions, or rather a single proposition with alternative truth values.
Every proposition describes a theory. Every proposition expresses an idea, regardless of whether the idea is true or false. Furthermore, every nonempty set of propositions describes a theory. Every such set expresses a system of ideas that all together form a composite whole, regardless of whether the composite is true or false.

A theory does not have to be consistent. Every proof by contradiction constructs an inconsistent theory and thus shows that some inconsistent theory exists.

The two postulates of B together express a system of ideas, which, although false, is a part of a theory. If you have trouble accepting the existence of B, perhaps you would accept the existence of naive set theory or the inconsistent theory constructed in any proof by contradiction.

A theory is false does not imply it does not exist. A proposition is false does not imply it does not exist.
Londoner wrote: April 30th, 2018, 7:46 am We would understand 'what fact' according to what theory B was about. But you do not give the theory.
The two given postulates of B are arithmetical propositions about the real number 1.
Londoner wrote: April 30th, 2018, 7:46 am Within logic, we do not need to make propositions stand for anything, any more than 1 + 1 + 2 requires us to know 'One what?'
That may be correct, but in Example 3, I'm not taking that abstract of an approach. Example 3 was meant to be a basic, easy to understand example.
Londoner wrote: April 30th, 2018, 7:46 amSo the paradox is created not because the propositions contradict each other, or because "my proof is sound and unsound" is self-contradictory, but because you are mixing the purely formal system of logic with factual/empirical notions of 'soundness' and 'theories'.
I disagree. I’m not taking my logic and my arguments to the high level of abstraction that you are referring to.
Londoner wrote: April 30th, 2018, 7:46 am But if i say a theory is sound then I have to be able to explain why.
I have not said that B is sound. I have said that B exists.

Since B is inconsistent, the proof of my original post shows trivialism is true. Since trivialism is true, B is true and false. So, I have said B is sound, in the sense I have said B is true, and I have explained why.
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Re: Inconsistent Theories Metatheoretically Prove Trivialism

Post by Mosesquine »

paulemok wrote: May 1st, 2018, 3:14 am
Mosesquine wrote: April 30th, 2018, 5:27 am Huh???
Yes, that’s correct.

You think that your sentence "that's correct" is true and not true, according to your theory, huh???
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Re: Inconsistent Theories Metatheoretically Prove Trivialism

Post by Londoner »

paulemok wrote: May 1st, 2018, 3:14 am
Every proposition describes a theory. Every proposition expresses an idea, regardless of whether the idea is true or false. Furthermore, every nonempty set of propositions describes a theory. Every such set expresses a system of ideas that all together form a composite whole, regardless of whether the composite is true or false.
This is where we disagree. A proposition in logic does not express anything, just like a number on its own does not express anything. '2' does not express the idea 'there are two apples on the table'. 'P' does not assert 'Socrates is a man'. If these terms did stand for ideas, then we would no longer be free to simply assume truth values. Whether P was 'true' or not would be an empirical matter whereas T/F values in logic are purely formal.

Nor could we say what idea a proposition stood for. We express ideas and theories in ordinary language but ordinary language does not work in a logical way. If we try to give an ordinary language equivalent of a proposition we find it never has a simple binary true/false value.
The two given postulates of B are arithmetical propositions about the real number 1.
Then I would ask what you mean by 'the real number 1'. For example, if you meant 'the way that number is used in maths' then we would consult a mathematician, or look at books, and discover which one of those two contradictory propositions was correct.

I can write 'Mary is in the park' and 'Mary is at home' and there is no contradiction because they are just two sentences. But if they are presented as the answer to the question; 'Where is Mary? i.e. theories about the location of Mary, then both cannot be true. And because we have an understanding of what we mean when somebody asks 'Where?' we would know how to find out whether either sentence was true or not.
Me: But if i say a theory is sound then I have to be able to explain why.

I have not said that B is sound. I have said that B exists.
By 'sound' I would understand an argument that is not only logically valid but where the premises are actually true; not just given an assumed formal value 'true' but true in the sense of being a fact about the world.

If 'B' is not said to be 'sound', so not a claim about some fact, yet we also say 'B' does 'exist', then how can I understand what that claim 'exist' might mean? The word 'exist' must have some sense. There must be some reason why we assert 'B' 'exists' as opposed to 'does not exist'. If it is not some sort of factual claim, then what is it?

There are various ways we might say 'B exists'. For example, we might say 'as words on this web page'. That would be an empirical matter; we could all agree that 'B' existed in that sense, while not agreeing that those sentences made sense, or counted as what is normally understood to be a 'theory'. If instead we claim 'B exists as a theory' that would be a assertion about language; does 'B' correspond to the way people understand and use the word 'theory'? And so on.

As I write above, this sort of uncertainty about the meaning of 'exist' - a word that seems on first sight to be very simple - applies to all ordinary language. You can only treat logical propositions as simple binary T/F things because they are not in ordinary language. But I think you do not clearly distinguish between the two. You start by describing your sentences as 'propositions' but these become 'postulates' and 'theories' without it being made clear whether all those other words are meant to be synonyms for 'proposition' or claims of something beyond formal logic.
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Re: Inconsistent Theories Metatheoretically Prove Trivialism

Post by paulemok »

There are some additional things I might say, but some of the rules for this website are pushing me not to do so. I may have unfortunately chosen the wrong website for this discussion.
Mosesquine wrote: May 1st, 2018, 4:37 am You think that your sentence "that's correct" is true and not true, according to your theory, huh???
Yes, that's correct.
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