Inconsistent Theories Metatheoretically Prove Trivialism

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Eduk
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Re: Inconsistent Theories Metatheoretically Prove Trivialism

Post by Eduk »

Paulemok, in answer to mosesquine you mean that's incorrect?
Unknown means unknown.
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Mosesquine
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Re: Inconsistent Theories Metatheoretically Prove Trivialism

Post by Mosesquine »

paulemok wrote: May 1st, 2018, 8:08 pm 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.

You are made up of molecules, and you are not made up of molecules, Huh???
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paulemok
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Re: Inconsistent Theories Metatheoretically Prove Trivialism

Post by paulemok »

Eduk wrote: May 2nd, 2018, 3:18 am Paulemok, in answer to mosesquine you mean that's incorrect?
No, I mean that’s correct.
Mosesquine wrote: May 2nd, 2018, 6:34 am You are made up of molecules, and you are not made up of molecules, Huh???
Yes, that’s correct.
Eduk
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Re: Inconsistent Theories Metatheoretically Prove Trivialism

Post by Eduk »

Sorry paulemok I keep on having to correct you? Surely you mean no that is incorrect when replying to Mosesquine. You said so yourself?
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Mosesquine
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Re: Inconsistent Theories Metatheoretically Prove Trivialism

Post by Mosesquine »

paulemok wrote: May 2nd, 2018, 11:12 am
Eduk wrote: May 2nd, 2018, 3:18 am Paulemok, in answer to mosesquine you mean that's incorrect?
No, I mean that’s correct.
Mosesquine wrote: May 2nd, 2018, 6:34 am You are made up of molecules, and you are not made up of molecules, Huh???
Yes, that’s correct.

Your stupidity is inborn, and your stupidity is not inborn, Huh???
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Whyme
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Re: Inconsistent Theories Metatheoretically Prove Trivialism

Post by Whyme »

I stumbled across Trivialism accidentally, by merely considering the referent of a thought to be it's causes.

If a thought is interpreted as referring to its causes, then it must necessarily be correct, and the concept of truth is vacuous.

On the other hand, if a thought is interpreted as representing something other than its causes, then the concept of truth arises, but is merely a matter of translation.
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ReasonMadeFlesh
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Re: Inconsistent Theories Metatheoretically Prove Trivialism

Post by ReasonMadeFlesh »

Trivialism is retarded.
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paulemok
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Re: Inconsistent Theories Metatheoretically Prove Trivialism

Post by paulemok »

Trivialism itself is an inconsistent theory. Therefore, the mere existence of trivialism as a theory implies trivialism. So, trivialism is true.
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Mosesquine
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Re: Inconsistent Theories Metatheoretically Prove Trivialism

Post by Mosesquine »

paulemok wrote: July 31st, 2018, 6:57 am Trivialism itself is an inconsistent theory. Therefore, the mere existence of trivialism as a theory implies trivialism. So, trivialism is true.

Your claim is equivalent, in the sense of analogy, to the following:
Transparent dragon itself is a such-and-such-and-such dragon. Therefore, the mere existence of transparent dragon as a dragon implies transparent dragon. So, transparent dragon is there.
So, your argument for trivialism fails, since mere defining does not make things into existence.
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paulemok
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Re: Inconsistent Theories Metatheoretically Prove Trivialism

Post by paulemok »

Mosesquine:

My claim
paulemok wrote: July 31st, 2018, 6:57 am Trivialism itself is an inconsistent theory.
is not a formal definition. It is a description, but is not a formal definition. My quoted claim may not be any definition of trivialism, since there may be, for all I know, an inconsistent theory that is not trivialism.

Trivialism does exist. That's how trivialism can be talked about. Multiple paper publications discuss trivialism. Two such publications are On the Plenitude of Truth by Paul Kabay, copyright 2010, and Doubt Truth to be a Liar by Graham Priest, copyright 2006. Multiple paper publications thus indicate that in some sense, trivialism exists.
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Mosesquine
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Re: Inconsistent Theories Metatheoretically Prove Trivialism

Post by Mosesquine »

paulemok wrote: August 5th, 2018, 8:59 am Mosesquine:

My claim
paulemok wrote: July 31st, 2018, 6:57 am Trivialism itself is an inconsistent theory.
is not a formal definition. It is a description, but is not a formal definition. My quoted claim may not be any definition of trivialism, since there may be, for all I know, an inconsistent theory that is not trivialism.

Trivialism does exist. That's how trivialism can be talked about. Multiple paper publications discuss trivialism. Two such publications are On the Plenitude of Truth by Paul Kabay, copyright 2010, and Doubt Truth to be a Liar by Graham Priest, copyright 2006. Multiple paper publications thus indicate that in some sense, trivialism exists.

The description that trivialism is such and such and such is not the same as the evidence that trivialism does exist. Such an attempt is merely begging the question. Your proof that trivialism exists is still not successful.
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paulemok
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Re: Inconsistent Theories Metatheoretically Prove Trivialism

Post by paulemok »

Mosesquine has suggested that the existence of a thing is not necessary for it to have a description. I disagree. Anything that has a description does, in some sense, exist. It is the very existence of a thing that is described by a description of the thing.

Anything that can be mentally conceived is logically possible. The rationality of the square root of 2, for example, can be mentally conceived, even though it doesn't exist in a conventional sense. It is known what it means for the square root of 2 to be rational. Since the rationality of the square root of 2 can be mentally conceived, it exists in some sense.

A single description of a thing is sufficient evidence that the thing exists.

If trivialism can be described, then it exists. So by contraposition, if trivialism does not exist, then it can not be described. That is, trivialism does not exist only if it can not be described.

Describing trivialism is not necessarily begging the question. In any valid argument, the truth of any conclusion is already implicitly or explicitly expressed by a truth of the premise or premises. My argument does not have as an explicit premise "trivialism exists." That trivialism exists is implicitly expressed by my premise
paulemok wrote: July 31st, 2018, 6:57 am Trivialism itself is an inconsistent theory.
That implicit expression is not logically fallacious, and it is not begging the question.

The following argument displays that trivialism exists.

Argument. By the law of excluded middle, trivialism exists or it is not true that trivialism exists.

Assume trivialism exists. Then the proposition "trivialism exists" is a description of trivialism. So, trivialism can be described. Since trivialism can be described and if trivialism can be described, then it exists, it follows by modus ponens that trivialism exists. Discharge the assumption.

Assume it is not true that trivialism exists. Then the proposition "it is not true that trivialism exists" is itself a description of trivialism. So, trivialism can be described. Since trivialism can be described and if trivialism can be described, then it exists, it follows by modus ponens that trivialism exists. Discharge the assumption.

Therefore, by disjunction elimination, trivialism exists. The argument has ended.
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Mosesquine
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Re: Inconsistent Theories Metatheoretically Prove Trivialism

Post by Mosesquine »

paulemok wrote: August 6th, 2018, 5:55 am Therefore, by disjunction elimination, trivialism exists. The argument has ended.

Disjunction elimination is related to formal structures. Of course, it's not that formal considerations are irrelevant to factual aspects. For example:

If there is a building in Seoul, then the head of the nickname user 'paulemok' is not made up of molecules.
There is a building in Seoul
Therefore, the head of the nickname user 'paulemok' is not made up of molecules.

and proof:

1. (∃x)Fx → ~(∃x)((∀y)(Gy ↔ x = y) & Hx)
2. (∃x)Fx
∴ ~(∃x)((∀y)(Gy ↔ x = y) & Hx)
3. asm: (∃x)((∀y)(Gy ↔ x = y) & Hx)
4. ~(∃x)Fx 1, 3, MT
5. (∃x)Fx 2, R
∴ 6. ~(∃x)((∀y)(Gy ↔ x = y) & Hx) from 3; 4 contradicts 5.
Q.E.D.

Therefore, by modus tollens, the head of the nickname user 'paulemok' is not made up of molecules. The argument has ended.

...AND it's the bonus:

Trivialism is wrong.
Therefore, trivialism is wrong.

proof:

1. ~(∃x)(Fx & ~(∃y)(Fy & x ≠ y) & Gx)
∴ ~(∃x)(Fx & ~(∃y)(Fy & x ≠ y) & Gx)
2. asm: (∃x)(Fx & ~(∃y)(Fy & x ≠ y) & Gx)
3. ~(∃x)(Fx & ~(∃y)(Fy & x ≠ y) & Gx) 1, R
∴ 4. ~(∃x)(Fx & ~(∃y)(Fy & x ≠ y) & Gx) from 2; 2 contradicts 3.
Q.E.D.

Therefore, by formal structures, trivialism is wrong. The argument has ended.
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paulemok
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Re: Inconsistent Theories Metatheoretically Prove Trivialism

Post by paulemok »

Mosesquine wrote: August 7th, 2018, 4:46 am Disjunction elimination is related to formal structures. Of course, it's not that formal considerations are irrelevant to factual aspects.
What is meant by the preceding quoted material is unclear.
Mosesquine wrote: August 7th, 2018, 4:46 am If there is a building in Seoul, then the head of the nickname user 'paulemok' is not made up of molecules.
There is a building in Seoul
Therefore, the head of the nickname user 'paulemok' is not made up of molecules.
Mosesquine's argument is unsound because the premise
Mosesquine wrote: August 7th, 2018, 4:46 am If there is a building in Seoul, then the head of the nickname user 'paulemok' is not made up of molecules.
is false. There is a building in Seoul, but my head is made up of molecules.
Mosesquine wrote: August 7th, 2018, 4:46 am Trivialism is wrong.
Therefore, trivialism is wrong.
That argument is valid because the conclusion follows from the premise by reiteration. However, it's not clear what the premise of Mosesquine's argument means. For all I know, it could mean trivialism is not true, trivialism is unmoral, or something else.
Mosesquine wrote: August 7th, 2018, 4:46 am proof:

1. ~(∃x)(Fx & ~(∃y)(Fy & x ≠ y) & Gx)
∴ ~(∃x)(Fx & ~(∃y)(Fy & x ≠ y) & Gx)
2. asm: (∃x)(Fx & ~(∃y)(Fy & x ≠ y) & Gx)
3. ~(∃x)(Fx & ~(∃y)(Fy & x ≠ y) & Gx) 1, R
∴ 4. ~(∃x)(Fx & ~(∃y)(Fy & x ≠ y) & Gx) from 2; 2 contradicts 3.
Q.E.D.
The propositional functions for that proof have been neither defined nor sufficiently described. I'm unsure what they mean. I don't understand how the premise that trivialism is wrong is represented in Mosesquine's proof.
Mosesquine wrote: August 7th, 2018, 4:46 am Therefore, by formal structures, trivialism is wrong.
It's not clear what Mosesquine means by "formal structures," and it's not clear how the conclusion that trivialism is wrong follows by it.
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Mosesquine
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Re: Inconsistent Theories Metatheoretically Prove Trivialism

Post by Mosesquine »

paulemok wrote: August 16th, 2018, 1:52 pm
Mosesquine wrote: August 7th, 2018, 4:46 am Disjunction elimination is related to formal structures. Of course, it's not that formal considerations are irrelevant to factual aspects.
What is meant by the preceding quoted material is unclear.
Mosesquine wrote: August 7th, 2018, 4:46 am If there is a building in Seoul, then the head of the nickname user 'paulemok' is not made up of molecules.
There is a building in Seoul
Therefore, the head of the nickname user 'paulemok' is not made up of molecules.
Mosesquine's argument is unsound because the premise
Mosesquine wrote: August 7th, 2018, 4:46 am If there is a building in Seoul, then the head of the nickname user 'paulemok' is not made up of molecules.
is false. There is a building in Seoul, but my head is made up of molecules.
Mosesquine wrote: August 7th, 2018, 4:46 am Trivialism is wrong.
Therefore, trivialism is wrong.
That argument is valid because the conclusion follows from the premise by reiteration. However, it's not clear what the premise of Mosesquine's argument means. For all I know, it could mean trivialism is not true, trivialism is unmoral, or something else.
Mosesquine wrote: August 7th, 2018, 4:46 am proof:

1. ~(∃x)(Fx & ~(∃y)(Fy & x ≠ y) & Gx)
∴ ~(∃x)(Fx & ~(∃y)(Fy & x ≠ y) & Gx)
2. asm: (∃x)(Fx & ~(∃y)(Fy & x ≠ y) & Gx)
3. ~(∃x)(Fx & ~(∃y)(Fy & x ≠ y) & Gx) 1, R
∴ 4. ~(∃x)(Fx & ~(∃y)(Fy & x ≠ y) & Gx) from 2; 2 contradicts 3.
Q.E.D.
The propositional functions for that proof have been neither defined nor sufficiently described. I'm unsure what they mean. I don't understand how the premise that trivialism is wrong is represented in Mosesquine's proof.
Mosesquine wrote: August 7th, 2018, 4:46 am Therefore, by formal structures, trivialism is wrong.
It's not clear what Mosesquine means by "formal structures," and it's not clear how the conclusion that trivialism is wrong follows by it.

You first appealed to the concept 'formality'. Your argument for trivialism depends on formal reasoning. This follows that the success/failure of your argument depends on formal validity/inference rules, etc.
This means that the success/failure of the arguments against trivialism also depends on formal structures, inference rules, etc. The argument suggested by me above is the argument proved by inference rules, logical principles, formal structures, etc. So, all arguments for/against things are successful/failed by formal structures.
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