Re: Undecidability based on epistemological antinomies V2 --Mendelson--

On 4/26/24 1:15 PM, olcott wrote:

On 4/26/2024 11:38 AM, Ross Finlayson wrote:

On 04/26/2024 08:28 AM, olcott wrote:

On 4/26/2024 3:42 AM, Mikko wrote:

On 2024-04-25 14:27:23 +0000, olcott said:
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On 4/25/2024 3:26 AM, Mikko wrote:

epistemological antinomy

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It <is> part of the current (thus incorrect) definition
of undecidability because expressions of language that
are neither true nor false (epistemological antinomies)
do prove undecidability even though these expressions
are not truth bearers thus not propositions.

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That a definition is current does not mean that is incorrect.
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...14 Every epistemological antinomy can likewise be used for a similar
undecidability proof...(Gödel 1931:43-44)
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An epistemological antinomy can only be an undecidable sentence
if it can be a sentence. What epistemological antinomies you
can find that can be expressed in, say, first order goup theory
or first order arithmetic or first order set tehory?
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It only matters that they can be expressed in some formal system.
If they cannot be expressed in any formal system then Gödel is
wrong for a different reason.
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Minimal Type Theory (YACC BNF)
https://www.researchgate.net/publication/331859461_Minimal_Type_Theory_YACC_BNF
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I created MTT so that self-reference could be correctly represented
it is conventional to represent self-reference incorrectly. MTT uses
adapted FOL to express arbitrary orders of logic. When MTT expressions
are translated into directed graphs a cycle in the graph proves that
the expression is erroneous.
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Here is the Liar Paradox in MTT: LP := ~True(LP)
00 root (1)
01 ~    (2)
02 True (0) // cycle
Same as ~True(~True(~True(~True(...))))
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In Prolog
?- LP = not(true(LP)).
LP = not(true(LP)).
?- unify_with_occurs_check(LP, not(true(LP))).
false.
Indicates  ~True(~True(~True(~True(...))))
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In mathematical logic, a sentence (or closed formula)[1] of a predicate
logic is a Boolean-valued well-formed formula with no free variables. A
sentence can be viewed as expressing a proposition, something that must
be true or false.
https://en.wikipedia.org/wiki/Sentence_(mathematical_logic)
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By definition epistemological antinomies cannot be true or false thus
cannot be logic sentences therefore Gödel is wrong.
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Actually what results is that Goedel refers to a particular kind
of enforced, opinionated, retro-Russell ordinarity, that sees it
so that "logical paradox" of quantifier ambiguity or quantifier
impredicativity, resulting one of these one-way opinions, stipulations,
assumptions, non-logical axioms of restriction of comprehension,
makes it sort of like so for Goedel as "completeness, you know,
yet, incompleteness, ...".
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 ...14 Every epistemological antinomy can likewise be used for a similar undecidability proof...(Gödel 1931:43-44)
 epistemological antinomies cannot be true or false thus cannot
be propositions that must be true or false.
 

Right, and Godel doesn't claim it is.
Your problem is you just don't understand what he is saying here, because you don't understand the meaning of the phase "can likewise be used for a similar undecidability proof"
The Epistemological Antinomy is NOT used as the final proposition that needs to be proven, but provides the structural form, to construct ANOTHER syntactically similar, but semantically different statement, that is then used to build the Primitive Recursive Relationship based on it, that forms the statement G.
Your skipping those steps just makes your arguement incorrect.

An undecidable sentence of a theory K is a closed wf ℬ of K such that
neither ℬ nor ¬ℬ is a theorem of K, that is, such that not-⊢K ℬ and
not-⊢K ¬ℬ. (Mendelson: 2015:208)
 Undecidable(K, ℬ) ≡ ∃ℬ ∈ K ((K ⊬ ℬ) ∧ (K ⊬ ¬ℬ))
 To hazard a guess about what you mean, or to precisely state exactly
what I mean there is no such ℬ in K because such a ℬ in K could not
be a proposition of K.
 

Right, and since Godel's G isn't the Epistemological Antinomy you think he is using, your argument just fails. G is, in fact, a statement that MUST have a truth value, as it is about the lack of existance of a finite number that matches a property, and such a number MUST either exist or not. If it exists, that existance proves G wrong via a finite sequence of steps to evaluate that property, and if the number doesn't exist, it is proven by the infinite number of steps of testing every one of the countably infinte numbers and testing them with a finite number of steps, and seeing that none of the satisfy the relationship.
But, if the number doesn't exist, that method can NOT be used as a "Proof", as a proof must use a finite number of steps, and you can't individually check an infinite number of values in a finite proof. (You would need to find a induction or recussion method to reduce the steps to something finite, which might not exist).
The fact that in the META theory, we can show that no such number can exist, using facts not present in the theory, so that proof doesn't move down to there, so we can demonstrate that G must be true, and by knowledge also in the meta-theory, we can show that no proof can exist in the theory to show it.
Of course, since the concept of what a meta-theory seems to be foreign to you, as even what a "Formal System" is, this is beyond your understanding.