On 4/22/23 4:47 PM, olcott wrote:
> On 4/22/2023 3:43 PM, Richard Damon wrote:
>> On 4/22/23 4:39 PM, olcott wrote:
>>> On 4/22/2023 3:14 PM, Richard Damon wrote:
>>>> On 4/22/23 4:10 PM, olcott wrote:
>>>>> On 4/22/2023 3:06 PM, Richard Damon wrote:
>>>>>> On 4/22/23 4:02 PM, olcott wrote:
>>>>>>> On 4/22/2023 3:00 PM, Richard Damon wrote:
>>>>>>>> On 4/22/23 3:54 PM, olcott wrote:
>>>>>>>>> On 4/22/2023 2:44 PM, Richard Damon wrote:
>>>>>>>>>> On 4/22/23 3:34 PM, olcott wrote:
>>>>>>>>>>> On 4/22/2023 2:15 PM, Richard Damon wrote:
>>>>>>>>>>>> On 4/22/23 3:11 PM, olcott wrote:
>>>>>>>>>>>>> On 4/22/2023 1:01 PM, Richard Damon wrote:
>>>>>>>>>>>>>> On 4/22/23 1:13 PM, olcott wrote:
>>>>>>>>>>>>>>> On 4/22/2023 11:56 AM, Richard Damon wrote:
>>>>>>>>>>>>>>>> On 4/22/23 12:45 PM, olcott wrote:
>>>>>>>>>>>>>>>>> On 4/22/2023 11:36 AM, Richard Damon wrote:
>>>>>>>>>>>>>>>>>> On 4/22/23 12:27 PM, olcott wrote:
>>>>>>>>>>>>>>>>>>> On 4/22/2023 11:12 AM, Richard Damon wrote:
>>>>>>>>>>>>>>>>>>>> On 4/22/23 11:39 AM, olcott wrote:
>>>>>>>>>>>>>>>>>>>>> On 4/22/2023 9:57 AM, Richard Damon wrote:
>>>>>>>>>>>>>>>>>>>>>> On 4/22/23 10:48 AM, olcott wrote:
>>>>>>>>>>>>>>>>>>>>>>> On 4/22/2023 9:38 AM, Richard Damon wrote:
>>>>>>>>>>>>>>>>>>>>>>>> On 4/22/23 10:28 AM, olcott wrote:
>>>>>>>>>>>>>>>>>>>>>>>>> On 4/22/2023 6:17 AM, Richard Damon wrote:
>>>>>>>>>>>>>>>>>>>>>>>>>> On 4/21/23 11:40 PM, olcott wrote:
>>>>>>>>>>>>>>>>>>>>>>>>>>> On 4/21/2023 9:45 PM, Richard Damon wrote:
>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 4/21/23 9:41 PM, olcott wrote:
>>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 4/21/2023 7:49 PM, Richard Damon wrote:
>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 4/21/23 8:33 PM, olcott wrote:
>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> ∃G ∈ F (G ↔ (G ⊬ F))
>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> There exists a G such that G is logically
>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> equivalent to its own unprovability in F
>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> *If we assume that there is such a G in F
>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> that means that*
>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> G is true means there is no sequence of
>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> inference steps that satisfies G in F.
>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> G is false means there is a sequence of
>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> inference steps that satisfies G in F.
>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> *Thus the above G simply does not exist
>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> in F*
>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> So?
>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>>>>>>>> I finally learned enough model theory to
>>>>>>>>>>>>>>>>>>>>>>>>>>>>> correctly link provability to
>>>>>>>>>>>>>>>>>>>>>>>>>>>>> truth in the conventional model theory way.
>>>>>>>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>>>>>>> Doesn't seem so, you don't seem to
>>>>>>>>>>>>>>>>>>>>>>>>>>>> understand the difference. You seem to
>>>>>>>>>>>>>>>>>>>>>>>>>>>> confuse Truth with Knowledge.
>>>>>>>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>>>>>>>> I finally approximated {G asserts its own
>>>>>>>>>>>>>>>>>>>>>>>>>>>>> unprovability in F}
>>>>>>>>>>>>>>>>>>>>>>>>>>>>> using conventional math symbols in their
>>>>>>>>>>>>>>>>>>>>>>>>>>>>> conventional way.
>>>>>>>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>>>>>>> Except that isn't what G is, you only think
>>>>>>>>>>>>>>>>>>>>>>>>>>>> that because you can't actually understand
>>>>>>>>>>>>>>>>>>>>>>>>>>>> even the outline of Godel's proof, so you
>>>>>>>>>>>>>>>>>>>>>>>>>>>> take pieces out of context.
>>>>>>>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>>>>>>> G never asserts its own unprovability.
>>>>>>>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>>>>>>> The statement that we now have a statement
>>>>>>>>>>>>>>>>>>>>>>>>>>>> that asserts its own unprovablity, as a
>>>>>>>>>>>>>>>>>>>>>>>>>>>> simplification describing a statment DERIVED
>>>>>>>>>>>>>>>>>>>>>>>>>>>> from G, and that derivation happens in
>>>>>>>>>>>>>>>>>>>>>>>>>>>> Meta-F, and is about what can be proven in F.
>>>>>>>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> Since Godel's G isn't of that form, but
>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> only can be used to derive a statment IN
>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> META-F that says that G is not provable in
>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> F, your argument says nothing about
>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> Godel's G.
>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>>>>>>>> F ⊢ GF ↔ ¬ProvF (┌GF┐).
>>>>>>>>>>>>>>>>>>>>>>>>>>>>> https://plato.stanford.edu/entries/goedel-incompleteness/#FirIncTheCom
>>>>>>>>>>>>>>>>>>>>>>>>>>>>> I have finally created a G that is
>>>>>>>>>>>>>>>>>>>>>>>>>>>>> equivalent to
>>>>>>>>>>>>>>>>>>>>>>>>>>>>> Panu Raatikainen's SEP article.
>>>>>>>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>>>>>>> So?
>>>>>>>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>>>>>>> Did you read that article?
>>>>>>>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> Also, you don't understand what those
>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> terms mean, because G being true doesn't
>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> mean there is no sequence of inference
>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> steps that satisfies G in F, but there is
>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> no FINITE sequence of inference steps that
>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> satisfies G in F.
>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>>>>>>>> ∃G ∈ F (G ↔ (G ⊬ F))
>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>>>>>>>> Because we can see that every finite or
>>>>>>>>>>>>>>>>>>>>>>>>>>>>> infinite sequence in F that
>>>>>>>>>>>>>>>>>>>>>>>>>>>>> satisfies the RHS of ↔ contradicts the LHS
>>>>>>>>>>>>>>>>>>>>>>>>>>>>> a powerful F can infer that G
>>>>>>>>>>>>>>>>>>>>>>>>>>>>> is utterly unsatisfiable even for infinite
>>>>>>>>>>>>>>>>>>>>>>>>>>>>> sequences in this more
>>>>>>>>>>>>>>>>>>>>>>>>>>>>> powerful F.
>>>>>>>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>>>>>>> Nope. Show the PROOF.
>>>>>>>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>>>>>>> You don't know HOW to do a proof, you can
>>>>>>>>>>>>>>>>>>>>>>>>>>>> only do arguement.
>>>>>>>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>>>>>> ∃G ∈ F (G ↔ (G ⊬ F))
>>>>>>>>>>>>>>>>>>>>>>>>>>> There exists a G in F such that G is
>>>>>>>>>>>>>>>>>>>>>>>>>>> logically equivalent to its own unprovability
>>>>>>>>>>>>>>>>>>>>>>>>>>> in F
>>>>>>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>>>>>> A proof is any sequence of steps that shows
>>>>>>>>>>>>>>>>>>>>>>>>>>> that its conclusion is a
>>>>>>>>>>>>>>>>>>>>>>>>>>> necessary consequence of its premises.\
>>>>>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>>>>> Boy are you wrong.
>>>>>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>>>>> A proof is a FINITE sequence of steps that
>>>>>>>>>>>>>>>>>>>>>>>>>> shows that a given statement is a necessary
>>>>>>>>>>>>>>>>>>>>>>>>>> consequence of the defined system.
>>>>>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>>>>> "Proof" doesn't have a  "Premise", it has a
>>>>>>>>>>>>>>>>>>>>>>>>>> system.
>>>>>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>>>>> The statement may have conditions in it
>>>>>>>>>>>>>>>>>>>>>>>>>> restricting when
>>>>>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>>>>>> ∃G ∈ F (G ↔ (G ⊬ F))
>>>>>>>>>>>>>>>>>>>>>>>>>>> There exists a G in F such that G is
>>>>>>>>>>>>>>>>>>>>>>>>>>> logically equivalent to its own unprovability
>>>>>>>>>>>>>>>>>>>>>>>>>>> in F
>>>>>>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>>>>>> If G is true then there is no sequence of
>>>>>>>>>>>>>>>>>>>>>>>>>>> inference steps that satisfies G in F making
>>>>>>>>>>>>>>>>>>>>>>>>>>> G untrue.
>>>>>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>>>>> no FINITE sequence, making G UNPROVABLE, and
>>>>>>>>>>>>>>>>>>>>>>>>>> there IS an INFINITE sequence making it TRUE.
>>>>>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>>>>> This is possible.
>>>>>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>>>>>> If G is false then there is a sequence of
>>>>>>>>>>>>>>>>>>>>>>>>>>> inference steps that satisfies G in F making
>>>>>>>>>>>>>>>>>>>>>>>>>>> G true.
>>>>>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>>>>> If G is false, then there is a finite sequence
>>>>>>>>>>>>>>>>>>>>>>>>>> proving G, which forces G to be true, thus
>>>>>>>>>>>>>>>>>>>>>>>>>> this is a contradiction.
>>>>>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>>>>>> Because the RHS of ↔ contradicts the LHS
>>>>>>>>>>>>>>>>>>>>>>>>>>> there is no such G in F.
>>>>>>>>>>>>>>>>>>>>>>>>>>> Thus the above G simply does not exist in F.
>>>>>>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>>>>> Nope, because we can have an infinite sequence
>>>>>>>>>>>>>>>>>>>>>>>>>> that isn't finite, G can be True but not
>>>>>>>>>>>>>>>>>>>>>>>>>> Provable.
>>>>>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>>>> If G is false and ↔ is true this makes the RHS
>>>>>>>>>>>>>>>>>>>>>>>>> false which negates the RHS making it say (G ⊢
>>>>>>>>>>>>>>>>>>>>>>>>> F) which makes G true in F.
>>>>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>>> Right, G can't be false, but it can be True.
>>>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>> Thus ↔ cannot be satisfied thus no such G exists
>>>>>>>>>>>>>>>>>>>>>>> in F.
>>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>> Why do you say that?
>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>> I don't think you know what you terms mean.
>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>> There exists a G in F such that G is true if and
>>>>>>>>>>>>>>>>>>>>>> only if G is Unprovable.
>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>> Logical equality
>>>>>>>>>>>>>>>>>>>>> p q p ↔ q
>>>>>>>>>>>>>>>>>>>>> T T   T // G is true if and only if G is Unprovable.
>>>>>>>>>>>>>>>>>>>>> T F   F //
>>>>>>>>>>>>>>>>>>>>> F T   F //
>>>>>>>>>>>>>>>>>>>>> F F   T // G is false if and only if G is Provable.
>>>>>>>>>>>>>>>>>>>>> https://en.wikipedia.org/wiki/Truth_table#Logical_equality
>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>> Row(1) There exists a G in F such that G is true if
>>>>>>>>>>>>>>>>>>>>> and only if G is
>>>>>>>>>>>>>>>>>>>>> unprovable in F making G unsatisfied thus untrue in F.
>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>> Row(4) There exists a G in F such that G is false
>>>>>>>>>>>>>>>>>>>>> if and only if G is
>>>>>>>>>>>>>>>>>>>>> provable in F making G satisfied thus true in F.
>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>> If either Row(1) or Row(4) are unsatisfied then ↔
>>>>>>>>>>>>>>>>>>>>> is false.
>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>> But if neither row values can ACTUALLY EXIST, then
>>>>>>>>>>>>>>>>>>>> the equality is true.
>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>> If either Row(1) or Row(4) cannot have the same value
>>>>>>>>>>>>>>>>>>> for p and q
>>>>>>>>>>>>>>>>>>> (for whatever reason) then ↔ is unsatisfied and no
>>>>>>>>>>>>>>>>>>> such G exists in F.
>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>> So, you don't understand how truth tables work.
>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>> You don't need to have all the rows with true being
>>>>>>>>>>>>>>>>>> possible, you need all the rows that are possible to
>>>>>>>>>>>>>>>>>> be True.
>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>> To the best of my knowledge
>>>>>>>>>>>>>>>>> ↔ is also known as logical equivalence meaning that the
>>>>>>>>>>>>>>>>> LHS and the RHS
>>>>>>>>>>>>>>>>> must always have the same truth value or ↔ is not true.
>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>> Right, and for that statement, the actual G found in F,
>>>>>>>>>>>>>>>> the ONLY values that happen is G is ALWAYS true, an
>>>>>>>>>>>>>>>> Unprovable is always true.
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>> Thus the equivalence is always true.
>>>>>>>>>>>>>>> I don't think that is the way that it works.
>>>>>>>>>>>>>>> We must assume that the RHS is true and see how that
>>>>>>>>>>>>>>> effects the LHS
>>>>>>>>>>>>>>> We must assume that the RHS is false and see how that
>>>>>>>>>>>>>>> effects the LHS
>>>>>>>>>>>>>>> ((True(RHS) → True(LHS)) ∧ (False(RHS) → False(LHS))) ≡
>>>>>>>>>>>>>>> (RHS ↔ LHS)
>>>>>>>>>>>>>>> False(RHS) → True(LHS) refutes (RHS ↔ LHS)
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> Nope, that isn't how it works.
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> Can you show me something that says that is how it works?
>>>>>>>>>>>>>
>>>>>>>>>>>>> p ↔ q would seem to mean ((p → q) ∧ (q → p))
>>>>>>>>>>>>> Here is a much clearer and conventional way of showing that
>>>>>>>>>>>>>
>>>>>>>>>>>>> Logical implication derives logical equivalence
>>>>>>>>>>>>> p---q---(p ⇒ q)---(q ⇒ p)---(q ↔ p)
>>>>>>>>>>>>> T---T------T----------T---------T
>>>>>>>>>>>>> T---F------F----------T---------F
>>>>>>>>>>>>> F---T------T----------F---------F
>>>>>>>>>>>>> F---F------T----------T---------T
>>>>>>>>>>>>>
>>>>>>>>>>>>>
>>>>>>>>>>>>
>>>>>>>>>>>> So, why does the fact that the last line is never used in
>>>>>>>>>>>> this case cause a problem.
>>>>>>>>>>>>
>>>>>>>>>>>
>>>>>>>>>>> ∃G ∈ F (G ↔ (G ⊬ F))
>>>>>>>>>>>
>>>>>>>>>>> I am just saying that according to the conventional rules of
>>>>>>>>>>> logic the
>>>>>>>>>>> above expression is simply false. There is no G that is
>>>>>>>>>>> logically
>>>>>>>>>>> equivalent to its own unprovability in F.
>>>>>>>>>>>
>>>>>>>>>>
>>>>>>>>>> But Godel's G satisfies that.
>>>>>>>>>>
>>>>>>>>>> Remember, G is the statement that there does not exist a
>>>>>>>>>> number g such that g statisifes a particular Primative
>>>>>>>>>> Recursive Relationship (built in Meta-F, but using only
>>>>>>>>>> operations defined in F).
>>>>>>>>>>
>>>>>>>>> There is no such G in F says the same thing, yet does not
>>>>>>>>> falsely place
>>>>>>>>> the blame on F.
>>>>>>>>>
>>>>>>>>
>>>>>>>> Yes, but can you PROVE your statement? If not, you are just
>>>>>>>> making unsubstantiated false claims, just like DT.
>>>>>>>>
>>>>>>>
>>>>>>> I just proved it. The only gap in the proof was your lack of
>>>>>>> understanding (an honest mistake not a lie) about how ↔ works.
>>>>>>>
>>>>>>>
>>>>>>
>>>>>> Nope, how did you prove that no such G exists? You claims that row
>>>>>> 4 can't be satisfied? it doesn't need to ever be used.
>>>>>
>>>>> Try and prove that with a source, in the mean time I will tentatively
>>>>> assume that you are wrong. I proved that I am correct with the above
>>>>> truth table yet this assumes: p ↔ q means ((p → q) ∧ (q → p))
>>>>>
>>>>
>>>>
>>>> WRONG, YOU are making the claim, so YOU need to prove it.
>>>>
>>> I may have been mistaken when I thought that more than one row of the
>>> truth table needed to be satisfied. Furthermore in retrospect this looks
>>> like a dumb mistake that I did not notice as a dumb mistake until I
>>> looked at the truth table for ∧. So we are back to row one.
>>>
>>> ∃G ∈ F (G ↔ (F ⊬ G))
>>> If the RHS is satisfied then this means that there are no inference
>>> steps in F that derive G, thus G cannot be shown to be true in F.
>>>
>>>
>>
>> Nope, there is no FINITE series of infernece steps in F that derive G.
>>
>
> This G cannot be shown to be true in F.

It can't be PROVEN in F, but it can be PROVEN to be true in F with a
proof in Meta-F

You just don't seem to understand how these Meta-systems work.

>
>> There can be an INFINITE series of inference steps in F that derive G,
>> making it True but unprovable.
>>
>
> You already said that there cannot be infinite inference steps in F.

Only for PROOFS. You don't seem to know the difference between Proof,
Knowledge, and Truth.

>
>> You are just continuing to show that you don't understand what "Proof"
>> means.
>
>