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.

> 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.

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

--
Copyright 2023 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer