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.

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