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).

It turns out that it is absolutely true that no such number g exists
that satisfies it, so G is true. (This can be proven in Meta-F, and in a
way that shows that G is true in F)

It also turns out, that it is absolutely true that no proof in F can
exist for this fact, thus G ⊬ F is true. (And this is proven in Meta-F,
showing that there can not exist a proof IN F for G).

True is equivalent to True, so the statment is True.

The fact that you don't beleive it, is your own problem.

Until you can show what "Rule of Logic" that it defies, you are just
proving your self to be an ignorant liar, and totally unknowledgeable
about how logic actually works.