Re: Within Proof Theoretic Semantics Gödel's G h as no meaning in PA

On 03/07/2026 18:38, olcott wrote:
> On 7/3/2026 4:28 AM, Mikko wrote:
>> On 02/07/2026 17:49, olcott wrote:
>>> On 7/2/2026 1:55 AM, Mikko wrote:
>>>> On 01/07/2026 18:16, olcott wrote:
>>>>> On 7/1/2026 2:24 AM, Mikko wrote:
>>>>>> On 30/06/2026 16:58, olcott wrote:
>>>>>>> On 6/30/2026 3:18 AM, Mikko wrote:
>>>>>>>> On 29/06/2026 16:29, olcott wrote:
>>>>>>>>> On 6/29/2026 1:14 AM, Mikko wrote:
>>>>>>>>>> On 29/06/2026 05:52, olcott wrote:
>>>>>>>>>>> On 6/28/2026 3:39 AM, Mikko wrote:
>>>>>>>>>>>> On 27/06/2026 17:50, polcott wrote:
>>>>>>>>>>>>> On 6/27/2026 1:53 AM, Tristan Wibberley wrote:
>>>>>>>>>>>>>> On 20/06/2026 18:32, olcott wrote:
>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> A proof theoretic expression is known to be true when
>>>>>>>>>>>>>>> it is fully grounded in its atomic base. Only two
>>>>>>>>>>>>>>> PTS semantics researchers deal with true Dag Prawitz
>>>>>>>>>>>>>>> is the one that began this. PTS previously only dealt
>>>>>>>>>>>>>>> with semantic meaning and never got around to true(L,x).
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> That's surprising, disregard for axioms?
>>>>>>>>>>>>>
>>>>>>>>>>>>> If there is no sequence of inference steps in Q from
>>>>>>>>>>>>> ~∃x x=S(x) to the axioms of Q then ~∃x x=S(x) is
>>>>>>>>>>>>> ungrounded in the PTS atomic base of Q.
>>>>>>>>>>>>>
>>>>>>>>>>>>> This does not mean undecidable or incomplete
>>>>>>>>>>>>> it means that ~∃x x=S(x) is out-of-scope for Q.
>>>>>>>>>>>>
>>>>>>>>>>>> It comes close. If ∃x x=S(x) is likewise "ungrounded" but in 
>>>>>>>>>>>> the
>>>>>>>>>>>> language of Q then ~∃x x=S(x) and ∃x x=S(x) are both 
>>>>>>>>>>>> undecidable
>>>>>>>>>>>> and Q is incomplete, bcause that is what the words mean.
>>>>>>>>>>>
>>>>>>>>>>> Q also can't bake a birthday cake, this does not make
>>>>>>>>>>> Q in any way "incomplete" relative to what it was
>>>>>>>>>>> defined to do. Incomplete only counts relative to
>>>>>>>>>>> its intended purpose. A car without an engine is
>>>>>>>>>>> incomplete relative to a mode of transportation.
>>>>>>>>>>
>>>>>>>>>> Irrelevant. The definition of completeness 
>>>>>>>>>
>>>>>>>>> It a misnomer and does not literally mean (as it implies)
>>>>>>>>> that something is missing that could be added to make
>>>>>>>>> it complete.
>>>>>>>>
>>>>>>>> It does mean that something is missing that could be added to
>>>>>>>> enabe a proof of an unprovable sentence. 
>>>>>>>
>>>>>>> Base-Extension Semantics (B-eS) allows that.
>>>>>>> It never was incomplete. It always did what it was defined to do.
>>>>>>> When Q is extended to become PA it stops being Q and becomes PA.
>>>>>>
>>>>>> However, there are theories that reamain incomplete even when
>>>>>> more postolates are added, as long as there is a way to know
>>>>>> which sentences are included in the added postulates. Important
>>>>>> examples include Peano arithmetic and ZFC set theory.
>>>>>
>>>>> Base-Extension Semantics (B-eS) seems to be essentially a cheat.
>>>>> When we ask what is grounded in an atomic base of Q and we
>>>>> add axioms to Q to become PA we cheated in that we changed
>>>>> the original question rather than answered it.
>>>>
>>>> Yes, in a sense. But sometimes it is better to have a partial answer
>>>> rather than no answer at all. Of course Q with any additional
>>>> postulate is not Q but if the additional postulates are true about
>>>> natural numbers then the strengthened theory is still a theory of
>>>> natural numbers. PA is one such strengthened Q but still incomplete
>>>> and can be strengthened further.
>>>
>>> Is  (∀x, S(x) ≠ x) provable or refutable in Q?
>>> Yes if you cheat, no if you don't cheat.
>>
>> As I already pointed out in another message, which you apparently
>> missed, it is neither. 
> 
> Thus PTS would say that (∀x, S(x) ≠ x) is semantically
> undefined in Q.

You don't need any PTS to see that (∀x, S(x) ≠ x) is semantically
undefined in Q. That everything is semantically undefined in Q is
sufficient to determine that so is (∀x, S(x) ≠ x).

-- 
Mikko