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Topic: Matheology § 224
Replies: 1   Last Post: Apr 15, 2013 7:59 AM

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Alan Smaill

Posts: 757
Registered: 1/29/05
Re: Matheology § 224
Posted: Apr 15, 2013 7:59 AM
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Nam Nguyen <namducnguyen@shaw.ca> writes:

> On 12/04/2013 8:23 AM, Alan Smaill wrote:
>> Nam Nguyen <namducnguyen@shaw.ca> writes:
>>

>>> On 12/04/2013 7:50 AM, Alan Smaill wrote:
>>>> Nam Nguyen <namducnguyen@shaw.ca> writes:
...
>>>>>
>>>>> I've used FOL ( _First Order Logic_ ) definitions that one should be
>>>>> familiar with.
>>>>>
>>>>> If anything, notation like "this" is defined in term of FOL
>>>>> terminologies.

>>>>
>>>> ie, you use a first-order meta-logic.

>>>
>>> OK. Why don't we just call it First Order Logic as everyone is supposed
>>> to understand already?

>>
>> Because you now have two logics --
>> good old statements like cGC are not statements about syntax,
>> they are statements about numbers. The language structures for
>> your theory of syntax is different from language structures for
>> natural numbers.

>
> I don't know what (or why) you're talking about _two_ logic. Over the
> years and in many threads what I've been presenting is about FOL
> definition of language structure. And there's only _one_ of such a thing
> referred to as _the_ concept FOL language structure.


But we also have FOL proofs *in* arithmetic to consider.
You at least recognise that there are two ways to come about
judgements about arithmetic statements -- via old-fashioned
proofs in PA, and by reasoning about language structures.

You may use FOL in both cases, but you are using different
axioms, and you will cause confusion at least by not making
the distinction.

>>>>> So I don't see all that historical context of "meta-logic" would
>>>>> have anything to do with the issue of, say, whether or not it's
>>>>> impossible to construct the naturals as a language model.

>>>>
>>>> The strength of the meta-logic does have a bearing.
>>>> Do you allow proof by induction over the syntax of formulas,
>>>> for example?

>>>
>>> So you're talking the strength (and weakness) of First Order Logic.

>>
>> Not only -- see above.
>> But do you allow proof by induction over the syntax of formulas?

>
> Your use of "proof" is too loose to make any sense: _exactly what_ is
> being proven by induction "over the syntax of formulas"? Iow, proof
> of _what_ "over the syntax of formulas"?


For example, suppose you want to show that for every FOL
formula, there is a logically equivalent formula, where all
the quantifiers appear at the start of the formula.
(This is a standard result about FOL.)

How could you show that this is in fact the case?

...
>>>> Suppose you have a language structure for the language of Peano
>>>> Arithmetic where all the axioms of PA are true, and suppose
>>>> that the underlying set is X, and "0" is the constant used
>>>> for the number zero.

>>>
>>> "Suppose" is hypothetical but sure, I could hypothesize such.

>>
>> How could it not be the case?

>
> Listen. I merely repeated (to emphasize) what you've stated "Suppose".


I am curious as to whether you think the opposite is also possible.
No need to commit yourself if you do not want to.

>>>> Is there a way of characterising the element of Z correponding to "0"?
>>>> Or is that a matter of opinion?
>>>>

>>> But, FOL definition of language structure already provides that, e.g.:
>>>
>>> ('0', {}).
>>>
>>> Right?

>>
>> If you're happy with that, fine.

>
> What an odd observation. It's not the issue of "happiness": it's the
> issue of my presentation follows FOL definition of language structure!


As I said, fine.

>> Now can you characterise the element of X that corresponds to "S(0)",
>> with "S" the symbol for successor?

>
> What does that have anything to do with my presenting the issue under
> debate, using FOL definition of language structure?


It will clarify which steps you permit in reasoning about language
structures.

Can you characterise the element of X that corresponds to "S(0)",
with "S" the symbol for successor?

--
Alan Smaill




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