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From: Jesse F. Hughes on 23 Mar 2010 08:58
Aatu Koskensilta <aatu.koskensilta(a)uta.fi> writes:

> "Jesse F. Hughes" <jesse(a)phiwumbda.org> writes:
>
>> Thus. the sentence
>>
>> ~(E a,b)( a^2/b^2 = 2 and gcd(a,b) = 1 )
>>
>> is meaningless, right?
>
> Newberry said that (x)(Px --> Qx) is meaningless if ~(Ex)Px is
> necessarily true. How do you get from this the meaninglessness of
>
> ~(E a,b)( a^2/b^2 = 2 and gcd(a,b) = 1 )
>
> which is not of the form (x)(Px --> Qx)?

He also says that

~(Ex)(Px & Qx)

is meaningless in exactly the same situations that

(Ax)(Px -> Qx)

is meaningless. He wants the two formulas to remain equivalent.

(Surely, you're not taking issue with the fact that I've used two
existential statements rather than one?)

As usual, Newberry can correct me if I'm mistaken on his claims.

--
"We are happy that you agree that customers need to know that Open
Source is legal and stable, and we heartily agree with that sentence
of your letter. The others don't seem to make as much sense, but we
find the dialogue refreshing." -- Linus Torvalds to Darl McBride
From: Aatu Koskensilta on 23 Mar 2010 09:07
"Jesse F. Hughes" <jesse(a)phiwumbda.org> writes:

> He also says that
>
> ~(Ex)(Px & Qx)
>
> is meaningless in exactly the same situations that
>
> (Ax)(Px -> Qx)
>
> is meaningless.

I see your knowledge in this field far surpasses mine!

--
Aatu Koskensilta (aatu.koskensilta(a)uta.fi)

"Wovon man nicht sprechan kann, dar�ber muss man schweigen"
- Ludwig Wittgenstein, Tractatus Logico-Philosophicus
From: Daryl McCullough on 23 Mar 2010 09:08
Aatu Koskensilta says...
>
>"Jesse F. Hughes" <jesse(a)phiwumbda.org> writes:
>
>> Thus. the sentence
>>
>> ~(E a,b)( a^2/b^2 = 2 and gcd(a,b) = 1 )
>>
>> is meaningless, right?
>
>Newberry said that (x)(Px --> Qx) is meaningless if ~(Ex)Px is
>necessarily true. How do you get from this the meaninglessness of
>
> ~(E a,b)( a^2/b^2 = 2 and gcd(a,b) = 1 )
>
>which is not of the form (x)(Px --> Qx)?

Well, Jesse's statement can be rewritten in the form

(forall a,b)( a^2/b^2 = 2 --> gcd(a,b) > 1 )

--
Daryl McCullough
Ithaca, NY

From: Aatu Koskensilta on 23 Mar 2010 09:10
stevendaryl3016(a)yahoo.com (Daryl McCullough) writes:

> Well, Jesse's statement can be rewritten in the form
>
> (forall a,b)( a^2/b^2 = 2 --> gcd(a,b) > 1 )

But does classical equivalence preserve meaninglessness?

--
Aatu Koskensilta (aatu.koskensilta(a)uta.fi)

"Wovon man nicht sprechan kann, dar�ber muss man schweigen"
- Ludwig Wittgenstein, Tractatus Logico-Philosophicus
From: Daryl McCullough on 23 Mar 2010 09:16
Aatu Koskensilta says...
>
>stevendaryl3016(a)yahoo.com (Daryl McCullough) writes:
>
>> Well, Jesse's statement can be rewritten in the form
>>
>> (forall a,b)( a^2/b^2 = 2 --> gcd(a,b) > 1 )
>
>But does classical equivalence preserve meaninglessness?

To be honest, I'm not an expert on meaninglessness.

--
Daryl McCullough
Ithaca, NY

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