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From: Jan Burse on 29 Jun 2010 18:36
Jan Burse schrieb:
Oops, should read:

> Herbrand Theorem for a certain class of Formulas A:
>
> |- forall x exists y A(x,y)
> ==> there are terms t1,..,tn such that
> |- forall x (A(x,t1) v .. v A(x,tn))
From: Charlie-Boo on 29 Jun 2010 21:22
On Jun 29, 6:36 pm, Jan Burse <janbu...(a)fastmail.fm> wrote:
> Jan Burse schrieb:
> Oops, should read:
>
>
>
> > Herbrand Theorem for a certain class of Formulas A:
>
> > |- forall x exists y A(x,y)
> > ==> there are terms t1,..,tn such that
> > |- forall x (A(x,t1) v .. v A(x,tn))-

If A(x,y) is x<y then I would think that the top line is true (do you
think so?) but the bottom line certainly is not. Does this help solve
either problem that I posed? I deal with only a 1-place relation and
you deal with a 2-place relation.

C-B

Hide quoted text -
>
> - Show quoted text -

From: Charlie-Boo on 29 Jun 2010 21:24
On Jun 29, 5:25 pm, MoeBlee <jazzm...(a)hotmail.com> wrote:
> On Jun 28, 7:04 pm, Charlie-Boo <shymath...(a)gmail.com> wrote:
>
>
>
>
>
> > On Jun 28, 12:44 pm, MoeBlee <jazzm...(a)hotmail.com> wrote:
>
> > > On Jun 26, 9:19 pm, Charlie-Boo <shymath...(a)gmail.com> wrote:
>
> > > > It would be cool if the following 3 things were equivalent:
>
> > > > 1. |- (allX)P(X)
> > > > 2. (allX) |- P(X)
> > > > 3. ~ |- (existsX)~P(X)
>
> > > I can help you here, if you're interested in understanding this.
>
> > > (1) is well formed.
>
> > > (2) is not well formed as you've given it. The reason is that you've
> > > mixed meta-language and object language in an incorrect way.
>
> > Did you read the definition of the syntax and semantics in the
> > previous post?
>
> Sorry, I made the mistake that your first post was intelligible
> standalone.
>
> > "P(x)" is a wff that is said to be
> > provable.  So it expresses the proposition that for all values of X,
> > the wff P(X) is provable.
>
> Then your formulations as given make even LESS sense.

What is wrong with the proposition that for all values of X the wff
P(X) is provable?

As a 3rd problem, what does it prove to prove that these are not all
equivalent?

C-B

> I'm out of time for you. I can't do what no one else in these threads
> has ever done: get you to understand ANYTHING.
>
> MoeBlee- Hide quoted text -
>
> - Show quoted text -

From: Jan Burse on 30 Jun 2010 17:49
Charlie-Boo wrote:
> On Jun 29, 6:36 pm, Jan Burse<janbu...(a)fastmail.fm> wrote:
>> Jan Burse schrieb:
>> Oops, should read:
>>
>>
>>
>>> Herbrand Theorem for a certain class of Formulas A:
>>
>>> |- forall x exists y A(x,y)
>>> ==> there are terms t1,..,tn such that
>>> |- forall x (A(x,t1) v .. v A(x,tn))-
>
> If A(x,y) is x<y then I would think that the top line is true (do you
> think so?) but the bottom line certainly is not. Does this help solve
> either problem that I posed? I deal with only a 1-place relation and
> you deal with a 2-place relation.

If A(x,y) is x<y, then forall x exists y A(x,y) does not
hold. The above theorem talks about tautologies and the
like.

If you have a preorder theory PO or something, then
maybe something like |- PO -> forall x exists y A(x,y)
might hold, but this is not the form where the
theorem applies.

B y e
From: Charlie-Boo on 30 Jun 2010 23:49
On Jun 29, 5:25 pm, MoeBlee <jazzm...(a)hotmail.com> wrote:
> On Jun 28, 7:04 pm, Charlie-Boo <shymath...(a)gmail.com> wrote:
>
>
>
>
>
> > On Jun 28, 12:44 pm, MoeBlee <jazzm...(a)hotmail.com> wrote:
>
> > > On Jun 26, 9:19 pm, Charlie-Boo <shymath...(a)gmail.com> wrote:
>
> > > > It would be cool if the following 3 things were equivalent:
>
> > > > 1. |- (allX)P(X)
> > > > 2. (allX) |- P(X)
> > > > 3. ~ |- (existsX)~P(X)
>
> > > I can help you here, if you're interested in understanding this.
>
> > > (1) is well formed.
>
> > > (2) is not well formed as you've given it. The reason is that you've
> > > mixed meta-language and object language in an incorrect way.
>
> > Did you read the definition of the syntax and semantics in the
> > previous post?
>
> Sorry, I made the mistake that your first post was intelligible
> standalone.

If you Google “made the mistake” and check a few dozen entries, you
will see 3 patterns following it:

1. End of the sentence.
2. of [verb]ing . . .
3. of [adj] [verb]ing . . .

verb: accuse, believe, contact, download, drink, drive, fall, get, go,
move
adj: binge, drunk

Examples:

“I made the mistake of falling in love.”
“I made the mistake of accusing him.”
“They made the mistake of binge drinking the same night.”

You say, “I made the mistake that your post was intelligible
standalone.” So after “made the mistake” you have “that your post was
intelligible standalone”.

You don’t use “of” but let us say that “that” means “of”. Then it
means, “I made the mistake of your post was intelligible standalone”
The “verb” is “your post was intelligible standalone”. But instead of
an active verb like the above, you have the verb to be. And it does
not apply to you, the subject of the sentence, it actually applies to
the post. So your mistake was the same mistake that the post made?

Are you trying to say something about something not being
intelligible?

What do they call a sentence that applies to itself?

Does it get any better than this?? LOL

C-B

> > "P(x)" is a wff that is said to be
> > provable.  So it expresses the proposition that for all values of X,
> > the wff P(X) is provable.
>
> Then your formulations as given make even LESS sense.
>
> I'm out of time for you. I can't do what no one else in these threads
> has ever done: get you to understand ANYTHING.
>
> MoeBlee- Hide quoted text -
>
> - Show quoted text -

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