Re: What's Wrong With...
From: num num (num_at_num.num)
Date: 11/15/04
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Date: Mon, 15 Nov 2004 10:13:54 -0500
On Sun, 14 Nov 2004 23:34:00 -0800, William Elliot <marsh@privacy.net>
wrote:
>On Mon, 15 Nov 2004, num num wrote:
>
>> What's wrong with this proof?
>> Ax(Px -> Qx) |- ~Ex(Px & ~Qx)
>>
>> 1. Ax(Px -> Qx) Premise
>> 2. Pa -> Qa UI 1
>> 3. Ex(Px -> Qx) EG 2
>> 4. Ex(~Px v Qx) Contra 3
>> 5. Ex~(Px & ~Qx) DeMorg 4
>> 6. ~Ex(Px & ~Qx) ??
>>
>> While the argument is valid (e.g, 6 can be derived from 1) the move in
>> the proof from 5 to 6 must be illegal, for if not, then we could also
>> conclude by a rule of quantifier negation
>>
>The argument 1-6 is not valid.
Why and where? Are the moves from 1 to 5 legal?
>The theorem as stated from the start is
>valid.
>
>> .
>> 5. Ex~(Px & ~Qx) DeMorg 4
>> 6. ~Ax(Px & ~Qx) QN 5
>>
>> since Ex~Px == ~AxPx; yet the two conclusions are not equivalent; so
>> num num puzzled.
>>
>Just because A -> B and A -> C you expect B <-> C ?
>No, you have shown counterexample.
I didn't suggest it was anything else, I'm just asking WHY and where
the first example is in error.
Thanks.
>What you can say is
> A -> (B <-> C)
>which is weaker than
> A -> (B & C)
>
>Since Bush is president, unemployment soars
>Since Bush is president, the deficit soars
>Therefore
> unemployment soars
>iff
> the deficit soars?
>
>> This valid proof of the same argument no puzzle num num:
>>
>> Ax(Px -> Qx) |- ~Ex(Px & ~Qx)
>>
>> 1. Ax(Px -> Qx) Premise
>> 2. Ax(~Px v Qx) Contra 1
>> 3. Ax~(Px & ~Qx) DeMorg 2
>> 4. ~Ex(Px & ~Qx) QN 3
>>
>> Yet the UI and EG in the former proof are legal, right? So what rule
>> of syntax is violated in the former proof? Please cite a source if
>> possible.
>>
>>
>> num num
>>
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