Re: if the barber cuts everyones beard who doesn't shave himself....


"Chris Menzel" <cmenzel@xxxxxxxxxxxxxxxxxxxx> wrote in message
news:slrnd99hj6.117.cmenzel@xxxxxxxxxxxxxxxxxxxx
> On 25 May 2005 10:39:18 -0700, ken quirici <kquirici@xxxxxxxxx> said:
>> Owen wrote:
>>> ~EyAx(Rxy <-> ~(Rxx)).
>>>
>>> Proof:
>>>
>>> 1. Ax(Rxy <-> ~(Rxx)) -> (Ryy <-> ~(Ryy))
>>> 2. (Ryy <-> ~(Ryy)) is a contradiction, because ~(p <-> ~p) is a
theorem.
>>> 3. Ax(Rxy <-> ~(Rxx)) -> (contradiction).
>>> 4. ~Ax(Rxy <-> ~(Rxx)).
>>> 5. Ay~Ax(Rxy <-> ~(Rxx)).
>
> A little better, I think, as it avoids Owen's metatheoretic step 2:
>
> 1. Ax(Rxy <-> ~Rxx) -> (Ryy <-> ~Ryy) Universal Instatiation
> 2. ~(Ryy <-> ~Ryy) Theorem of propositional logic
> 3. ~Ax(Rxy <-> ~Rxx) 1,2 Modus Tollens
> 4. Ay~Ax(Rxy <-> ~Rxx) Universal Generalization
> 5. ~EyAx(Rxy <-> ~Rxx) Negation/Quantifier Exchange
>
>> Hmmm. This might be more subtle than I thought.
>
> It's just basic predicate logic.
>
>> y is in fact a particular y, not the unbound variable I had thought.
>
> "y is ... a particular y" doesn't make any clear sense. What is clear
> is that all of the occurrences of the variable "y" in lines 1, 2, and 3
> are unbound.
>
>> But that means that its that same y in 5 [line 4, in the revised
>> proof],
>
> "It's the same y" doesn't have any clear sense either. The occurrences
> of "y" lines 4 and 5 are bound. The only question is whether it is
> legitimate to apply Universal Generalization with respect to "y" in line
> 4.
>
>> so you can't universalize it when you go from 4 to 5 [3 to 4].
>
> Yes you can. "y" does not occur in any undischarged assumptions, so its
> occurrence in line 3 can be universally generalized in line 4.
>
> Chris Menzel
>

Your explantion and your proof is much better than mine, thanks.


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