Re: Inside or Outside ?
- From: Nam Nguyen <namducnguyen@xxxxxxx>
- Date: Fri, 21 Jul 2006 06:16:50 GMT
Rupert wrote:
Nam Nguyen wrote:
P(x) df= Ayz[((x /= 0) /\ (x /= S0) /\ (S0 < y) /\ (y < x)) -> ~(x=y*z)]
How about
P(x) df= ((x /= 0) /\ (x /= SO)) /\ Ayz[((S0 < y) /\ (y < x)) ->
~(x=y*z)]
Sure. I don't see anything wrong with your def. (Did you see anything
mine?)
"2 is a prime" then is just P(SS0). The key question is then can we
prove P(SS0), without depending on a) the assumed PA's consistency
Yes.
and b) proof by contradiction?
Define when a proof does not use proof by contradiction.
Assuming T is consistent, a proof that "does not use proof by
contradiction" is a proof that is not a proof-by-contradiction.
--
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What we call 'I' is just a swinging door which moves
when we inhale and exhale.
Shunryu Suzuki
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