fom
Posts:
1,968
Registered:
12/4/12


Re: A finite set of all naturals
Posted:
Aug 12, 2013 8:17 PM


On 8/12/2013 6:04 AM, Ben Bacarisse wrote: > fom <fomJUNK@nyms.net> writes: > <snip> >> Here is the proof he provides when pressed to produce: >> >> https://groups.google.com/d/msg/comp.ai.philosophy/AVnJgA5ZIk0/fk1sQQNSYVgJ > > I could not resist taking just a little peek...
Please accept my apologies. You should have resisted.
> That post claims to be > "written in the language of arithmetic L(PA)" but it contains this > definition early on: > > Def03a: even1(x) <> Ey[x=y+y] > Def03b: even2(x) <> Ey[x=2*y] > Def03c: even(x) <> (even1(x) \/ even2(x)) > > Does this mean that Ax[even1(x)=even2(x)] is not a theorem of the system > being used, or does it mean that the author is inclined to use overly > complex definitions? Either way, I didn't find it an encouraging start. >
After arriving at a similar emotional state, I went looking for the alleged proof and only looked for definitions relevant to its analysis...
> Maybe I don't understand the notation,
Does anyone? You are not alone.
But, I am not one to speak. My own statements seem to puzzle others.
> because just a few lines further > on I see: > > Def05c: aGC(x) <> > (even(x) /\ (SS0<x)) > Ap1p2[prime(p1) /\ prime(p2) /\ > (p1+p2<x \/ x<p1+p2)] > > but isn't Ap1p2[prime(p1) /\ prime(p2) /\ (p1+p2<x \/ x<p1+p2)] simply > false? >
Nam probably wanted restricted quantifications in this definition,
Ap1[prime(p1) > Ap2[prime(p2) > [(p1+p2<x \/ x<p1+p2)]
Then, the apparent intent would be that aGC(x) holds if every sum of primes is different from x.
So, suppose aGC(x) holds.
Then, if (even(x) /\ (SS0<x)), the apparent intent would be realized for x.
Suppose the antecedent fails because (SS0>=x). Then doesn't x<(p1+p2) hold for all (correctly) defined primes? The antecedent fails, but the consequent does not. So, the conditional would appear to hold. If I am correct in this, then
aGC(0)
aGC(1)
aGC(2)
hold.
Suppose the antecedent fails because ~even(x) holds. Then, presumably odd(x) holds (although I have not proved this from Nam's definitions).
Consider prime pairs. One has, for example,
2 + 3 = 5
So
(2+3<5 \/ 5<2+3)
does not hold. Since 3 is prime,
prime(3)
holds. so
prime(3) > (2+3<5 \/ 5<2+3)
does not hold and
Ap2[prime(p2) > (2+p2<5 \/ 5<2+p2)]
does not hold. Similarly,
Ap1[prime(p1) > Ap2[prime(p2) > (p1+p2<5 \/ 5<p1+p2)]
does not hold. So, since both the antecedent and the consequent fail, the conditional holds. Therefore, for the example,
aGC(5)
holds.
Since the given analysis should apply to every prime pair, one should have
aGC(x) = {0, 1, 2} union {x  prime(x) /\ Ey[prime(y) /\ x=y+2]} union {x  (even(x) /\ (SS0<x)) /\ Ap1[prime(p1) > Ap2[prime(p2) > [(p1+p2<x \/ x<p1+p2)]}
Given my recent lack of practice, I have probably made a mistake here.
But, the important thing is how one must remember that one must interpret Nam's incorrect definitions so that they make sense with respect to his claims.
Suppose you actually consider his definition as given.
Then since, as you observed,
Ap1p2[prime(p1) /\ prime(p2) /\ (p1+p2<x \/ x<p1+p2)]
cannot hold, the conditional is true whenever
(even(x) /\ (SS0<x))
does not hold. Hence,
aGC(x) = {0, 1, 2} union {x  odd(x)}
Once again, perhaps I have made some mistake.
Yet, since you have motivated me to actually make sense of his definition of aGC(x) to the best of my ability, the question is whether or not I can resist seeing how he actually uses it.
chuckle

