@Zetherin,
Zetherin;122604 wrote:Oh, I see. I misrepresented my position. I wasn't clear, my apologies. What I meant was that the toddlers wouldn't understand and come to sophisticated conclusions; not that, through random and childish action, they wouldn't eventually write Newton's Principia.
I should have phrased that differently, my bad.
No problem mate. I agree that a toddler would not be able to understand something very sofisticated.
---------- Post added 01-26-2010 at 03:03 PM ----------
Owen;122694 wrote:Emil,
"Some people think that that something is logically possible logically implies that it has a nonzero chance of happening [◊P⇒Pr(P)>0] but that's not true."
Of course it is true.
Pr(p)=0 <=> ~<>P.
~(Pr(p)=0) <=> <>p.
therefore,
<>p => ~(Pr(p)=0).
Previous correspondence with you have been a waste of time, so I will answer you only because Z. may find what you write convincing. That is,
if he can guess what your symbolism means.
P1 of the argument above is also false. But it is the case that not possible P logically implies Pr(P)=0, just not conversely.
To see why this and the other proposition is false, consider a thought example. An arrow will be shot at a disc. The arrow's head is 1 point large. The head will hit the disc somewhere at random. The disc has an infinite number of points. The chance of the arrow hitting any single point on the disc is thus 1/∞=0. But if we then accept that for all points, it is impossible that the arrow hits there we get a contradiction. Thus, the implication from probability 0 to impossibility is false.
[INDENT]1. (∀x)(Pr(Hx)=0) [Premise]
For all points, the probability of the arrow hitting a point is zero.
2. (∃x)(Hx) [Premise]
There exists a point such that the arrow will hit that point.
3. (∀x)((Pr(Hx)=0)⇒?◊Hx) [assumption for RAA]
For all points, that the probability of the arrow hitting a point is zero logically implies that it is not logically possible that the arrow will hit that point.
Thus, 4. (∀x)(?◊Hx) [1, 3 MP]
For all points, it is not possible that the arrow will hit a point.
Thus, 5. (∀x)(?Hx) [4, modal logic]
For all points, the arrow will not hit that point.
Thus, 6. ?(∃x)(Hx) [5, predicate logic equivalence]
It is not the case that there exists a point such that the arrow will hit that point.
Thus, 7. ?(∀x)((Pr(Hx)=0)⇒?◊Hx) [RAA, 1-6]
It is not the case that for all points, that the probability of the arrow hitting a point is zero logically implies that it is not logically possible that the arrow will hit that point.[/INDENT]
