On 15 Jun., 17:41, William Hughes <wpihug...@hotmail.com> wrote: > On Jun 15, 10:38 am, WM <mueck...@rz.fh-augsburg.de> wrote: > > > On 15 Jun., 16:01, William Hughes <wpihug...@hotmail.com> wrote: > > > > On Jun 15, 9:23 am, WM <mueck...@rz.fh-augsburg.de> wrote: > > > > > On 15 Jun., 00:19, William Hughes <wpihug...@hotmail.com> wrote: > > > > Your claim is that "no possibility exists to construct or to > > > distinguish by one or many or infinitely many nodes > > > of the tree another path." > > You are trying a proof by contradiction of > " Actually infinite paths exist" > > We have > If actually infinite paths exist > there is a path p that can be distinguished > from every path of P. > > We need > > If actually infinite paths exist > there is no path p that can be distinguished > from every path of P.
Please check that again. Compare: We want to prove by contradiction that pi is not rational. We have, if pi is rational, then pi is p/q. We need, according to you, if pi is rational, then pi is not p/q.
> My proof is as follows: If Actual then Distinct and not Distinct. > > > Proof: You > > cannot distinguish p from the tree, > > Irrelevant. You do not distinguish p from the tree. > You distinguish p from every element of P.
You cannot distinguish p from every path of the tree. Every path of the tree is is from P. > > > P contains every subset of nodes of the binary tree. > > The union of the paths in P contains every subset of > nodes. However, p is a subset of nodes that is not > contained in any single element of P.
Then you would be able to distinguish p from every path of the tree that is constructed by P. But you aren't.
