On 18 Jun., 20:22, William Hughes <wpihug...@hotmail.com> wrote: > On Jun 18, 1:49 pm, WM <mueck...@rz.fh-augsburg.de> wrote: > > > > > > > On 18 Jun., 14:44, 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." > > > > A: actually infinite paths exist, > > > B: the infinite tree contains a path p that can be > > > distinguished from every path of P. > > > > You agree > > > > A ==> B > > > > So if A is true then B is true > > > and your claim is false. > > > > You want to show > > > > ~A [Follows from ( A ==> B, ~B) ==> ~A ] > > > > by proving ~B > > > > (Note that assuming ~A is circular) > > > > <snip evasion> > > > > Please answer yes or no > > > > t is not an element of P > > > The actually infinite path t is not an element of P > > and not a path of T, because t does not exist. > > Note it would be circular to simply assume that actually > infinite paths do not exist.
It is not assumed. It is proved by the fact that t cannot be distinguished from the paths of P and T. > > > Proof by the fact that t cannot be distinguished from the set P > > Nope: you have agreed that t can be distintuished from the every > element > of the set P > > (Recall, You have agreed. > > A subset of nodes is distinguished from every > element of P if and only if it is not contained in > a single element of P.)
All of t, that does exist, is contained in a path of P. It is only your delusion that opposes to this fact. You say that for every path p of P there is a bit of t that does not belong to p. And you think this would exclude t from P. But that is wrong, because "every" always only concerns to paths of P that you can specifically address. The important fact is: Every bit of t and all its predecessors are in infinitely many paths of P. When P is ordered as a tree, your delusion can no longer deceive you.