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.
> 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.)