On Jun 20, 8:45 am, WM <mueck...@rz.fh-augsburg.de> wrote: > On 20 Jun., 13:19, William Hughes <wpihug...@hotmail.com> wrote: > > > > > On Jun 20, 4:25 am, WM <mueck...@rz.fh-augsburg.de> 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) > > > > I prove ~B, for instance by your inability to distinguish t from T, > > > with no regard to the truth of A. > > > Nope. You need to distinguish the subset of nodes t from every > > element of P, not from the tree, > > The paths of the tree are the same as the subsets of P.
Nope. You have agreed that if infinite paths actually exist then t is not contained in any element of P. So you cannot argue "The paths of the tree are the same as the subsets of P" without assuming ~A.
We are left we the result that you cannot prove A or ~A.