On 19 Jun., 17:56, 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) > > > > you have agreed that t can be distinguished from > > > every element of P. > > It does not matter whether I have agreed > > it matters whether it is > > true. > > As you see from the tree, > > the subset of nodes t is in the tree
You cannot distinguish it from all paths of the tree. > > And we agree that > > the subset of nodes t is not in > a single element of the list P
if actual infinity axists. > > so it is true that the subset of nodes t is > in the tree and can be distinguished from > every element of P.
If you were right. But that would imply the existence of actual infinity and the creation of paths by dropping bits. As the latter is wrong, you are not right.