On 19 Jun., 19:37, William Hughes <wpihug...@hotmail.com> wrote: > On Jun 19, 1:17 pm, WM <mueck...@rz.fh-augsburg.de> wrote: > > > > > > > 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 > > > > Outside of Wolkenmuekenheim it does. > > > > > it matters whether it is > > > > true. > > > > As you see from the tree, > > > > the subset of nodes t is in the tree > > <snip> > > > > And we agree that > > > > the subset of nodes t is not in > > > a single element of the list P > > > if actual infinity axists.
Otherwise your alleged "proof" is even more obviously wrong. > > So you cannot prove ~B without making > the assumption ~A which is what you are trying > to prove.
Wrong. ~B is proven by the impossibility to distinguish t from every path of the binary tree and the fact that the tree does not contain paths that are missing in P.
