On Tuesday, June 16, 2009 12:17:11 AM UTC-7, WM wrote: > On 15 Jun., 22:04, William Hughes <wpihug...@hotmail.com> wrote: > > > > > You are trying a proof by contradiction of > > > > " Actually infinite paths exist" > > > > > > We have > > > > If actually infinite paths exist > > > > there is a path p that can be distinguished > > > > from every path of P. > > > > > > We need > > > > > > If actually infinite paths exist > > > > there is no path p that can be distinguished > > > > from every path of P. > > > You know modus tollens? > ((A ==> B) & ~B) ==> ~A > > A: actually infinite paths exist. > B: p can be distinguished from every path of P. > > > > > > > You cannot distinguish p from every path of the tree. > > > > Irrelevant, you distinguish p from every element of P > > The tree is nothing but every path of P written in some unconvential > form. > > > > > Every path of the tree is is from P. > > > > Nope. Every *node* of the tree is from P. > > However, there is a *subset of nodes* in the tree that is not > > contained in one element of P. > > Every subset of nodes, that can form a path of the tree, is in the > tree. It stems from writing down the paths of P and no bit more. > > (Even if the tree is considered to be a union of sets {p_n} with p_n > in P, then the union could not contain more than is contained in at > least one of the united sets.) > > Regards, WM
