```Date: Jun 17, 2009 3:20 PM
Author: William Hughes
Subject: Re: Answer to Dik T. Winter

On Jun 17, 2:53 pm, WM <mueck...@rz.fh-augsburg.de> wrote:> On 17 Jun., 20:19, William Hughes <wpihug...@hotmail.com> wrote:>Your claim is that "no possibility exists to construct or todistinguish by one or many or infinitely many nodesof 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 ==> BSo if A is true then B is trueand your claim is false.You want to show~A  [Follows from ( A ==> B, ~B) ==> ~A ]by proving ~B(Note that assuming ~A is circular)> > > What do you understand by "all nodes"?>> >  "all nodes of t are in the tree"> >  There is no node that is in t but is not in the tree>> That is correct.>>>> >  "all nodes of t are in P"> >   There is no node that is in t but is not in an element of P>> Also correct.<snip evasion>Which is the first statement you disagree with   all nodes of t are in the tree   t is in the tree   all nodes of t are in P   t is not an element of P> > WM: the list of paths P is the same as the tree>> > Nope.  You have agreed that the tree contains a subset of> > nodes that is not contained in one element of the list of paths P.>> All subsets of nodes that are in the tree also are in the list of> paths.But, as you have agreed, not in a single path.> Why should the subset belong to one element of the list of> pathsIt doesn't."the tree contains a subset of nodes that is*not* contained in one element of the list of paths"                        - William Hughesi
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