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 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.
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)
> > > 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.
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
But, as you have agreed, not in a single path.
> Why should the subset belong to one element of the list of
"the tree contains a subset of nodes that is
*not* contained in one element of the list of paths"
- William Hughes