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.

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)

> > > 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

> paths

It doesn't.

"the tree contains a subset of nodes that is

*not* contained in one element of the list of paths"

- William Hughes

i