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"
