On Jun 17, 12:29 pm, WM <mueck...@rz.fh-augsburg.de> wrote: > On 17 Jun., 18:09, 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 your claim is false. >
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 your claim is false.
You want to show
~A [Follows from ( A ==> B, ~B) ==> ~A ]
by proving ~B
(Note that assuming ~A is circular)
<snip>
> > 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 >
<snip evasion>
Please answer the question
Note: The list of paths of P is not the tree in slightly different notation. If all the nodes of a path h are in the tree then h is in the tree, however, if all the nodes of h are in P, h may or may not be an element of P.
