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.
The claim is false if and only if you can distingusih a path p from all pats from the set P when written in form of the tree. > > You want to show > > ~A [Follows from ( A ==> B, ~B) ==> ~A ] > > by proving ~B >
> > To create a path requires to add that path. > > Nope I give a counterexample. > > > > To do this [create a path] we > > > need to add a subset of paths. > > > > E.g. The subset of paths added is > > > > 1000... > > > 11000... > > > 111000... > > > ... > > > > The subset of nodes that is now in the tree is > > > > t={1,11,111, ...}
There is no sequence of 1's added that is longer than every such sequence of paths in P. > > > > Note that t is not an element of P, so the subset > > > t is never added to the tree. > > 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
I disagree with the statement: More nodes of t are in tree than are in any single path of P. Proof: The tree is nothing but a another way to write the paths of P. So why do you think (or rather pretend to think) that there are longer sequences of 1's in P than in P? > > Note: 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.
Wrong. The list of paths of P is the tree in slightly different notation. When changing from list to tree, then several bits are mapped on one node, not the other way round. That number does decrease, not increase. Again you are trying to rape logic.
