On 17 Jun., 19:01, William Hughes <wpihug...@hotmail.com> wrote:
> 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.
That is nonsense. If (A) pi has a rational presentation, then (B) it can be written as p/ q. B is false. And A cannot be true without B. > > 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
What do you understand by "all nodes"? > > Note: The list of paths of P is not > the tree in slightly different > notation.
Note the list of paths P is the same as the tree in slightly different notation. Proof: Every pair of sequences is written in a form such that their common initial sequence is written only once. This is continued as far as possible.
>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.
That is a self-contradiction, because the tree is not different from the list other than that some bits are written only once. In particular there cannot be more in the tree than is already in the list. (With exception of mathemagicians' tricks, of course.)