In article <firstname.lastname@example.org>, WM <email@example.com> wrote:
> On 18 Mrz., 18:32, fom <fomJ...@nyms.net> wrote: > > On 3/18/2013 6:43 AM, WM wrote: > > > > > On 18 Mrz., 07:26, fom <fomJ...@nyms.net> wrote: > > > > >> You turn to an outdated strategy directed to > > >> a situation that no longer exists rather > > >> than do the hard work of grounding your > > >> claims. You do this to say that just > > >> because you do not believe a particular > > >> axiom, > > > > > Wrong. I prove that the axiom is nonsense like the axiom that a > > > triangle with four edges exists. > > > > That would be more forceful if you used the > > term 'trilateral'. > > Then it would be trivial. My example requires a little bit deeper > thought.
Not even a micron deeper. > > > > Once again. You have *proven* nothing. > > As your foregoing hint shows, you seem to welcome trivialities, but > you seem to be not able to understand more difficult ideas. >
FOM's relatively short list of posts frequently deal with far more difficult ideas than WM's endless rants here ever have dealt with.
WM has frequently claimed that HIS mapping from the set of all infinite binary sequences to the set of paths of a CIBT is a linear mapping.
In order to show that such a mapping is a linear mapping, WM would first have to show that the set of all binary sequences is a linear space (which he has not done and apparently cannot do) and that the set of paths of a CIBT is also a vector space (which he also has not done and apparently cannot do) and then show that his mapping, say f, satisfies the linearity requirement that f(ax + by) = af(x) + bf(y), where a and b are arbitrary members of the field of scalars and x and y and f(x) and f(y) are arbitrary members of suitable linear spaces.
While this is possible, and fairly trivial for a competent mathematician to do, WM has not yet been able to do it.