On Jun 30, 8:22 am, WM <mueck...@rz.fh-augsburg.de> wrote: > On 30 Jun., 16:36, "Dik T. Winter" <Dik.Win...@cwi.nl> wrote: > > > In article <5c70e8d7-67e1-48e7-8dee-ec3d853bd...@r34g2000vba.googlegroups.com> WM <mueck...@rz.fh-augsburg.de> writes: > > > On 22 Jun., 15:11, "Dik T. Winter" <Dik.Win...@cwi.nl> wrote: > > ... > > > > > Every meaning of every word is defined by a language. > > > > > Every language is a finite definition. > > > > > The number of finite definitions is countable. > > > > > > > > For each word the meaning can indeed be countable. But that does not mean > > > > that the set of meanings for all the words in a language is countable. > > > > > > No? How can that be accomplished? > > > I think by some negation of the axiom of choice. > > I did not ask how to "prove" that. I asked how to *do* it. > > Regards, WM
We know every language contains words, or symbolic means or presentation. Factor this into the equationa nd begin to produce a creative execution.
Here is a clever take:
Languages and Finite Automata is finite. Proof: We will reduce the halting problem. to this problem ... construct. YES. NO. NO. YES. halting problem decider. finite language. problem ...
The language is contained and finite if to every input there is only Yes or No. The system then only ask the right question to achieve the right answer in the language.