On 19/10/2013 2:58 PM, Peter Percival wrote: > Nam Nguyen wrote: >> On 19/10/2013 2:44 PM, Peter Percival wrote: >>> Nam Nguyen wrote: >>> >>>> [...], I'll have to work with you on your understanding >>>> first, and we'll see if you'd still have the same protest. >>> >>> Surely that would be a "cruel and unusual punishment"? >> >> I don't know what you're talking about. > > It's a quotation from some amendment to the US constitution. Don't ask > me, I know nothing. > >> Surely other posters (e.g., fom, Alan, ...) and I on the occasions >> _when needed_ did clarify technical matters. Why can't you? > > Is this clear enough?
No: it's not clear enough that you do - or do not - understand the definition. > > From your definition: > > > _A meta truth_ is said to be impossible to know if it's not in the > > collection of meta truths, resulting from all available definitions, > > permissible reasoning methods, within the underlying logic framework > > [FOL(=) in this case].
Right. That's my definition of the phrase here. > > this follows:
No. _It does NOT follow from my definition_ . > > We don't yet know if PA|-cGC or PA|-~cGC, so we don't know if "PA|-cGC" > or "PA|-~cGC" is in the collection of meta truths. So we don't know if > it's impossible to know cGC (or ~cGC). Why, then, do you claim that > it's impossible to know cGC (or ~cGC)? > > Do you know that both cGC and ~cGC are not in the collection of meta > truths? If so you must know that neither PA|-cGC nor PA|-~cGC. You > should publish your proof. And stop claiming that Gödel's > incompleteness theorem is invalid, because if neither PA|-cGC nor > PA|-~cGC, then that is an example of incompleteness. > > Also if you know that neither PA|-cGC nor PA|-~cGC, then you've proved > PA consistent. So you should stop claiming that its consistency is > unprovable. >
-- ----------------------------------------------------- There is no remainder in the mathematics of infinity.