> This implies second order logic. No, it does not. The second incompleteness theorem is provable in first-order Peano Arithmetic. > Of course I'm in no posit

> This implies second order logic. No, it does not. The second incompleteness theorem is provable in first-order Peano Arithmetic. > Of course I'm in no position to say a similar thing about Goedels incompleteness theorem, and I even referred to it's result in higher order logics, but I still doubt the relevance, as many seem to be ignorant of his former completeness theorem. I have no idea what you're trying to say, but I assure you that people who do this kind of research are aware of the completeness theorem.