3 ms·
Not in first order logic.
by danharaj 11y ago
Not in first order logic.
- wolfgke 11y agoWhy doesn't one simply use higher order logic to solve this problem? (serious question, I'm a mathematician, but not a set theorist).
- danharaj 11y agoA lot of properties that you'd want in a logic stop holding once you hit second-order logic: https://en.wikipedia.org/wiki/Second-order_logic#Metalogical_results https://en.wikipedia.org/wiki/Second-order_logic#Metalogical...
- John_Baez 11y agoIt turns out this only pushes back the problem. You can write down second-order axioms for arithmetic which have a unique model, but in a model of a second-order theory predicates are interpreted as sets. So, you need to choose a version of set theory, to know what a model actually is. Suppose you choose ZFC (the usual axioms of set theory). Unfortunately this itself has infinitely many models! I explained this in a bit more detail here: https://johncarlosbaez.wordpress.com/2016/04/02/computing-the-uncomputable/#comment-79002 https://johncarlosbaez.wordpress.com/2016/04/02/computing-th...
- wolfgke 11y agoThanks, this looks plausible.