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It 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-ord
by John_Baez 10y ago
It 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 10y agoThanks, this looks plausible.