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Some naive questions from an interested layman regarding the cardinality of the set of all math theorems (discovered and undiscovered).

1) What branches of math are not contained in ZF set theory + the axiom of choice + the continuum hypothesis?

2) What is the cardinality of all discovered and undiscovered theorms under ZF set theory?

3) Cardinality of theorems under ZF + axiom of choice?

4) Cardinality of theorems under ZF + continuum?

5) Cardinality of theorems under ZF + axiom of choice + continuum?

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1. non standard analysis, but the beauty of it is that any assertion contained in ZFC is true in ZFC even if the proof uses non standard arguments 2,3,4,5 : if you think of theorems that can be proved in a finite number of applications of axioms (which I understand is the definition of "provable") then there are only countable many of those – Glougloubarbaki Nov 30 '11 at 17:48

Each of these theories has a countable infinity of theorems. It is immediate that they have at least that many theorems, because $n=n$ is a theorem of each of them for any numeral $n$. On the other hand, every theorem can, by definition, be represented as a finite sequence of symbols from a countable (or finite) alphabet, and there are only countably many such sequences.

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Follow up question: are the theorems in ZFC not in ZF also countably infinite? – Jack Dec 1 '11 at 16:25
Yes, because $AC \land (n=n)$ is a theorem for every numeral $n$ and cannot be proved from ZF. – Henning Makholm Dec 1 '11 at 17:33
This answer is good, because it tells OP that even things that seem obvious can be classified as theorems. – Alfred Yerger May 13 '15 at 8:18

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