Skip to main content

You are not logged in. Your edit will be placed in a queue until it is peer reviewed.

We welcome edits that make the post easier to understand and more valuable for readers. Because community members review edits, please try to make the post substantially better than how you found it, for example, by fixing grammar or adding additional resources and hyperlinks.

Required fields*

6
  • 8
    $\begingroup$ ZFC actually has infinitely many axioms, and one can prove that you cannot find an equivalent finite set of axioms. This doesn't change much of what you said, but it's just a correction to "ZFC has (about) 10 axioms" $\endgroup$ Commented Nov 25, 2017 at 8:43
  • 2
    $\begingroup$ @Max Nice catch, but the "infinitely many" are all dictated by the single axiom schemata of set-building axioms, so this laziness is not too harmful here I think. $\endgroup$ Commented Nov 25, 2017 at 16:36
  • 1
    $\begingroup$ By comparing the size of encodings into Automath, Freek Wiedijk gives quantitative estimates of how much more complex MLTT is than ZFC (including the cost of formalizing FOL). There's roughly a factor of two difference across a few metrics (e.g. number of primitive concepts, compressed size of the encoding). Of course, the "cognitive load" is unable to be quantified. It's not completely surprising as MLTT is constructive which means some symmetry is lost requiring additional primitives. Also, MLTT is more usable out-of-the-gate than ZFC. $\endgroup$ Commented Nov 26, 2017 at 2:21
  • 16
    $\begingroup$ I find the argument that "in set theory we can prove different statements about isomorphic group" is sort of like saying "hey, how about a bleach enema for cleaning your colon?". Yes, technically you're right, there is a trivial group where $\omega$ is the only element, and there is another where $\mathcal P(\omega)$ is the only element, and these are different. But this ignores context entirely, since you're not proving something about the group, you prove it about the underlying set. When you say "prove something about the group", [...] $\endgroup$ Commented Nov 27, 2017 at 10:02
  • 13
    $\begingroup$ [...] it implies (to me anyway) that you're talking about a statement in the language of groups, using some first/second/third/etc. order logic. But these various logics do not have access to the actual underlying set. So this is not really a statement about the group anymore. $\endgroup$ Commented Nov 27, 2017 at 10:03