Minimal axiom system for a set of provable statements - MathOverflow most recent 30 from http://mathoverflow.net2013-05-18T12:02:10Zhttp://mathoverflow.net/feeds/question/3736http://www.creativecommons.org/licenses/by-nc/2.5/rdfhttp://mathoverflow.net/questions/3736/minimal-axiom-system-for-a-set-of-provable-statementsMinimal axiom system for a set of provable statementsMartyguy2009-11-02T03:50:44Z2009-11-02T06:22:08Z
<p>I am not a mathematician, so forgive me if this question is trivial. The basic idea of my question is: For a given set of provable statements, can we find an axiom system with the smallest number of true statements (smallest by inclusion)? So here we go with my question.</p>
<p>Suppose we have a set of axioms A and a logic system L that we use to prove theorems arising from A. Let P(A) be the set of provable statements from A using L. Let S be the set of all other sets axiom systems B for which P(A) is a subset of P(B).</p>
<p>Now for any axiom system B in S, let T(B) be the set of all true statements arising from B and L. Partially order the set of axioms in S such that for B1, B2 in S, B1 < B2 if T(B1) is a subset of T(B2).</p>
<p>Question: Does there exist an axiom system M in S such that M < B for all B in S?</p>
http://mathoverflow.net/questions/3736/minimal-axiom-system-for-a-set-of-provable-statements/3749#3749Answer by Andrew Critch for Minimal axiom system for a set of provable statementsAndrew Critch2009-11-02T05:36:22Z2009-11-02T06:22:08Z<p>This question certainly makes sense to ask, and the answer is no, because of how flexible the definition of "logical system" is.</p>
<p>First, your definition of B1 < B2 can be simplified: "T(B1) is a subset of T(B2)" is equivalent to "B2 proves B1".</p>
<p>Now, suppose we form a logical system whose only statement symbols are x<sub>n</sub>, where n can be an integer, and (countably many) rules of inference stating that x<sub>m</sub> implies x<sub>n</sub> when m > n. If A is the set of axioms {x<sub>1</sub>,x<sub>2</sub>,x<sub>3</sub>...}, then there is no minimal set of axioms implying A (and P(A)). This is because a set of axioms B will imply A if and only if it contains a sequence of axioms with subscripts approaching infinity, in which case we can always remove finitely many axioms from B so that it still implies A.</p>
<p>(FYI, systems with infinitely many axioms and/or rules of inference are typical of modern mathematics; ZFC, the most commonly accepted foundation, is itself not finitely axiomatizable.)</p>
http://mathoverflow.net/questions/3736/minimal-axiom-system-for-a-set-of-provable-statements/3752#3752Answer by Harrison Brown for Minimal axiom system for a set of provable statementsHarrison Brown2009-11-02T06:14:58Z2009-11-02T06:14:58Z<p>Andrew's correct, but there is something similar which is a serious research area -- it's called <a href="http://en.wikipedia.org/wiki/Reverse%5Fmathematics" rel="nofollow">reverse mathematics</a>. The thing is, most basic theorems of mathematics can be formulated and proved in second-order arithmetic (which is kind of like the Peano axioms on hypersteroids -- you get to quantify over sets, which gives you lots of extra room to work). But a lot of the stuff that people care about in practice doesn't need the full power of second-order arithmetic. Reverse mathematics tries to find the largest "fragments" that you need to prove these sorts of things.</p>