-2
$\begingroup$

In classical logic, any given theorem implies any other given theorem (the soundness of classical logic comes as one way to help realize this). I realize the term "mathematics" has a certain vagueness to it, but I would think that the idea of provable propositions in number theory, group theory, analysis, or any other field of mathematics doesn't have a similar vagueness to it (perhaps I'm naive in this belief). In number theory, group theory, or any other mathematical field, does any given theorem imply any other given theorem? If not, does it make sense to use classical logic when talking about any of these fields? Or does the notion of a theorem concerning the integers depend on the system of logic one takes for granted, and thus we don't really have a single field which talks about all the theorems concerning the integers, but rather number theories distinct in some way? Does there exist any sort of consensus with respect to these questions?

  • 1
    Well, $A \implies B$ is true if $A$ and $B$ are true (but not only if), so in a sense any theorem trivially "implies" any other theorem. But in general mathematicians speak of how "strong" certain statements are. Strength of assertions isn't rigorously defined as far as I know, but it's typically related to how general a statement is, or otherwise how "deep" it is embedded into a theory (roughly how much effort it takes to get to it from the starting definitions). If there is an easy way to go from $A$ to $B$ but not vice versa, then $A$ is probably considered stronger than $B$.2011-08-22
  • 2
    Also, I'd believe most number theorists don't really pay attention too much attention to precisely which axioms they're taking for granted. Separating "number theories" based on which axioms are present and which aren't is mostly something that logicians do.2011-08-22
  • 0
    @anon: I wouldn't say logicians so much as people working in [reverse mathematics](http://en.wikipedia.org/wiki/Reverse_mathematics).2011-08-22
  • 0
    @Anon Besides assuming completeness as you did, there exists another way to see that any theorem p in classical logic implies any other theorem q, given a rich enough set of rules. First, we take p as our first hypothesis. Since q comes as a zero-premise conclusion (theorem), a proof of it can get transposed into a sub-proof "anywhere". So, we then write out the formulas used to prove q with p as the first hypothesis before those used to derive q. So, p and q now have the same scope of the hypothesis p, with q after p in the proof. By conditional introduction we have Cpq as a theorem.2011-08-22
  • 0
    @Doug: I was talking about boolean logic, where $\text{True}\implies\text{True}$ is basically true by definition, and theorems are true statements. I don't think I assumed completeness of any theory. I don't have enough background in proof theory or formal logic to understand the rest of your comment.2011-08-22
  • 0
    @anon By "completeness" I meant "semantical completeness". This means that if a proposition p is true, then p qualifies as a theorem. So, to deductively get from "(a->b) is true if "a" and "b" are true" to "(a->b) is a theorem if "a" and "b" are theorems" or equivalently "any theorem "implies" any other theorem" you need semantical completeness. But, you can get "any theorem "implies" any other theorem" without taking semantical completeness for granted, basically if you have conditional introduction as a valid rule of inference.2011-08-22

1 Answers 1