Looking for proof of conditional probability

Your question doesn't make any sense to me. I don't know what it means to "prove" a definition. Did you mean to ask for an (informal) argument that the concept is useful?

My confusion is compounded by the fact that I find P(A|B) = P(A∩B)/P(B) pretty self-explanatory. What seems to be the difficulty?

I agree with the OP: simply defining a probability concept doesn't by itself map it to our intuitions about it. For example, if we defined P(A|B) = P(AB) / 2P(B), it wouldn't correspond to our intuitions, and here's why.

Intuitively, P(A|B) is the probability of A happening if we know that B already happened. In other words, the entirety of the elementary outcome space we're taking into consideration now are those that correspond to B. Of those remaining elementary outcomes, the only ones that can lead to A are those that lie in AB. Their measure in absolute terms is equal to P(AB); however, their measure in relation to the elementary outcomes in B is equal to P(AB)/P(B).

Thus, P(A|B) is P(A) as it would be if the only elementary outcomes in existence were those yielding B. P(B) here is a normalizing coefficient: if we were evaluating the conditional probability of A in relation to a set of exhaustive and mutually exclusive experimental outcomes, as it is done in Bayesian reasoning, dividing by P(B) means renormalizing the elementary outcome space after B is fixed.

Syntactically, adding conditional probability doesn't do anything new, besides serving as short-hand for expressions that look like "P(A∩B)/P(B)".

The issue you're recognizing is that, semantically, conditional probabilities should mean something and behave coherently with the system of probability you've built up to this point.

This is not really a question of probability theory but instead a question of interpretations of probability. As such, it has a very philosophical feel and I'm not sure it's going to have the solid answers you're looking for.

Short answer: The Kolmogorov axioms are just mathematical. They have nothing inherently to do with the real world. P(A|B)=P(A∩B)/P(B) is the definition of P(A|B). There is a compelling argument that the beliefs of a rational agent should obey the Kolmogorov axioms, with P(A|B) corresponding to the degree of belief in A when B is known.

Long answer: I have a sequence of posts coming up.

Umm...I don't know how rigorous this explanation this is, but it might lead you in the right direction...because if you consider the Venn Diagram with probability spaces A and B, the probability space of A within B is given by the overlap of the two circles, or P(A∩B). Then you get the probability of landing in that space out of all the space in B...as in, the probability that if you choose circle B, you land in the overlap between A and B.

That's probably not what you were looking for, but hope it helps.

I'm not sure if this is what you're looking for, but I believe one way you can derive it is via a dutch book argument.

I think you need the common-sense axioms P(B|B) = 1 and P(A|all possibilities) = P(A). Given these, the Venn diagram explanation is pretty straightforward.

Why is P(A∩B)/P(B) called conditional probability? Or, let's turn it the other way round (which is your question), why would conditional probability be given by P(A∩B)/P(B)? I think I was able to develop a proof, see below. Of course, double-checking by others would be required.

First, I would define conditional probability as the “Probability of A knowing that B occurs”, which is meaningful and I guess everybody would agree on (see also wikipedia).

Starting from there, “Probability of A knowing that B occurs” means the probability of A in a restricted space where B is known to occur. This restricted space is simply B. Now, in this restricted space (forget about the larger space Ω for a minute), we want to know the probability of A. Well, the occurrence of A in B is given by A∩B. In case of equiprobability of elementary events, the probability of A∩B in B is card(A∩B)/card(B). Now, this can be tweaked as card(A∩B)/card(Ω)×card(Ω)/card(B)=P(A∩B)/P(B), where P is the probability in the original space Ω. The latter result being Kolmogorov’s definition of conditional probability!

Note that we can also think in terms of areas, writing that the probability of A∩B in B is the area of A∩B divided by the area of B: area(A∩B)/area(B), which can be tweaked the same way, and works without requiring equiprobability (and the probability space could also be non-discrete).

I'm sorry, the comments on this post all seem to miss the point. Bayes' Theorem can be proven from basic logic, look at places like the Khan Academy, or Lukeprog's Intuitive Explanation of Yudkowsky's Intuitive Explanation. Once you understand that, the Kolmogorov axioms will be obvious. It's not assumed,

Even Wikipedia notes that Cox's Theorem makes another approach possible -- that seems like the place to start looking if you want a mathematical proof. So I think Larks came close to the right question (though it may or may not address your concerns).

Cox and Jaynes show that we can start by requiring probability or the logic of uncertainty to have certain features. For example, our calculations should have a type of consistency such that it shouldn't matter to our final answer if we write P(A∩B) or P(B∩A). This, together with the other requirements, ultimately tells us that:

P(A∩B) = P(B)P(A|B) = P(A)P(B|A)

Which immediately gives us a possible justification for both the Kolmogorov definition and Bayes' Theorem.

Maybe you'd find it more intuitive that P(AnB) = P(A|B)*P(B) ?