Hardness Guarantees for AES Many public-key cryptosystems have some kind of provable security. For example, the Rabin cryptosystem is provably as hard as factoring.

I wonder whether such kind of provable security exists for secret-key cryptosystems, such as AES. If not, what is the evidence that breaking such cryptosystems is hard? (other than resistance to trial-and-error attacks)

Remark: I'm familiar with AES operations (AddRoundKey, SubBytes, ShiftRows, and MixColumns). It seems that the hardness of AES stems from the MixColumns operation, which in turn must inherit its difficulty from some hard problem over Galois Fields (and thus, algebra). In fact, I can restate my question as: "Which hard algebraic problem guarantees the security of AES?"
As David said, we do not have such reductions for AES. However, this does not mean that Rabin or RSA cryptosystem are more secure than AES. In fact, I'd trust the (at least one-way, probably also pseudorandomess as well) security of block ciphers such as  AES/DES etc.. (perhaps with a bit more rounds than standardly used) more than the assumption that factoring is hard, precisely because there is no algebraic structure and so it's harder to imagine that there will be some kind of breakthrough algorithm.

One can construct block ciphers directly from one-way functions, which is a minimal assumption for much of crpyotgraphy, but the resulting construction will be terribly inefficient and hence not used.