An excerpt from the wikipedia article on slide attacks states:
...The only requirements for a slide attack to work on a cipher is that it can be broken down into multiple rounds of an identical F function. This probably means that it has a cyclic key schedule. The F function must be vulnerable to a known-plaintext attack...
and:
...Once a slid pair is identified, the cipher is broken because of the vulnerability to known-plaintext attacks. The key can easily be extracted from this pairing....
The statements:
The next step is to collect 2^{n/2} plaintext-ciphertext pairs. Depending on the characteristics of the cipher fewer may suffice, but by the birthday paradox no more than 2^{n/2} should be needed.
lead me to think that even a truly random oracle would produce slid pairs, so it would appear that eliminating the existence of slid pairs is not a possibility.
My question is relatively simple. Suppose the round function of a cipher was resistant to known plaintext attack (i.e. known plaintext does not facilitate recovery of key information). Could such a cipher claim provable resistance to slide attacks? If not, what advantage would a slid pair offer an attacker?