Proof that if a relation R is in 3NF and if every key in R is simple, then R is in BCNF?
Interesting question (I happen to be reading the Ullman book right now). I think the answer is something like this:

Assume that X is not a superkey. Then, each attribute in Y is a member of a key (is prime). Since each key consists of one attribute, y -&gt; {every other attribute}, where y is a member of Y. Then by transitivity, X -&gt; {every other attribute}, which implies that X is a superkey. We contradicted our original assumption, that X is not a superkey, hence X must be a superkey. 

The most important part of the proof is the transitivity part.