一个关于关系数据库的关系模式证明

虽然觉得楼主好二,但是还是把答案给你吧,英文的哦Exercise 19.24 Prove that,if R has only one key,it is in BCNF if and only if it is in
3NF.
Answer 19.24 Let F (F+) denote the (closure of the) set of functional dependencies
satisfied by the schema R which is assumed to be in 3NF.We need to show that for
each nontrivial dependency X → A in F+,X is a superkey.To this end,consider such
a dependency.If X is not a superkey,the 3NF property guarantees that the attribute
A is part of a key.Since all keys are simple by assumption,we have that A is a key.
This last fact together with the dependency X → A implies that X is a superkey (this
follows,from the transitivity axiom) which is a contradiction.