I read a statement "Relation R in BCNF with at-least one simple candidate key is also in 4NF"
I don't think that it is always true but I am not able to prove it.
Can someone please help ?
The statement is true and this is the sketch of the proof taken from the paper "Simple Conditions for Guaranteeing Higher Normal Forms in Relational Databases", by C.J.Date and R.Fagin, ACM TODS, Vol.17, No. 3, Sep. 1992.
A relation is in 4NF if, for every nontrivial multivalued dependency X →→ Y in F+, X is a superkey for R. So, if a relation is in BCNF, but not in 4NF, then there must exists a nontrivial multivalued dependency (MVD) X →→ Y such that X is not the key. We will show that this is in contradiction with the fact that the relation is in BCNF and has a candidate key K constituted by a unique attribute (simple candidate key).
Consider the fact that, in a relation R(T), when we have a nontrivial MVD X →→ Y, (assuming, without loss of generality that X and Y are disjoint), then also the MVD dependency X →→ Z must hold in the same relation, with Z = T - X - Y (that is Z are all the other attributes of the relation). We can now prove that each candidate key must contain at least an attribute of Z and an attribute of Y (so it must contain at least 2 attributes!).
Since we have X →→ Y and X →→ Z, and X is not a candidate key, assume that the hypothesis is false, that is that there is a candidate K which does not contain a member of Y (and for symmetry, neither a member of Z). But, since K is a key, we have that K → Y, with K and Y disjoint.
Now, there is an inference rule that says that, in general, if V →→ W and U → W, where U and W are disjoint, then V → W.
Applying this rule to our case, since X →→ Y, and K → Y, we can say that X → Y. But this is a contradiction, since we have said that R is in BCNF, and X is not a candidate key.
In other words, if a relation is not in 4NF, than each key must have at least 2 attributes.
And given the initial hypothesis, that we have a relation in BCNF with at least a simple candidate key, for the previous lemma, the relation must be in 4NF (otherwise every key should be constituted by at least 2 attributes!).