Let us take R , S , T be decomposed subrelations of A.So to check lossless property , we do:
a) Let us start by taking R ∩ S.If common attribute exist in this intersection and that is the superkey of either R or S or both , then we take natural join of R and S and then proceed.We take now intersection of T and join of R and S and hence find common attribute.If that is not superkey of T or R join S , then we have to start with other pair and repeat the same procedure.
Else we can terminate and can conclude that it is a lossless join decomposition.For the 1st case , that is failure of finding superkey , we have to start with other pair ,e.g. we start with intersection of S and T.
b) If in the above case ,where we have started taking intersection of R and S,suppose we get a common set of attribute which is not a superkey for either R or S,then we stop right there and start with other pair , e.g. start taking the common attribute of S and T first and so on.
So in total we have to consider 3 cases here :
i) start with R,S
ii) start with S,T
iii) start with R,T
If we fail on getting common attribute at last following the procedure described above which is superkey , then we can conclude that the decomposition is lossy.
I hope I have addressed your doubt.