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Consider the schema $R=(S,T, U, V)$ and the dependencies $S \rightarrow T, T \rightarrow U, U \rightarrow V$ and $V \rightarrow S$. Let $R = (R1\text{ and } R2)$ be a decomposition such that $R1 \cap R2 \neq \phi$. The decomposition is

  1. not in $\textsf{2NF}$
  2. in $\textsf{2NF}$ but not $\textsf{3NF}$
  3. in $\textsf{3NF}$ but not in $\textsf{2NF}$
  4. in both $\textsf{2NF}$ and $\textsf{3NF}$
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Considering the given functional dependencies, it is observed that S,T,U,V, are prime attributes. Given, R1∩R2≠ϕ Let’s consider R1(S,T,U) and R2(U,V). Since, every attribute is a prime attribute thus all functional dependencies are satisfying the condition of 3NF, that X→Y, where Y should be a prime attribute or X should be a super key.
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SINCE EVERY ATTRIBUTE IS PRIMARY KEY IT IS IN 3NF . WITH LOSSLESS JOIN AND FD PRESERVED 

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The condition is R1 ∩ R2 ≠ ϕ ,So they have something in common and every variable over here is a key ,so however we divide there will be a intersection and that intersection will be a key   ,so the  decomposition will be lossless and every variable is key  so the highest possible is BCNF
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