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the statement is true, but analyzing completely this question, there are so many cases, therefore it depends upon the relation

if there is no common attributes ===>it is in 1 NF

if common attributes are keys in the relation ===>it is in 3 NF

R1(A,B,C) and R2(A,B,D) and AB is the only key in both R1 and R2 ===> R(A,B,C,D) IS IN 3NF, if R1 and R2 are in 3NF

if common attributes are keys in only one of the relation ===>it is in 2 NF

and many more cases....

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