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SUPERKEY

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My answer is 11 but in book given answer is 10.if i am wrong please give proper explanation.

as i am getting

2^3+2^2-2^0=11

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Didn't use any formula, instead went by the naive method.

Superkeys for A : A, AB, AC, AD, ABC, ABD, ACD, ABCD. So n(A) = 8

Superkeys for BC : BC, BCD, ABC, ABCD. So n(BC) = 4

Here, n(A $\cap$ BC) = 2 as ABC and ABCD are common superkeys

So taking union, we get

n(A $\cup$ BC) = 8 + 4 - 2 = 10
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