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A relation $R$ with $5$ attributes $A_1,A_2,A_3,A_4,A_5$. Given the following FDs

              $A_1\rightarrow A_2$

              $A_2A_3\rightarrow A_5$

              $A_4A_5\rightarrow A_1$

the number of candidates keys that includes attribute $A_3$ is ________________.
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3, A3A4A1,A3A4A2,A3A4A5
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A3 and A4 is independent attribute.
So A3 and A4 must be in Candidate Key.

(A3A4)= A3A4

(A1A3A4)= A1A2A3A4A5

(A2A3A4)= A1A2A3A4A5

​​​​​​​(A5A3A4)= A1A2A3A4A5

​​​​​​​Answer is 3.

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