If A and B are two sets,then A cap (A cup B)' | vedclass

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(C) We are given the expression $A \cap (A \cup B)'$. Using De Morgan's Law,we know that $(A \cup B)' = A' \cap B'$. Substituting this into the expres

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$\phi$

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(C) We are given the expression $A \cap (A \cup B)'$. Using De Morgan's Law,we know that $(A \cup B)' = A' \cap B'$. Substituting this into the expression,we get $A \cap (A' \cap B')$. By the associative property of intersection,this is equal to $(A \cap A') \cap B'$. Since $A \cap A' = \phi$ (the empty set),the expression becomes $\phi \cap B'$. Since the intersection of any set with the empty set is the empty set,we have $\phi \cap B' = \phi$. Therefore,$A \cap (A \cup B)' = \phi$.

If $A$ and $B$ are two sets,then $A \cap (A \cup B)'$ is equal to

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