Answer
Left-hand side (LHS) of the equation is as follows:
\[\begin{align}
& \text{LHS}=A'\cap \left( B\cup C \right) \\
& =\left\{ \text{III,VI,VII,VIII} \right\}\cap \left( \left\{ \text{II,III,V,VI} \right\}\cup \left\{ \text{IV,V,VI,VII} \right\} \right) \\
& =\left\{ \text{III,VI,VII,VIII} \right\}\cap \left\{ \text{II,III,IV,V,VI,VII} \right\} \\
& =\left\{ \text{III,VI,VII} \right\}
\end{align}\]
Now, right-hand side (RHS) of the equation is as follows:
\[\begin{align}
& \text{RHS}=\left( A'\cap B \right)\cup \left( A'\cap C \right) \\
& =\left[ \left\{ \text{III,VI,VII,VIII} \right\}\cap \left\{ \text{II,III,V,VI} \right\} \right]\cup \left[ \left\{ \text{III,VI,VII,VIII} \right\}\cap \left\{ \text{IV,V,VI,VII} \right\} \right] \\
& =\left\{ \text{III,VI} \right\}\cup \left\{ \text{VI,VII} \right\} \\
& =\left\{ \text{III,VI,VII} \right\}
\end{align}\]
Since, the LHS of the equation is equal to RHS.
Thus, \[A'\cap \left( B\cup C \right)=\left( A'\cap B \right)\cap \left( A'\cap C \right)\]is a theorem.
Hence, \[A'\cap \left( B\cup C \right)=\left( A'\cap B \right)\cap \left( A'\cap C \right)\]is a theorem.
Work Step by Step
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