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Let $A$, $B$ and $C$ be three non empty sets. Consider the two relations given below:

$$\begin{array}{lll} A-(B-C)=(A-B) \cup C & & (1) \\ A – (B \cup C) = (A -B)-C & & (2) \end{array}$$

- Both $(1)$ and $(2)$ are correct
- $(1)$ is correct but $(2)$ is not
- $(2)$ is correct but $(1)$ is not
- Both $(1)$ and $(2)$ are incorrect

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$LHS=A-(B-C)= A-(B \cap C') = A \cap (B \cap C')'=A\cap(B' \cup C)=(A \cap B') \cup (A \cap C)$

$RHS=(A-B)\cup C= (A \cap B') \cup C= (A \cup C)\cap (B' \cup C)$

Thus $(1)$ is wrong.

$LHS=A-(B\cup C)= A \cap (B \cup C)' = A \cap (B' \cap C')=A \cap B' \cap C'$

$RHS=(A-B)-C= (A \cap B') - C = (A \cap B') \cap C'=A \cap B' \cap C'$

Thus $(2)$ is correct.

$\therefore$ Option $C.$ is correct.

$RHS=(A-B)\cup C= (A \cap B') \cup C= (A \cup C)\cap (B' \cup C)$

Thus $(1)$ is wrong.

$LHS=A-(B\cup C)= A \cap (B \cup C)' = A \cap (B' \cap C')=A \cap B' \cap C'$

$RHS=(A-B)-C= (A \cap B') - C = (A \cap B') \cap C'=A \cap B' \cap C'$

Thus $(2)$ is correct.

$\therefore$ Option $C.$ is correct.