MHB Proving $A\cap (B - C) = (A\cap B) - (A\cap C)$

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The discussion focuses on proving the set identity $A\cap (B - C) = (A\cap B) - (A\cap C)$. The proof involves demonstrating both inclusions: first, showing that $A\cap (B - C)$ is a subset of $(A\cap B) - (A\cap C)$ by establishing that elements in $A$ and $B$ are not in $C$. The second part confirms that $(A\cap B) - (A\cap C)$ is contained in $A\cap (B - C)$, utilizing the complement of $C$. The approach is validated through set operations and logical reasoning, confirming the identity holds true. This proof effectively illustrates the relationship between the sets involved.
Dustinsfl
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$A\cap (B - C) = (A\cap B) - (A\cap C)$For the identity, we will show $A\cap (B - C) \subseteq (A\cap B) - (A\cap C)$ and $A\cap (B - C) \supseteq (A\cap B) - (A\cap C)$.
Let $x\in A\cap (B - C)$.
Then $x\in A$ and $x\in B - C$.
So $x\in A$ and $x\in B$ and $x\notin B\cap C$.

Is this the right approach? I know $B-C = $ some other expression but I can't remember it.
 
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B-C = B \cap C^c
C^c it is the complement of C, About you work it is right since x in A and in B so it is in the intersection and x it is not in C so it is not in the intersection of A and C

now the other direction
 
dwsmith said:
$A\cap (B - C) = (A\cap B) - (A\cap C)$
$\begin{align*}(A\cap B)-(A\cap C)&= (A\cap B)\cap(A^c\cup C^c)\\&=(A\cap B\cap A^c)\cup (A\cap B\cap C^c) \\&=\emptyset \cup (A\cap B\cap C^c)\\&= (A\cap B\cap C^c)\\&=(A\cap B)-C\end{align*}$
 

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