Proof that $A \cap (B -C)=(A \cap B)-(A \cap C)$ I'm trying to prove the following:
$A \cap (B -C)=(A \cap B)-(A \cap C)$
My line of reasoning is going astray, can someone help point me to where I'm making my mistake(s)? I'm new to creating mathematical proofs so I suspect I my error may be pretty basic.
My line of reasoning is as follows:
$$\begin{align}A \cap (B -C)
&= \{ x \in \varepsilon:x \in A \wedge(x\in B \wedge x \notin C) \}
\\&= \{x \in \varepsilon: x \in A \wedge (x \in B \wedge x \in C') \}
\\&= \{x \in \varepsilon: (x \in A \wedge x \in B) \wedge x \in C' \}
\\&= \{x \in \varepsilon: (x \in A \cap B) \wedge x \in C' \}
\\&=(A\cap B) - C\end{align}$$
 A: A different approach from the proposed on the comments.
We are interested in proving that $A\cap(B - C) = (A\cap B) - C$.
Let us prove the inclusion $\subseteq$ first.
If $x\in A\cap(B - C)$, this means that $x\in A$ and $x\in B - C$, which implies that $x\in B$ e $x\not\in C$. Hence we can conclude that $x\in A\cap B$ and $x\not\in C$, that is to say, $x\in(A\cap B) - C$.
Conversely, let us prove the inclusion $\supseteq$ this time.
If $x\in(A\cap B) - C$, then $x\in A\cap B$ and $x\not\in C$. Such claim means that $x\in A$, $x\in B$ and $x\not\in C$. Consequently, we conclude that $x\in A$ and $x\in B - C$, which is exactly what we are interested in.
EDIT
Based on the modification of the proposed question, the new solution can be given by:
\begin{align*}
(A\cap B) - (A\cap C) & = (A\cap B)\cap(A\cap C)^{c}\\\\
& = (A\cap B)\cap(A^{c}\cup C^{c})\\\\
& = ((A\cap B)\cap A^{c})\cup((A\cap B)\cap C^{c})\\\\
& = \varnothing\cup(A\cap B\cap C^{c})\\\\
& = A\cap(B - C)
\end{align*}
and we are done.
Hopefully this helps!
A: $$\begin{align}
(A \cap B)-(A \cap C)&=(A \cap B)\cap(A \cap C)^c\\
\\
&=(B \cap A)\cap(A^c \cup C^c)\\
\\
&=B \cap (A\cap(A^c \cup C^c))\\
\\
&=B \cap (( A\cap A^c ) \cup (A\cap C^c))\\
\\
&=B \cap (\emptyset\cup(A\cap C^c))\\
\\
&=B \cap (A\cap C^c)\\
\\
&=A \cap (B\cap C^c)\\
\\
&=A \cap (B-C)\\
\\
\end{align}$$
