# Prove that $(A\cap C)-B=(C-B)\cap A$

$$\mathbf{Question:}$$ Prove that $$(A\cap C)-B=(C-B)\cap A$$

$$\mathbf{My\ attempt:}$$

Looking at LHS, assuming $$(A\cap C)-B \neq \emptyset$$

Let $$x\in (A\cap C)-B$$

This implies $$x\in A$$ and $$x\in C$$ and $$x\notin B$$

Looking at RHS, assuming $$(C-B)\cap A \neq \emptyset$$,

Let $$y \in (C-B)\cap A$$

This implies $$y\in C$$ and $$y\notin B$$ and $$y\in A$$

By comapring the LHS and RHS, we find that: $$x,y\in A$$

$$x,y\in C$$

$$x,y\notin B$$

Thus LHS = RHS.

Is this correct?

• I think you made a small error in the first step of the LHS: you say $x\in A$ and $x\in C$ and $x\notin C$, which is a contradiction. I think you meant $x\notin B$. – scoopfaze Jun 26 at 21:44
• Ah yes, fixed the error. Thanks. – Anubis Jun 26 at 21:46
• in the fifth line it should be $x\notin B$ not $C$ – alphaomega Jun 26 at 21:48
• and yes it's correct – alphaomega Jun 26 at 21:48
• You should convince yourself that you can prove this without writing a word down. – ε-δ Jun 26 at 21:50

\begin{align}&(A\cap C)\smallsetminus C \\ =~&\{x:(x\in A\wedge x\in C)\wedge x\notin B\}\\=~&\{x:(x\in C\wedge x\notin B)\wedge x\in A\}\\=~&(C\smallsetminus B)\cap A\end{align}
\begin{align*} (A\cap C) - B = (A\cap C)\cap\overline{B} = (C\cap\overline{B})\cap A = (C - B)\cap A \end{align*}