$\mathbf{Question:}$ 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?

  • $\begingroup$ 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 $x\notin B$. $\endgroup$
    – scoopfaze
    Jun 26 '20 at 21:44
  • $\begingroup$ Ah yes, fixed the error. Thanks. $\endgroup$
    – Anubis
    Jun 26 '20 at 21:46
  • $\begingroup$ in the fifth line it should be $x\notin B$ not $C$ $\endgroup$
    – alphaomega
    Jun 26 '20 at 21:48
  • 1
    $\begingroup$ and yes it's correct $\endgroup$
    – alphaomega
    Jun 26 '20 at 21:48
  • $\begingroup$ You should convince yourself that you can prove this without writing a word down. $\endgroup$ Jun 26 '20 at 21:50

\begin{align*} (A\cap C) - B = (A\cap C)\cap\overline{B} = (C\cap\overline{B})\cap A = (C - B)\cap A \end{align*}


Aside from the typo, yes. In short.

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


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