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I want to prove that $(A\setminus B)\setminus(A \setminus C)=(A \cap C)\setminus B $, so I did this:

$ x \in (A \setminus B)\setminus(A \setminus C) \iff $

$ (x \in A \setminus B) \wedge (x \notin A \setminus C) \iff $

$ (x \in A \wedge x \notin B) \wedge (x \notin A \vee x \in C) \iff $

$ (x \in A \wedge x \notin B \wedge x \notin A) \vee (x \in A \wedge x \notin B \wedge x \in C) \iff $

(Now, the statement in the first bracket is always false, so the truth value of the whole disjunction in the line above depends only on the second bracket)

$x \in A \wedge x \notin B \wedge x \in C \iff $
$ x \in (A \cap C) \setminus B $

Is the method correct? Thanks in advance.

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  • 1
    $\begingroup$ Looks good to me. $\endgroup$
    – gt6989b
    Oct 7 '19 at 20:29
  • $\begingroup$ I can't find any errors. $\endgroup$ Oct 7 '19 at 20:32
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It's correct. A (not so) different way could be to set $U=A\cup B\cup C$ and denoting complementation with respect to $U$ by $X^c$: \begin{align} (A\setminus B)\setminus(A\setminus C) &=(A\cap B^c)\cap(A\cap C^c)^c \\ &=(A\cap B^c)\cap(A^c\cup C)\\ &=(A\cap B^c\cap A^c)\cup(A\cap B^c\cap C)\\ &=\emptyset\cup(A\cap B^c\cap C)\\ &=(A\cap C)\cap B^c\\ &=(A\cap C)\setminus B \end{align} One could also go in a different direction: $$ (A\setminus B)\setminus(A\setminus C)=(A\cap B^c\cap C)=(A\setminus B)\cap C $$

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