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Does the above mean:

  1. $x$ is in $X$ but [$x$ is not in $Y$ or $x$ is not in $Z$]

OR

  1. $x$ is in $X$ but [$x$ is not in $Y$ and $x$ is not in $Z$]

?

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    $\begingroup$ The second one is right. $\endgroup$ Oct 23, 2023 at 8:28
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    $\begingroup$ [$x$ is in $X-(Y\cup Z)$] means [$x$ is in $X$ and not ($x$ is in $Y$ or $x$ is in $Z$)] $\endgroup$ Oct 23, 2023 at 10:01

3 Answers 3

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See the image of your expression in a Venn diagram: Illustration of a union of two sets subtracting from a third set

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Recall that

$\cup$ is the set-theoretic counterpart of the logical operator $\vee$ (OR)

$\cap$ is the set-theoretic counterpart of the logical operator $\wedge$ (AND)

You can also use this identity $ A-B =A\cap B^C $ which just spells out the meaning of the set difference ( to be in the first one AND not in the second one) and De Morgan's identity to translate it to intersection, unions and complements:

$E:=X-(Y \cup Z)=X \cap(Y \cup Z)^C=X \cap (Y^C \cap Z^C)=X\cap Y^C \cap Z^C$

so $x \in E$ means $x $is in $X$ but not in $Y $ and not in $ Z $

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    $\begingroup$ Why the down vote? I don't think there is anything incorrect here $\endgroup$ Oct 23, 2023 at 9:03
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    $\begingroup$ I didn't downvote, but one pedagogical issue I see is that in most modern axiomatizations of set theory such as ZF(C) there is no such thing as "the complement of a set" (or at least such a thing, if defined, cannot itself be a set). While the notation $X^C$ is sometimes used as a shorthand for $U \setminus X$, where $U$ is some "universal set" containing all the elements of all sets that we're interested in, this shorthand really only makes sense in contexts where such a universal set $U$ is provided. And you haven't done that here. $\endgroup$ Oct 23, 2023 at 22:13
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When in doubt, simplify.

Let's give the set $Y \cup Z$ a name; I'll call it $S$, just to pick some arbitrary letter. The set $X - (Y \cup Z)$ is then equal to $X - S$. Thus, an element $a$ is in $X - S$ if and only if $a$ is in $X$ and $a$ is not in $S$.

Meanwhile, given that $S$ is the union of $Y$ and $Z$$a$ is in $S$ if and only if $a$ is in $Y$ or $a$ is in $Z$.

Putting these together, we can see that $a$ is in $X - (Y \cup Z)$ (which is the same as $X - S$) if and only if $a$ is in $X$ and not ($a$ is in $Y$ or $a$ is in $Z$). Or, in more compact notation: $$\begin{aligned} a \in X - (Y \cup Z) &\iff a \in X - S \\ &\iff a \in X \land \lnot (a \in S) \\ &\iff a \in X \land \lnot (a \in Y \lor a \in Z). \end{aligned}$$


Note that, while this is the simple answer to your question, there are also other correct answers. For example, if you like, you can apply De Morgan's law $\lnot (P \lor Q) \iff \lnot P \land \lnot Q$ to the answer above to convert the subexpression $\lnot (a \in Y \lor a \in Z)$ into the equivalent form $\lnot (a \in Y) \land \lnot (a \in Z)$, written more compactly as $a \notin Y \land a \notin Z$, giving $$a \in X - (Y \cup Z) \iff a \in X \land a \notin Y \land a \notin Z$$ or, using English words instead of logical symbols, "$a$ is in $X - (Y \cup Z)$ if and only if $a$ is in $X$ and $a$ is not in $Y$ and $a$ is not in $Z$."

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