# What exactly does $X - (Y ∪ Z)$ mean?

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 $$Z$$]

?

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

See the image of your expression in a Venn diagram:

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 $$Z$$

• Why the down vote? I don't think there is anything incorrect here Oct 23, 2023 at 9:03
• 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. Oct 23, 2023 at 22:13

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 $$Z$$."