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does $B - (A \cup C) = B \cup (A' \cup C')$?

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    To render the math you use dollar signs `$` and symbols like setminus are used with backslash `\`. I've edited your post - is this what you originally wanted to write?2012-10-30

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Hint: To see that this is not true, take $M=\{a,b,c,d,e,f,g\}$ as the mother set and $A=\{a,b\}, B=\{a,c,d\}, C=\{c,e\}$ and evaluate both sides of your so-called equation. For giving a formal set theatrical fact use @Martin's answer.

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    @MathildaPitt: The union of the mother set $U$ and any subset of it, say $B$ is always the whole set $U$, but the intersection is always the small set.2012-10-30
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$B-(A\cup C)=B\cap(A\cup C)'$ (this is basically the definition)

$(A\cup C)'=A'\cap C'$ (by de Morgan)

Hence $B-(A\cup C)=B\cap A'\cap C'$.

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    $X\setminus Y$ denotes the [relative complement](http://en.wikipedia.org/wiki/Complement_%28set_theory%29) or difference of two sets. It consists of elements which belong to $X$ but not to $Y$. In this case $x$ belongs to $B\setminus(A\cup C)$ if it belongs to $B$ and it doesn't belong to $A\cup C$. Which is the same thing as you wrote. An it is the same thing as saying that $x$ belongs to $B$ and to the complement of $A\cup C$, i.e. $x\in B\cap (A\cup C)'$.2012-10-30