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So the textbook uses a counter example to show this which is pretty simple. I tried playing around with the algebra. Ie.

$(A \cup B)-B$ is equal to $(A \cup B)\cap \bar{B}$ and associative law says this is equal to $A \cup (B\cap \bar{B})$. This results in $A \cup \emptyset$ which equals to $A$.

What am I doing wrong here?

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    $\begingroup$ Let $A=B\ne\emptyset.$ $\endgroup$
    – user296113
    May 13, 2016 at 19:40
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    $\begingroup$ associative law? $\endgroup$ May 13, 2016 at 19:40
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    $\begingroup$ Associativity applies when the operations are all unions or all intersections. Things are trickier when you mix them. $\endgroup$ May 13, 2016 at 19:46

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The "associative law" you use doesn't actually hold.

For example, take $A=\{1\},B=\{1\}, C=\{2\}$. Then $(A\cup B)\cap C=\emptyset$, but $A\cup (B\cap C)=\{1\}$.


EDIT: Even though $\cup$ and $\cap$ are individually associative, that doesn't mean you can combine the associative laws. For a more concrete example of this, consider $+$ and $\times$: they're each associative, but $$(1+2)\times 3=9\color{red}{\not=}7=1+(2\times 3).$$

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$(A \cup B)-B = (A \cup B)\cap \bar{B} = (A \cap \bar{B})\cup(B\cap \bar{B}) = A \cap \bar{B}$

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  • $\begingroup$ That's not the associative law, it's the distributive law. $\endgroup$ May 13, 2016 at 19:43
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That $\cup$ is associative means that for any $A, B, C$ you have $(A \cup B) \cup C = A \cup (B \cup C)$.

That $\cap$ is associative means that for any $A, B, C$ you have $(A \cap B) \cap C = A \cap (B \cap C)$.

Neither of these laws of associativity means that for any $A, B, C$ you get $(A \cup B) \cap C = A \cup (B \cap C)$. In fact, this equality is wrong in general, it is true if and only if $A \subset C$.

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consider space{1,2,3,4,5) A = {1,2,3}, B = {1,2,3,4}.$(A \cup B)$={1,2,3,4}=B $(A \cup B)-B = (A \cup B)\cap \bar{B} = {1,2,3,4}\cap \bar{5}=$ {}

in short this doesn't hold true when the intersection of A and B is empty.

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