Show that $(A \cup B)-B=A$ is false. Why is my method wrong? 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?
 A: 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 $$(1+2)\times 3=9\color{red}{\not=}7=1+(2\times 3).$$
A: $(A \cup B)-B = (A \cup B)\cap \bar{B} = (A \cap \bar{B})\cup(B\cap \bar{B}) = A \cap \bar{B}$
A: 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 $A \subset C$.
A: 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. 
