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"When is the set statement:

(A\B)⊕(A ∩ B) = A

true? Is it sometimes true, never true, or always true? If sometimes, state the specific cases where it is. A & B are arbitrarily selected sets."

I said it was sometimes, as, it seems if A is empty and B is empty, then A\B is empty, A ∩ B is empty, and so the entire left side is then empty, and so is A. Would that be correct? Further, are there are other cases where this is a true statement?

Thank you for your time. ^^

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    Alrighty. I'll do so.2012-09-10

3 Answers 3

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If $A=\varnothing$, it doesn’t matter what $B$ is: $A\setminus B=\varnothing\setminus B=\varnothing$, and $A\cap B=\varnothing\cap B=\varnothing$, so $(A\setminus B)\oplus(A\cap B)=\varnothing\oplus\varnothing=\varnothing=A$.

But in fact it’s always true, for any sets $A$ and $B$. If $x\in A\cap B$, then $x\in B$, so $x\notin A\setminus B$. Conversely, if $x\in A\setminus B$, then $x\notin B$, so $x\notin A\cap B$. Thus,

$\begin{align*} (A\setminus B)\oplus(A\cap B)&=\Big((A\setminus B)\setminus(A\cap B)\Big)\cup\Big((A\cap B)\setminus(A\setminus B)\Big)\\ &=(A\setminus B)\cup(A\cap B)\\ &=A\;. \end{align*}$

If you’re in doubt about that last step, notice that $A\setminus B$ consists of the things that are in $A$ but not in $B$, while $A\cap B$ consists of those things that are in both $A$ and $B$, so between them they pick up every element of $A$.

More generally, the symmetric difference of two disjoint sets is always their union: if $X\cap Y=\varnothing$, then $X\oplus Y=X\cup Y$. Here the disjoint sets are $A\setminus B$ and $A\cap B$, and their union is $A$.

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    Thank you very much. It makes perfect sense, now. ^^2012-09-10
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We can easily demonstrate this by starting with the most complex (left hand) side, expanding the definitions (see related answer), and simplifying: for every $\;x\;$, \begin{align} & x \in (A \setminus B) \oplus (A \cap B) \\ \equiv & \;\;\;\;\;\text{"definition of $\;\oplus\;$"} \\ & x \in A \setminus B \;\not\equiv\; x \in A \cap B \\ \equiv & \;\;\;\;\;\text{"definitions of $\;\setminus, \cap\;$"} \\ & x \in A \land x \not\in B \;\not\equiv\; x \in A \land x \in B \\ \equiv & \;\;\;\;\;\text{"move $\;\lnot\;$ to the outside -- to prepare for the next step"} \\ & \lnot(x \in A \land x \not\in B \;\equiv\; x \in A \land x \in B) \\ \equiv & \;\;\;\;\;\text{"factor $\;x \in A\;$ out of $\;\equiv\;$"} \\ & \lnot(x \in A \;\Rightarrow\; (x \not\in B \;\equiv\; x \in B)) \\ \equiv & \;\;\;\;\;\text{"logic: contradiction"} \\ & \lnot(x \in A \;\Rightarrow\; \text{false}) \\ \equiv & \;\;\;\;\;\text{"logic: simplify"} \\ & x \in A \\ \end{align} By set extensionality, this proves the statement in question.

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let $A,B \subset X$ and denote $(\forall Y), \bar Y=X \setminus Y$, then $(A\setminus B)\oplus(A\cap B) = (A \cap \bar B)\oplus(A\cap B) = A \cap(\bar B \oplus B) = A \cap X = A $

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    took a while to get that right! dodgy trying to do math whilst on the phone. but it is a fairly simple derivation that is worth considering2013-12-17