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The definition of a symmetric difference of two sets, that my book provides, is: Set containing those elements in either $A$or $B$, but not in both $A$ and $B$.

So, in set builder notation, I figured that it would be $A⊕B = \{x| (x \in A \vee x \in B) \wedge (x \notin A \wedge x \notin B)\}$

Is this correct? I have some properties of symmetric differences of sets to prove, and I wanted to prove them by subsets, and by using definitions and laws of logic.

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up vote 6 down vote accepted

As written, you say that $x$ is in at least one of them, and $x$ is in neither of them. In other words, you've defined the empty set. To fix it, you could change $$x\notin A\wedge x\notin B$$ to $$x\notin A\vee x\notin B.$$

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Oh, yes. I see the mistake, now. – Mack Mar 4 '13 at 15:17

Taking your first sentence and putting it in symbols we have $A\oplus B=A\cup B- A\cap B$. Now for any sets $X$ and $Y$, the set difference $X-Y$ put in set-builder notation is $\{x\in X: x\not\in Y\}$. So $$A\oplus B=\{x\in A\cup B:x\not\in A\cap B\}=\{x\in A\cup B:x\not\in A\vee x\not\in B\}.$$

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Notation $\oplus$ is not apropriate for symmetric difference of sets we must use $\Delta$ or $\ominus$ $$A\Delta B=(A\cup B)\setminus(A\cap B)=(A\setminus B)\cup(B\setminus A $$

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I have seen $\oplus$ used for symmetric difference. I'm pretty sure it's fine. – Andrew Salmon Mar 4 '13 at 15:21
according Wikipedia we can use $\ominus$ – Adi Dani Mar 4 '13 at 15:27

no. $A \oplus B = \{ x \in A \cup B \mid x \in A \setminus B \text{ or } x \in B \setminus A \}$.

The definition you gave could be fixed by changing $x \notin A \wedge x \notin B$ to $\lnot (x \in A \wedge x \in B)$

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Really, that's one way of defining it? It seems odd that the book wants me to prove that $A⊕B=(A−B)∪(B−A)$, seeing as it is the definition... – Mack Mar 4 '13 at 15:15
if a definition makes your proof trivial then it's probably a good definition.. on the other hand it might make something else harder to prove. – user58512 Mar 4 '13 at 15:20
So, if I wanted to prove that $A⊕B=(A−B)∪(B−A)$ is true, I could state that, by definition, it is true; and then give a little rationale on why it is true? – Mack Mar 4 '13 at 15:25

The most simple definition of set difference I know is $$ x \in A ⊕ B \;\;\equiv\;\; x \in A \;\;\not\equiv\;\; x \in B $$ for all $x$, $A$, and $B$, where $\equiv$ is logical equivalence (often written as $\Leftrightarrow$) and $\not\equiv$ is inequivalence ('exclusive or', occasionally written as $\not\Leftrightarrow$).

See another answer of mine for a practical application of this definition (but using the different symbol $\Delta$).

(Note, by the way, that $\equiv$ and $\not\equiv$ are both and mutually associative: $(P \equiv Q) \not\equiv R$ is equivalent to $P \equiv (Q \not\equiv R)$, so it is perfectly safe to leave out the parentheses.)

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