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I'm studying for my exam and I came up with this little proof, but I'm wary because the professor took a much longer approach. Am I right in saying that a symmetric difference is the same as the difference between a union and an intersection? Thanks in advance.

Suppose $A$ and $B$ are disjoint.

This means $A \cap B = \varnothing$

Since $A \bigtriangleup B$ is the set of all elements that belong to $A$ or $B$ but not to both, $$A \bigtriangleup B = A \cup B - A \cap B$$

$A \cap B = \varnothing$, therefore $A \bigtriangleup B = A \cup B$.

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    $\begingroup$ This is half of the proof. You are not done. And if the symmetric difference was defined differently (for instance, as $(A\setminus B)\cup(B\setminus A)$), then you still need to prove that it is union minus intersection. $\endgroup$ – Andrés E. Caicedo Dec 5 '14 at 6:26
  • $\begingroup$ @AndresCaicedo what would be the second half? $\endgroup$ – epon Dec 5 '14 at 8:59
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$A \Delta B=(A-B)\cup(B-A)$ $=(A\cap B^C)\cup(B\cap A^C)$ $=(A\cup B)\cap(A^C\cup B^C)$ $=(A\cup B)\cap(A\cap B)^C=(A\cup B)-(A\cap B)$

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You can just see the proprieties of definition of symmetrical difference If $A\cap B= \emptyset$ then $A\backslash B = A$ and $B\backslash A=B$, then you can say that the definition of symmetrical difference is $A\bigtriangleup B = (A\backslash B) \cup (B\backslash A)$ there for the initial condition you can say $A\bigtriangleup B = A\cup B$

The thing the professor might do it longer than yours is because of the arguments, but in essence its the same, but be sure to argument every set its a big deal in mathematics.

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$ \newcommand{\calc}{\begin{align} \quad &} \newcommand{\calcop}[2]{\\ #1 \quad & \quad \unicode{x201c}\text{#2}\unicode{x201d} \\ \quad & } \newcommand{\endcalc}{\end{align}} \newcommand{\ref}[1]{\text{(#1)}} \newcommand{\true}{\text{true}} \newcommand{\false}{\text{false}} \newcommand{\empty}{\varnothing} \newcommand{\diff}{\mathbin \triangle} $This is not a direct answer to your question, but an alternative proof in a different style.

Here is (the beginning of) a 'logical' way to prove your $$ \tag 0 A \cap B = \empty \;\equiv\; A \cup B = A \diff B $$ i.e., by using the definitions to go to the element level, and using the laws of logic to simplify and complete the proof.

We will use the following not-too-well-known definition of $\;\diff\;$: for all $\;x\;$, $$ \tag 1 x \in A \diff B \;\equiv\; x \in A \not\equiv x \in B $$ and we use the fact that $\;\equiv\;$ and $\;\not\equiv\;$ are symmetric, associative, and even mutually associative.


We start at the most complex side of $\ref 0$, and calculate

$$\calc A \cup B = A \diff B \calcop\equiv{set extensionality, definition of $\;\cup\;$, definition $\ref 1$ of $\;\diff\;$} \langle \forall x :: x \in A \lor x \in B \;\equiv\; x \in A \not\equiv x \in B \rangle \calcop\equiv{logic: extract $\;\lnot\;$ -- to prepare for $\ref 2$, see below} \langle \forall x :: \lnot (x \in A \lor x \in B \;\equiv\; x \in A \;\equiv\; x \in B) \rangle \calcop\equiv{...} \endcalc$$


Now finish it yourself, by applying the 'golden rule' $$ \tag 2 P \;\equiv\; Q \;\equiv\; P \lor Q \;\equiv\; P \land Q $$ and the definitions of $\;\cap\;$ and $\;\empty\;$ to reach the left hand side of $\ref 0$.

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