Prove that |A⊕B|≥|A−B||A \oplus B| \geq |A-B|

Let AA and BB be finite sets. Prove that |A⊕B|≥|A−B|.|A \oplus B| \geq |A-B|.

I first found that A⊕B=(A∪B)−(A∩B)=(A∪B)∩(¯A∪¯B)=((A∪B)∩¯A)∪((A∪B)∩¯B)=((¯A∩B)∪∅)∪((¯B∩A)∪∅)=(¯A∩B)∪(¯B∩A)\begin{align*}A \oplus B &= (A \cup B)-(A \cap B)\\&=(A \cup B) \cap (\overline{A} \cup \overline{B})\\&=((A \cup B) \cap \overline{A}) \cup ((A \cup B) \cap \overline{B})\\&=((\overline{A} \cap B) \cup \emptyset) \cup ((\overline{B} \cap A) \cup \emptyset)\\&=(\overline{A} \cap B) \cup (\overline{B} \cap A)\end{align*} and so |A⊕B|≥|(A∩¯B)|=|A−B||A \oplus B| \geq |(A \cap \overline{B})| =|A-B|.

Is there a simpler way of thinking about it or is this the correct way?

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1 Answer
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If you want to prove it algebraically like you did, there is (I am pretty sure) no essentially simpler way. However, it is trivial to show directly from the definitions
A−B={a∣a∈A∧a∉B}A⊕B={a∣(a∈A∧a∉B)∨(a∉A∧a∈B)}
A – B =\{a \mid a \in A \land a \notin B\} \qquad\qquad A \oplus B = \{a \mid (a \in A \land a \notin B) \lor (a \notin A \land a \in B)\}

that A−B⊆A⊕BA – B \subseteq A \oplus B; namely, if a∈A−Ba \in A – B, then a∈Aa \in A and a∉Ba \notin B, whence a∈A⊕Ba \in A \oplus B.