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Let A and B be sets. Define the symmetric difference of A and B as A∆B= (A ∪ B) − (A ∩ B). (a) Prove that A∆B = (A − B) ∪ (B − A)

I tried to start this but am getting really lost. if someone could try to help that would be great

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  • $\begingroup$ Do you know of DeMorgan's Laws ? $\endgroup$ – babemcnuggets Feb 3 at 17:49
  • $\begingroup$ I do know DeMorgan Laws $\endgroup$ – Sascha816 Feb 3 at 17:51
  • $\begingroup$ Then use it on $(A\cup B) \setminus (A\cap B)$. $\endgroup$ – babemcnuggets Feb 3 at 17:53
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If you are new to elementary set-theory the best way to prove this was given by user247327. However, if not then normally you should not make this any more difficult than it is.

Let $ A∆B=(A \cup B) \setminus (B \cap A) $. Remember DeMorgan: $A\setminus(B \cup C)= (A\setminus B) \cap (A\setminus C) $

We come down to :

$$ (A\cup B)\setminus (B \cap A)=((A\cup B)\setminus B)\cup ((A\cup B)\setminus A) $$
If you see what $(A\cup B)\setminus B$ and $(A\cup B)\setminus A$ is you are done.

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Hint: If $x \in A\Delta B$, then x is an element that’s in A or B, but not both. Compare this to $(A - B)\cup(B-A)$. Would $x$ be in this set as well?

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  • $\begingroup$ I did the proof up to the point to show that x is an element of A or B but not an element of A and B. I am now stuck on how to use that to show that its A-B or B-A $\endgroup$ – Sascha816 Feb 3 at 18:05
  • $\begingroup$ Since x is in A, but not B (or B, but not A), it’s in A-B (respectively B-A). The possibilities are that x is in A and not in A (a contradiction), x is in A and not in B (which is fine), or the dual case where x is in B. $\endgroup$ – user458276 Feb 3 at 18:12
  • $\begingroup$ that was very helpful thank you. Now I am trying to go the other way and prove AΔB. I got that x is not an element of A or x is an element of B by DeMorgan's Law and am stuck from there. $\endgroup$ – Sascha816 Feb 3 at 18:45
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The most direct way (not always the simplest) to prove that "$X= Y$", for sets, is to prove both $X\subseteq Y$ and $Y\subseteq X$. And to prove "$X\subseteq Y$" start with "if $x\in X$" then use the properties of sets X and Y to conclude "then $x\in Y$".

Here the two sets are $X= (A\cup B)- (A\cap B)$ and $Y= (A- B)\cup (B- A)$. If $x\in (A\cup B)- (A\cap B)$ then $x\in A\cup B)$ and $x\notin A\cap B$. That means x is in either A or B but not both.
Case 1: Suppose x is in A but not in B. Then x is in $A- B$ so in $(A- B)\cup (B- A)$. Case 2: Suppose x is in B but not in A. Then x is in $B- A$ so in $(A- B)\cup (B- A)$. In either case, $(A\cup B)- (A\cap B)\subseteq (A- B)\cup (B- A)$.

Now the other way- Suppose $x\in (A- B)\cup (B- A)$. Then either Case 1: $x\in A- B$. Then $x\in A\cup B$ but not in $A\cap B$ so $x\in (A\cup B)- (A\cap B)$. Case 2: $x\in B= A$. Then, again, $x\in A\cup B$ but not in $A\cap B$ so $x\in (A\cup B)- (A\cap B)$. In either case, $\cup (B- A)\subseteq (A\cup B)- (A\cap B)$.

Therefore, $(A\cup B)- (A\cap B)= (A- B)\cup (B- A)$.

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Using that $A - B = A \cap B^C$:

$A \Delta B = $

$(A \cup B) - (A \cap B) = $

$(A \cup B) \cap (A \cap B)^C =$

$ (A \cup B) \cap (A^C \cup B^C) =$

$ (A \cap (A^C \cup B^C)) \cup (B \cap (A^C \cup B^C)) = $

$(A \cap B^C) \cup (B \cap A^C) = $

$(A - B) \cup (B - A)$

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