Proving Symmetric Difference of A and B

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

• Do you know of DeMorgan's Laws ? – babemcnuggets Feb 3 at 17:49
• I do know DeMorgan Laws – Sascha816 Feb 3 at 17:51
• Then use it on $(A\cup B) \setminus (A\cap B)$. – babemcnuggets Feb 3 at 17:53

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.

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?

• 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 – Sascha816 Feb 3 at 18:05
• 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. – user458276 Feb 3 at 18:12
• 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. – Sascha816 Feb 3 at 18:45

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

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

$$A \Delta B =$$

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

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

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

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

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

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