(Discrete math) Prove sets Can someone help me prove that: $$((A \cap B)  \oplus  A)' = A' \cup B$$
$A'$ is $A$ complement.
Thank you.
 A: 
$$p⊕q\equiv (p\land\neg q)\lor (\neg p\land q)\equiv(p\lor q)\land(\neg p\lor\neg q)\tag*{aka Xor}$$

This set notation '$⊕$' corresponding to xor in logic.
Any $x\in ((A\cap B)⊕A)'$ if and only if:$$\neg((x\in A\land x\in B)⊕x\in A)$$
Since $\neg((p\land q)⊕ p)\leftrightarrow(\neg p\lor q)$ is a tautology,
It's clearly equivalent to $x\not\in A\lor x\in B$, hence proved$\dots$
But if you can't say this, here is a proof with Logical equivalence:
Apply def. of Xor:
$$\neg(((x\in A\land x\in B)\land x\not\in A)\lor(\neg(x\in A\land x\in B)\land x\in A))$$
Apply Commutative law & Associative law:
$$\neg(((x\in A\land x\not\in A)\land x\in B)\lor(\neg(x\in A\land x\in B)\land x\in A))$$
Apply Negation law:
$$\neg((\bot\land x\in B)\lor(\neg(x\in A\land x\in B)\land x\in A))$$
Apply Domination law:
$$\neg(\bot\lor(\neg(x\in A\land x\in B)\land x\in A))$$
Apply Identity law:
$$\neg(\neg(x\in A\land x\in B)\land x\in A)$$
Apply De Morgan's law:
$$\neg((x\not\in A\lor x\not\in B)\land x\in A)$$
Apply Distributive law:
$$\neg((x\not\in A\land x\in A)\lor (x\not\in B\land x\in A))$$
Apply Negation law:
$$\neg(\bot\lor (x\not\in B\land x\in A))$$
Apply Domination law:
$$\neg(x\not\in B\land x\in A)$$
Apply De Morgan's law:
$$x\in B\lor x\not\in A$$
Apply Commutative law:
$$x\not\in A\lor x\in B$$
This hold if and only if $x\in A'\cup B$ 
Hence we proved $((A\cap B)⊕A)'=A'\cup B\tag*{$\square$}$
A: Note that $\oplus$ denotes the symmetric difference between two sets. In other words, 

$$P \oplus Q = (P-Q) \cup (Q-P) = (P  \cap Q') \cup (Q \cap P')$$

Using this definition as well as set identities, we have the following proof:
$ \big( (A \cap B) \oplus A \big)' $
$= \bigg[ \big( (A \cap B) \cap A' \big) \cup \big( A \cap (A \cap B)' \big) \bigg]' $ ----- by definition of $\oplus$
$= \bigg[ \big( (B \cap A) \cap A' \big) \cup \big( A \cap (A \cap B)' \big) \bigg]' $ ----- by the commutative law
$= \bigg[ \big( B \cap (A \cap A') \big) \cup \big( A \cap (A \cap B)' \big) \bigg]' $ ----- by the associative law
$= \bigg[ ( B \cap \emptyset ) \cup \big( A \cap (A \cap B)' \big) \bigg]' $ ----- by the complement law
$= \bigg[ \emptyset \cup \big( A \cap (A \cap B)' \big) \bigg]' $ ----- by the domination law
$= \bigg[ A \cap (A \cap B)' \bigg]' $ ----- by the identity law
$= A' \cup (A \cap B)'' $ ----- by DeMorgan's law
$= A' \cup (A \cap B) $ ----- by the double complement law
$= (A' \cup A) \cap (A' \cup B) $ ----- by the distributive law
$= U \cap (A' \cup B) $ ----- by the complement law
$= A' \cup B $ ----- by the identity law
NOTE: As @Manx has noted, you may also use the definitions of various set operations to expand these statements about sets and then apply the rules of inference. For simplicity, I chose to preserve the set notation and utilize set identities, which I assume you are familiar with.
