How do you prove the statement: $A^c=(A\cup B)^c\cup (B \setminus A)$ I've been stuck on this question for a while and the problem is basically I just don't know how to prove it, I tried converting it to $\vee$ and $\wedge$ symbols then I did some research and found mathematically it doesn't make sense to compare them, so I'm currently stuck on how to prove such a statement.
Question
Determine whether the following statement is true or false, if true prove it, if false, provide a counterexample.
$A^c=(A\cup B)^c\cup (B \setminus A)$
Working
$A^c=$ ~$A$, $(A \cup B)^c=($~$A$ $\wedge$ ~$B$), $(B\setminus A)=$ ~$A$ 
$\implies A^c=(A\cup B)^c\cup (B \setminus A)=$~$A =($~$A$ $\wedge$ ~$B)$ $\wedge$ ~$A$ $\rightarrow$ I then went on to show this was equal to ~$A$ $\wedge $ ~$B$ which doesn't show anything additionally it's mathematically incorrect from what I found.
Note
I tried finding a counterexample and couldn't find one, so I'm assuming it is true, and as I said that's where I'm stuck, I don't know how to prove it. ANY help would GREATLY be appreciated, thanks! :)
 A: The standard way to show that two sets are equal is to show that one is contained in the other. On the left, you have everything that is not in $A$. On the right, everything that is not in ($A$ or $B$) together with everything in $B$ but not $A$. So if $x$ is not in $A$ then show that either $x \in (A \cup B)^c$ or $x \in B \setminus A$ (which depends on whether or not $x \in B$). For the reverse inclusion, show that if $x \in (A \cup B)^c$ then $x \not\in A$ and if $x \in B \setminus A$ then $x \not\in A$.
You can also show this algebraically if you can justify each of the following equalities:
$$
\begin{align*}
(A \cup B)^c \cup (B \setminus A) &= (A^c \cap B^c) \cup (B \setminus A) \\
&= (A^c \cap B^c) \cup (B \cap A^c) \\
&= A^c \cap (B^c \cup B) \\
&= A^c \cap X \\
&= A^c
\end{align*}
$$
where $X$ is the whole set.
A: I like Venn diagrams for showing equality of sets. $A^c$ is the pink section in the left figure. $(A\cup B)^c$ is the darker green in the right figure, and $B-A$ is the lighter green in the right figure.

A: If an element is in $(A\cup B)^\complement\cup(B\setminus A)$ then it is either in $(A\cup B)^\complement$ or it is in $(B\cap A^\complement)$.   If it is in the first, then it is not in $A$, and if it is in the second then it is not in $A$.    (Because...)
So $(A\cup B)^\complement\cup(B\setminus A)\subseteq A^\complement$.
Likewise if an element is in $A^\complement$ it is in neither $(A\cup B)^\complement$ nor is it in $(B\cap A^\complement)$.   (Because...)   So it will not be in their union.
So $(A\cup B)^\complement\cup(B\setminus A)\supseteq A^\complement$.
Therefore $(A\cup B)^\complement\cup(B\setminus A) = A^\complement$.

Don't try to convert set algebra straight into boolean logic.   Although they are related concepts they are not exactly the same thing.
However, you can express it like this: $$\begin{align}x\in \big((A\cup B)^\complement\cup(B\setminus A)\big) &\iff \lnot(x\in A\lor x\in B)\lor (x\in B\land x\notin A) \\ & \iff \\ &\iff  \\ &\iff x\notin A \\[2ex]\therefore\quad(A\cup B)^\complement \cup(B\setminus A)&~=~A^\complement\end{align}$$
(Now fill in the missing steps.)
A: $A^c $ is everything that that isn't in A.
$(A\cup B)^c $ is everything that isn't in $A $ or $B $.
$(A\cup B)^c \cup B\setminus A $ is everything that isn't in $A $ or $B $ with everything in $B $ that isn't in $A $ added back in.  So everything in $A$ is omitted and everything that isn't is included, whether it is in $B $ or not.
