Show that $(A\cap B)^c = (A^c \cap B) \cup (A^c \cap B^c) \cup (A \cap B^c)$ It is given that $\;A^c = (A^c \cap B) \cup (A^c \cap B^c),\;$ and $\;B^c = (A \cap B^c) \cup (A^c \cap B^c)$
Therefore we have,
$$(A\cap B)^c = A^c \cup B^c = ((A^c \cap B) \cup (A^c \cap B^c)) \cup ((A \cap B^c) \cup (A^c \cap B^c))$$
But I'm not sure how this leads to $(A^c \cap B) \cup (A^c \cap B^c) \cup (A \cap B^c)$.
As far as I can tell, it would be $(A^c \cap B) \cup 2(A^c \cap B^c) \cup (A \cap B^c)$.
What am I not seeing?
 A: The highlighted sets below are the same set:
\begin{align} (A\cap B)^c &= A^c \cup B^c \tag{DeMorgan's}\\ \\ &= ((A^c \cap B) \cup \color{blue}{(A^c \cap B^c)}) \cup ((A \cap B^c) \cup \color{blue}{(A^c \cap B^c)})\\ \\ &= (A^c\cap B) \cup \color{blue}{(A^c \cap B^c)}\cup \color{blue}{(A^c \cap B^c)} \cup (A\cap B^c)\tag{*}\end{align}
We need only count a set once; afterall, given any set $C,\;\,C\cup C = C.$     In this case we have $$(A^c \cap B^c) \cup (A^c \cap B^c) = A^c\cap B^c.$$
Hence, we arrive at 
$$(A\cap B)^c = A^c \cup B^c = (A^c \cap B)  \cup (A \cap B^c) \cup (A^c \cap B^c).$$
$(*)$  We used the associative property and the commutative property of set-union: $\cup$.

Note:  By the definition of the intersection of two sets, e.g. $A^c\cap B^c$, we have that $$A^c \cap B^c = \{x\mid x \not\in A \land x \not\in B\}.$$  
Just because we see $A^c \cap B^c$ appear twice in your expansion, there is no need to repeat the fact that $A^c\cap B^c = \{x\mid x\notin A \land x\notin B\}$.  
They both refer to one set of elements: those that are not in A and those that are not in B. Hence, we can simplify by mentioning it only once.
