Any hint in how to simplify a set theory expression: $(A \cap B) \cup (A \cap B \cap C' \cap D) \cup (A'\cap B)$ I'm having trouble simplifying this set theory expression
$$\begin{align}
(A \cap B) \cup (A \cap B \cap C' \cap D) \cup (A'\cap B)
& 
\end{align}
$$
In the books says that absorption law can help, but I do not understand why
$$\begin{align}
(A \cup A') \cap B = U \cap B = B
& 
\end{align}
$$
Can someone guide me in how do this please, just a hint please? I am stuck
 A: We have : 
$$(A \cap B) \cup (A \cap B \cap C' \cap D) \cup (A'\cap B)=((A \cap B)  \cup (A'\cap B))\cup (A \cap B \cap C' \cap D)$$
You just use here the associativity and comutatitivity of $\cup$. Now :
$$(A \cap B)  \cup (A'\cap B)=(A\cup A')\cap B=U\cap B=B$$
I used distribution of $\cap$ with respect to $\cup$. Then $A\cup A'=U$ because $A\cup A'$ consists by definition of elements of $U$ which are in $A$ or not in $A$ (clearly, all elements of $U$ verify this). Then $U\cap B=B$ because $B\subseteq U$. Now you have :
$$(A \cap B) \cup (A \cap B \cap C' \cap D) \cup (A'\cap B)=B\cup  (A \cap B \cap C' \cap D)$$
And I will let you finish...
A: To prove these, you need logic laws, which in turn are proved using truth tables.
Absorption law: $\color{#00F}X\cup (\color{#00F}X\cap \color{#F00}Y)=\color{#00F}X$, which is simple to prove: you prove $$(x\in X\cup(X\cap Y))\iff (x\in X)$$
In this case:   
$$(\color{#00F}{A \cap B}) \cup ((\color{#00F}{A \cap B}) \cap (\color{#F00}{C' \cap D})) \cup (A'\cap B)$$   
$$\stackrel{\text{Absorp}}=(\color{#00F}{A\cap B})\cup (A'\cap B)\stackrel{\text{Distr}}=(A\cup A')\cap B\stackrel{\text{Compl}}=U\cap B\stackrel{\text{Ident}}=B$$   
Distr - distributive law. Proof:
$$(x\in ((X\cup Y)\cap Z))\iff ((x\in (X\cup Y))\wedge (x\in Z))$$   
$$\iff (((x\in X)\lor (x\in Y))\wedge (x\in Z))$$ $$\iff (((x\in X)\wedge (x\in Z))\lor ((x\in Y)\wedge (x\in Z)))$$   
$$\iff ((x\in X\cap Z)\lor (x\in Y\cap Z))\iff (x\in ((X\cap Z)\cup (Y\cap Z)))$$   
Compl - complement law, Ident - identity law. They are both trivial: you prove $$(x\in A\cup A')\iff (x\in U), \ \ \ \ \ \ (x\in U\cap B)\iff (x \in B),$$  
respectively.
A: I assume that $A,B,C,D\subset U$ and $A':=U\setminus A$ (it should say something like that in the question, otherwise you have to clarify the notation you're using in the book). 
Let $A\subset U$ for any set $U\neq\emptyset$. Then we define the absolute complement of $A$ in $U$, meaning everything that is "outside of $A$ but still in $U$", as $A':=U\setminus A$.
As $A\subset U$ we can write $A=\{x\in U \wedge x\in A\}$ and $A'=\{x\in U\wedge x\notin A\}$. Thus if we look at the union $A\cup A'$ we get $$A\cup A'=\{x\in U \wedge x\in A\} \cup \{x\in U\wedge x\notin A\} =\{(x\in U\wedge x\in A)\vee (x\in U \wedge x\notin A\}.$$
Every $x$ that fulfills $x\in U \wedge x\in A$ lies in this union; these are all $x$ "inside of $A$". Every $x$ that fulfils $x\in U\wedge x\notin A$ lies in this union; these are all $x$ "outside of $A$".
As there are no other $x\in U$ (either $x\in A$ or $x\notin A$), we get $A\cup A'=U$.
A: Use distributivity and comuutativity of $\,\cap\,$ and $\,\cup$:
\begin{align}
(A \cap B) \cup &(A \cap B \cap C' \cap D) \cup (A'\cap B)=\bigl((A \cap B) \cup (A'\cap B)\bigr)\cup (A \cap B \cap C' \cap D) \\
& =\bigl((A  \cup A')\cap B\bigr)\cup (A \cap B \cap C' \cap D)= (U\cap B)\cup (A \cap B \cap C' \cap D)\\
&=B\cup (A \cap B \cap C' \cap D)\qquad\text{since}\enspace B\subset U.
\end{align}
A: Hint 1: Note that $P\cup (P \cap Q) = P$, can you find such P or Q in your expression? (possibly a larger set)
Hint 2: $(P\cap Q) \cup (R \cap Q) = (P \cup R)\cap Q$, do you see any such use in this equation (so you may then get to using the absorbtion law)?
A: (A intersection B) Union (A' intersection B)is B. Then what remains is to have the union of B and the content of the large bracket, that is B union(AnBnC'nD). This can still be expressed in other way round using Morgan's. 
A: Starting with your second question, recall that $A'$ is the set of all things that aren't in $A$. Well, $A \cup A'$ is all things in $A$ together with all things not in $A$. In other words, it's all things, since for any given thing, it's either in $A$ or it isn't. That's why $A \cup A' = U$.
Now to your first question, we simply rearrange:
\begin{align}
(A \cap B) \cup (A \cap B \cap C' \cap D) \cup (A'\cap B) &= (A \cap B) \cup (A'\cap B) \cup (A \cap B \cap C' \cap D) \\
&= ((A \cap B) \cup (A'\cap B)) \cup (A \cap B \cap C' \cap D) \\
&= ((A \cup A') \cap B) \cup (A \cap B \cap C' \cap D) \\
&= (U \cap B) \cup (A \cap B \cap C' \cap D) \\
&= B \cup (A \cap B \cap C' \cap D)
\end{align}
The first line is from commutativity of $\cup$. 
The second is from associativity of $\cup$. 
The third is from distributivity of $\cap$ over $\cup$. 
The fourth is from your second question.
The fifth is because $B \subseteq U$; since everything in $B$ is in $U$, the intersection (that is, the set of elements that both sets have in common) is just $B$. See if you can use this principle to finish.
