# 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

• I assume $A,B,C,D\subset U$ for a given set $U$ und $A':=U\setminus A$? Jun 17, 2015 at 12:11
• Is the third term $A'\cap B$ instead of $A'\cup B$? Jun 17, 2015 at 12:14
• The problem does not say anything about it, just say simplify and in the solution manual says that using that law Jun 17, 2015 at 12:14
• @ClémentGuérin yes sorry, I already edit Jun 17, 2015 at 12:15

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...

• Thank you for answering I will solve the other part Jun 17, 2015 at 13:03
• You seem to have a typo. In line 4 it should say "because $A\cup A'$ consists by definition...". Jun 17, 2015 at 13:06

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.

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 $A\cup A'=U$.

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}

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 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.

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.