Simplifying the Boolean expression $x'y + x(x+y')$

Question

Expression:

$$x'y + x(x+y')$$

My attempt:

1. $x'y + x(x+y')$
2. $x'y + xx + xy' \quad \textit{After applying second Distributive law.}$
3. $x'y + x + xy' \quad \textit{After applying second Idempotent law.}$
4. $x'y + x(1 + y') \quad \textit{Breaking out x}$
5. $x'y + x(1) \quad \textit{After applying Annihilation law}$
6. $x'y + x$

The answer is: $x + y$.

N.B. I am still new to this and learning, I am using the following book:

Discrete Mathematics for Computing / Edition 3
by Peter Grossman

Answer 1

For short, I'll abbreviate below Absorption, Associativity, Commutativity, Distributivity and Idempotence as abs, assoc, comm, distr and idemp, respectively.

Lemma. In a Boolean algebra, if $a+b=ab$, then $a=b$.

Proof. If $a+b = ab$ then \begin{align} a &= a (a + b) \tag{abs}\\ &= a (ab) \tag{hypothesis}\\ &= ab \tag{assoc, idemp}\\ &= b (ab) \tag{assoc, comm, idemp}\\ &= b (a + b) \tag{hypothesis}\\ &= b. \tag{abs} \end{align}

Now, \begin{align} (x + x'y) + (x + y) &= x + y + x'y \tag{assoc, comm, idemp}\\ &= x + (y + x'y) \tag{assoc}\\ &= x + y, \tag{abs} \end{align} and \begin{align} (x + x'y)(x+y) &= x + xy + x'yx + x'y \tag{distr}\\ &= x + (xy + x'y) \tag{assoc, $xx'=0$}\\ &= x + (x + x')y \tag{distr}\\ &= x + y. \tag{$x + x' = 1$} \end{align}

Hence, letting $a = x + x'y$ and $b = x + y$, we have $a + b = b = ab$, and so $a=b$, by the Lemma.
So $x + x'y = x + y$.

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Shorter version: \begin{align} x + x'y &= (x + xy) + x'y \tag{abs}\\ &= x + (xy + x'y) \tag{assoc}\\ &= x + (x + x')y \tag{distr}\\ &= x + y. \end{align} According to a deleted answer below, this law is called Redundancy law.

Answer 2

Simplest Solution

You can use venn diagram to solve create it use 2 different circles for x and y

x'y=y\x x(x+y')=x

sum of both=x+y

That is why