boolean algebra simple question

Question

Simplfy : $$ (x+y)\cdot(x+yz) $$ I have tried to solve the question through by evaluating the expressions $x(x+y)$ and $yz(x+y)$ but I didn't get the right answer which is: $$ (x+y)\cdot(x+yz)=x + yz $$

Answer 1

Hint:

$$a\cdot a=a$$

$$b+b= b$$

$$xyz+yz=yz(x+1)=?$$

Answer 2

Here you can use the negation trick and use DeMorgan-rules:

• Set $\bar E = \overline{(x+y)\cdot(x+yz)}$ $$\Rightarrow \bar E =\bar x \bar y + \bar x (\bar y + \bar z) = \bar x(\bar y + \bar z) \Rightarrow E = x + yz$$

Answer 3

Another way to look at this problem:

$Y\cap Z\subseteq Y$ so that $X\cup (Y\cap Z)\subseteq X\cup Y$ or equivalently:$$(X\cup Y)\cap(X\cup(Y\cap Z))=X\cup (Y\cap Z)$$

Translation gives:$$(x+y)\cdot(x+y\cdot z)=x+y\cdot z$$

Answer 4

With the use of the second distributivity law D2 we have $$x+yz=(x+y)\cdot (x+z),$$ from where $$\begin{aligned}(x+y)\cdot(x+yz)&=(x+y)\cdot (x+y)\cdot(x+z)\\&=(x+y)\cdot (x+z)\\&=x+yz, \end{aligned}$$ the last step again by D2.