Question

Use the laws of Boolean algebra to show that the two Boolean expressions in each pair are equivalent. Show that a. xy + xy = x. b.Show that xy + zx = (x + y)(z + x) c. x + xy = x

Answer 1

Answer:

a) b) and c) is proven :)

Step-by-step explanation:

a) $xy+x\bar{y}=x(y+\bar{y})=x*1=x$

( For a) we used Boolean law $a+\bar{a}=1$ like always true, and

a1=1a= a)

b)  $\overline{(xy+z\bar{x})}=\overline{(xy)}\overline{ (z\bar{x})}=(\bar{x}+\bar{y})(\bar{z}+\bar{\bar{x}})=(\bar{x}+\bar{y})(\bar{z}+x)$  

( For we used double negation $\bar{\bar{a}}=a$, and De Morgans law $\bar{(a+b)}=\bar{a}\bar{b}$ and $\bar{(ab)}=\bar{a}+\bar{b}$)

c) x+xy=x(1+y)=x1=x

(For c) we used Distribution law a(b+c)=ab+ac, and identity a1=1a=a)