Answer:
a) b) and c) is proven :)
Step-by-step explanation:
a) [tex]xy+x\bar{y}=x(y+\bar{y})=x*1=x[/tex]
( For a) we used Boolean law [tex]a+\bar{a}=1[/tex] like always true, and
a1=1a= a)
b) [tex]\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)[/tex]
( For we used double negation [tex]\bar{\bar{a}}=a[/tex], and De Morgans law [tex]\bar{(a+b)}=\bar{a}\bar{b}[/tex] and [tex]\bar{(ab)}=\bar{a}+\bar{b}[/tex])
c) x+xy=x(1+y)=x1=x
(For c) we used Distribution law a(b+c)=ab+ac, and identity a1=1a=a)
