Question

F(x,y,z)=xyz+x(yz)'+x'(y+z)+(xyz)'

Answer 1

The simplified Boolean expression is:

F(x,y,z)=xyz+x′ y+x′ z+x′ y′ z′

Let's simplify the given Boolean expression:

F(x, y, z) = xyz + x(yz)' + x' (y + z) + (xyz)'

Using Boolean algebra laws and De Morgan's theorem, we can simplify this expression step by step:

1. Distributive Law: A + AB = A xyz + x(yz)' + x(y + z) + (xyz)'

2. Apply the distributive law to the second term: xyz + xy + x(y + z) + (xyz)'

3. Apply the distributive law to the fourth term:

xyz + xy + xy + x'z + (xyz)'

4. Apply the distributive law to the first and fifth terms: xyz + xy + xy + x'z + x'y'z'

5. Combine like terms: xyz + xy + xy + x'z + x'y'z'

6. Apply the idempotent law: A+A'BA+B

xyz + x/y + x'z + x'y'z'

So, the simplified Boolean expression is:

F(x,y,z)=xyz+x′ y+x′ z+x′ y′ z′

Quesion

simplifying the expression f(x,y,z)=xyz+x(yz)'+x'(y+z)+(xyz)'