Question

Given the Boolean function F(x,y,z)=x'y + xyz'

a)Derive an algebraic expression for the complement of F. Express in sum-of-products form.
b)Show that FF' = 0.
c)Show that F + F' =1.

Answer 1

To find the complement F' of the Boolean function F(x,y,z)=x'y + xyz', De Morgan's laws and Boolean algebra rules are applied to simplify to sum-of-products form. It can be shown that FF' always equals 0 and F + F' always equals 1 because F' is the complement of F.

Step-by-step explanation:

The Boolean function provided is F(x,y,z)=x'y + xyz'. To find the complement of this function, denoted as F', we need to apply De Morgan's laws and the complement rules for Boolean algebra.

1. First, apply De Morgan's laws to get the complement of each term:
2. F' = (x'y)'+(xyz')' = (x+y')(x'+y+z)
3. Next, apply distributive laws to simplify:
4. F' = (x'y)'+(xyz')' = (x'+y)(x'+y+z)
5. Finally, apply Boolean algebra rules to further simplify:
6. F' = x'y' + x'z + yz

Now, to show that FF' = 0 and F + F' = 1,

1. FF' can be found by multiplying F and F'. As they are complements, they will not share any common terms when multiplied, resulting in a zero function.
2. FF' = (x'y + xyz')(x'y' + x'z + yz) = 0
3. F + F' is always 1 because it represents the sum of a function and its complement.
4. F + F' = (x'y + xyz') + (x'y' + x'z + yz) = 1