Exam 12: Boolean Algebra

Show that the Boolean function F given by $F ( x , y , z ) = x ( z + y z ) + y (\overline{ \overline { x z } x )}$ simplifies to $x z + \bar { x } y$ by using only the difinition of a booleab algebra .

In questions mark each statement TRUE or FALSE. - $\{ + , \cdot \} \text { is a functionally complete set of operators. }$

In questions mark each statement TRUE or FALSE. - $\overline { x \mid y } = x \downarrow y$

In questions determine whether the statement is TRUE or FALSE. Assume that x, y, and z represent Boolean variables. -(0 + x)(1 + x) = x x.

In questions mark each statement TRUE or FALSE. - $\bar { x } \bar { y } = ( ( x \mid x ) \mid ( y \mid y ) ) \mid ( ( x \mid x ) \mid ( y \mid y ) )$

Use the Quine-McCluskey method to simplify the Boolean expression x y z + x y z + x y z + x y z + x y z

There are Boolean functions with 4 variables.

In questions determine whether the statement is TRUE or FALSE. Assume that x, y, and z represent Boolean variables. - $\overline { x + y } = \bar { x } + \bar { y }$

In questions mark each statement TRUE or FALSE. - $x x x + x y + y y = x y$

In questions determine whether the statement is TRUE or FALSE. Assume that x, y, and z represent Boolean variables. - $\overline { y + x z } = \bar { y } \bar { x } + \bar { y } \bar { z }$

A circuit is to be built that takes the numbers 0 through 9 as inputs (1 = 0001, 2 = 0010, . . . , 9 = 1001). Let $G ( w , x , y , z )$ be the Boolean function that produces an output of 1 if and only if the input is an odd number. Find a Karnaugh map for G and use the map and don't care conditions to find a simple expression for G.

If $f ( w , x , y , z ) = \overline { ( \bar { x } + y \bar { z } ) } + ( \bar { w } x )$ , find $f ( 0,1,0,1 )$

Using only the five properties associative laws, commutative laws, distributive laws, identity laws, and complement laws, prove that $x + x = x$ is true in all Boolean algebras.

Prove that F=G , where $F ( x , y ) = \overline { ( x + \bar { x } y ) } \bar { y }$ and $G ( x , y ) = \overline { x + y }$

Write $( x + y ) ( \bar { x } + \bar { y } )$ as a sum-of-products in the variables x and y .

Write $x + y + z$ as a sum-of-products in the variables $x , y , \text { and } z$

If $f ( w , x , y , z ) = \overline { ( \bar { x } + y \bar { z } ) } + ( \bar { w } x )$ , find $f ( 1,0,1,1 )$

In questions mark each statement TRUE or FALSE. - $\text { When written as a sum of minterms in the variables } x \text { and } y , x + \bar { y } = x y + x \bar { y } + \bar { x } \bar { y } \text {. }$

Give a reason for each step in the proof that $x + 1 = x$ is true in Boolean algebras. Your reasons should come from the following: associative laws for addition and multiplication, commutative laws for addition and multiplication, distributive law for multiplication over addition and distributive law for addition over multiplication, identity laws, unit property, and zero property. $1 = x + \bar { x } = x + \bar { x } \cdot 1 = ( x + \bar { x } ) ( x + 1 ) = 1 \cdot ( x + 1 ) = x + 1$

The idempotent laws in a Boolean algebra state that and .