
## Section5.2Exercises

##### 2

Prove the commutative property expressed by Equation (5.1.5) and Equation (5.1.6).

Solution

For Equation (5.1.5):

 $x$ $y$ $x \cdot y$ $y \cdot x$ $\binary{0}$ $\binary{0}$ $\binary{0}$ $\binary{0}$ $\binary{0}$ $\binary{1}$ $\binary{0}$ $\binary{0}$ $\binary{1}$ $\binary{0}$ $\binary{0}$ $\binary{0}$ $\binary{1}$ $\binary{1}$ $\binary{1}$ $\binary{1}$

And for Equation (5.1.6):

 $x$ $y$ $x + y$ $y + x$ $\binary{0}$ $\binary{0}$ $\binary{0}$ $\binary{0}$ $\binary{0}$ $\binary{1}$ $\binary{1}$ $\binary{1}$ $\binary{1}$ $\binary{0}$ $\binary{1}$ $\binary{1}$ $\binary{1}$ $\binary{1}$ $\binary{1}$ $\binary{1}$
##### 4

Prove the complement property expressed by Equation (5.1.9) and Equation (5.1.10).

Solution

For Equation (5.1.9):

 $x$ $x'$ $x \cdot x'$ $\binary{0}$ $\binary{1}$ $\binary{0}$ $\binary{1}$ $\binary{0}$ $\binary{0}$

And for Equation (5.1.10):

 $x$ $x'$ $x + x'$ $\binary{0}$ $\binary{1}$ $\binary{1}$ $\binary{1}$ $\binary{0}$ $\binary{1}$
##### 5

Prove the idempotent property expressed by Equation (5.1.11) and Equation (5.1.12).

Solution
 $x$ $x$ $x \cdot x$ $\binary{0}$ $\binary{0}$ $\binary{0}$ $\binary{1}$ $\binary{1}$ $\binary{1}$

And for Equation (5.1.12):

 $x$ $x$ $x + x$ $\binary{0}$ $\binary{0}$ $\binary{0}$ $\binary{1}$ $\binary{1}$ $\binary{1}$
##### 6

Prove the distributive property expressed by Equation (5.1.13) and Equation (5.1.14).

Solution
 $x$ $y$ $z$ $y + z$ $x \cdot (y + z)$ $x \cdot y$ $x \cdot z$ $x \cdot y + x \cdot z$ $\binary{0}$ $\binary{0}$ $\binary{0}$ $\binary{0}$ $\binary{0}$ $\binary{0}$ $\binary{0}$ $\binary{0}$ $\binary{0}$ $\binary{0}$ $\binary{1}$ $\binary{1}$ $\binary{0}$ $\binary{0}$ $\binary{0}$ $\binary{0}$ $\binary{0}$ $\binary{1}$ $\binary{0}$ $\binary{1}$ $\binary{0}$ $\binary{0}$ $\binary{0}$ $\binary{0}$ $\binary{0}$ $\binary{1}$ $\binary{1}$ $\binary{1}$ $\binary{0}$ $\binary{0}$ $\binary{0}$ $\binary{0}$ $\binary{1}$ $\binary{0}$ $\binary{0}$ $\binary{0}$ $\binary{0}$ $\binary{0}$ $\binary{0}$ $\binary{0}$ $\binary{1}$ $\binary{0}$ $\binary{1}$ $\binary{1}$ $\binary{1}$ $\binary{0}$ $\binary{1}$ $\binary{1}$ $\binary{1}$ $\binary{1}$ $\binary{0}$ $\binary{1}$ $\binary{1}$ $\binary{1}$ $\binary{0}$ $\binary{1}$ $\binary{1}$ $\binary{1}$ $\binary{1}$ $\binary{1}$ $\binary{1}$ $\binary{1}$ $\binary{1}$ $\binary{1}$

And for Equation (5.1.14):

 $x$ $y$ $z$ $y \cdot z$ $x + y \cdot z$ $x + y$ $x + z$ $(x + y) \cdot (x + z)$ $\binary{0}$ $\binary{0}$ $\binary{0}$ $\binary{0}$ $\binary{0}$ $\binary{0}$ $\binary{0}$ $\binary{0}$ $\binary{0}$ $\binary{0}$ $\binary{1}$ $\binary{0}$ $\binary{0}$ $\binary{0}$ $\binary{1}$ $\binary{0}$ $\binary{0}$ $\binary{1}$ $\binary{0}$ $\binary{0}$ $\binary{0}$ $\binary{1}$ $\binary{0}$ $\binary{0}$ $\binary{0}$ $\binary{1}$ $\binary{1}$ $\binary{1}$ $\binary{1}$ $\binary{1}$ $\binary{1}$ $\binary{1}$ $\binary{1}$ $\binary{0}$ $\binary{0}$ $\binary{0}$ $\binary{1}$ $\binary{1}$ $\binary{1}$ $\binary{1}$ $\binary{1}$ $\binary{0}$ $\binary{1}$ $\binary{0}$ $\binary{1}$ $\binary{1}$ $\binary{1}$ $\binary{1}$ $\binary{1}$ $\binary{1}$ $\binary{0}$ $\binary{0}$ $\binary{1}$ $\binary{1}$ $\binary{1}$ $\binary{1}$ $\binary{1}$ $\binary{1}$ $\binary{1}$ $\binary{1}$ $\binary{1}$ $\binary{1}$ $\binary{1}$ $\binary{1}$