Solution Found!
Prove the following properties of Boolean algebras. Give a reason for each step. a. (x +
Chapter 8, Problem 10(choose chapter or problem)
Prove the following properties of Boolean algebras. Give a reason for each step.
a. (x + y) + (y \(\cdot\) x') = x + y
b. (y + x) \(\cdot\) (z + y) + x \(\cdot\)z \(\cdot\) (z + z') = y + x \(\cdot\) z
c. (y' \(\cdot\) x) + x + (y + x) \(\cdot\) y' = x + (y' \(\cdot\) x)
d. (x + y') \(\cdot\) z = [(x' + z') \(\cdot\) (y + z')]'
e. (x \(\cdot\) y) + (x' \(\cdot\) z) + (x' \(\cdot\) y \(\cdot\) z') = y + (x' \(\cdot\) z)
Questions & Answers
QUESTION:
Prove the following properties of Boolean algebras. Give a reason for each step.
a. (x + y) + (y \(\cdot\) x') = x + y
b. (y + x) \(\cdot\) (z + y) + x \(\cdot\)z \(\cdot\) (z + z') = y + x \(\cdot\) z
c. (y' \(\cdot\) x) + x + (y + x) \(\cdot\) y' = x + (y' \(\cdot\) x)
d. (x + y') \(\cdot\) z = [(x' + z') \(\cdot\) (y + z')]'
e. (x \(\cdot\) y) + (x' \(\cdot\) z) + (x' \(\cdot\) y \(\cdot\) z') = y + (x' \(\cdot\) z)
Step 1 of 5
a.
\(\begin{array}{l} (x+y)+\left(y \cdot x^{\prime}\right)=x+y \\ \text { LHS }=(x+y)+\left(y \cdot x^{\prime}\right) \\ =x+\left(y+\left(y \cdot x^{\prime}\right)\right) \quad \text { by associativity property } \\ =x+\left(\left(y \cdot x^{\prime}\right)+y\right) \quad \text { by commutative property } \\ =\left(x+\left(y \cdot x^{\prime}\right)\right)+y \quad \text { by associativity property } \\ =(x+y) \cdot\left(x+x^{\prime}\right)+y \quad \text { by distributive property } \\ =(x+y) \cdot 1+y \quad \text { by complement property } \\ =(x+y)+y \quad \text { by identity property } \\ =x+(y+y) \quad \text { by associativity property } \\ =x+y \quad \text { by idempotent property } \\ =\mathrm{RHS} \\ \text { LHS }=\text { RHS } \Rightarrow(x+y)+\left(y \cdot x^{\prime}\right)=x+y \\ \end{array}\)
Hence proved.