Section 11.1: Discrete Mathematics and Its Applications, Extra Examples

Rosen, Discrete Mathematics and Its Applications, 6th edition
Extra Examples
Section 11.1—Boolean Functions


p.754, icon at Example 10
#1. Prove the idempotent law x = x · x using the other identities of Boolean algebra listed in Table 5 of
Section 11.1 the textbook.
Solution:
x = x ·1 identity law
= x · (x + x) unit property
= x · x + x · x distributive law
= x · x + 0 zero property
= x · x. identity law


p.754, icon at Example 10
#2. Prove the domination law x · 0 = 0 using the other identities of Boolean algebra listed in Table 5 in
Section 11.1 of the textbook.
Solution:
x ·0 = x · (x · x) zero property
= (x · x) · x associative law
= x · x idempotent law
= 0. zero property


p.754, icon at Example 10
#3. Using the properties of Boolean algebra, prove that
yz + x(xz) + y(z + 1)+zx
can be simplified to give y + zx.
Solution:
yz + x(xz) + y(z + 1) + zx = yz + x(x + z) + y(z + 1) + zx De Morgan’s law
= yz + xx + xz + yz + y + zx distributive law; identity law
= yz +0+xz + yz + y + zx zero property
= yz + xz + yz + y + zx identity law
= y + yz + yz + xz + zx commutative law
= y + y(z + z) + xz + zx distributive law
= y + y1+ xz + zx unit property
= y + y + xz + zx identity law
= y + xz + zx idempotent law= y + zx + zx commutative law
= y + zx. idempotent law