• :00Days
• :00Hours
• :00Mins
• 00Seconds
A new era for learning is coming soon

Suggested languages for you:

Americas

Europe

Q3E

Expert-verified
Found in: Page 818

### Discrete Mathematics and its Applications

Book edition 7th
Author(s) Kenneth H. Rosen
Pages 808 pages
ISBN 9780073383095

# Show that $$(1 \cdot 1){\bf{ + }}(\overline {0 \cdot 1} {\bf{ + }}0){\bf{ = }}1$$.Translate the equation in part $$(a)$$ into a propositional equivalence by changing each $$0$$ into an $${\bf{F}}$$, each $$1$$ into a $$T$$, each Boolean sum into a disjunction, each Boolean product into a conjunction, each complementation into a negation, and the equals sign into a propositional equivalence sign.

a) The given $$(1 \cdot 1){\bf{ + }}(\overline {0 \cdot 1} {\bf{ + 0}}){\bf{ = 1}}$$ is proved.

b) The translate of the given equation $$(1 \cdot 1){\bf{ + }}(\overline {0 \cdot 1} {\bf{ + 0}}){\bf{ = 1}}$$ is $$({\bf{T}} \wedge {\bf{T}}) \vee (\neg ({\bf{F}} \wedge {\bf{T}}) \vee {\bf{F}}) \equiv {\bf{T}}$$

See the step by step solution

## Step 1: Definition

The complement of an element: $${\bf{\bar 0 = 1}}$$ and $${\bf{\bar 1 = }}0$$

The Boolean sum + or $$OR$$ is 1 if either term is 1.

The Boolean product $$\cdot$$ or $$AND$$ is 1 if both terms are 1.

## Step 2: (a) Using the Boolean product and sum

Using the Boolean product and sum

$$\begin{array}{c}(1 \cdot 1){\bf{ + }}(\overline {0 \cdot 1} {\bf{ + }}0){\bf{ = 1 + (\bar 0 + 0)}}\\{\bf{ = 1 + (1 + 0)}}\\{\bf{ = 1 + 1}}\\{\bf{ = 1}}\end{array}$$

Therefore, you get $$(1 \cdot 1){\bf{ + }}(\overline {0 \cdot 1} {\bf{ + 0}}){\bf{ = 1}}$$.

## Step 3: (b) Using the Boolean product and sum

$$(1 \cdot 1){\bf{ + }}(\overline {0 \cdot 1} {\bf{ + 0}}){\bf{ = 1}}$$

Replace 0 by F

Replace 1 by T

Replace + by $$\vee$$

Replace $$\cdot$$ by $$\wedge$$

Replace - by $$\neg$$ (placed in front of the expression that used to be underneath)

Replace = by $$\equiv$$

Therefore, you get $$({\bf{T}} \wedge {\bf{T}}) \vee (\neg ({\bf{F}} \wedge {\bf{T}}) \vee {\bf{F}}) \equiv {\bf{T}}$$.