Algebra of Statements: Mathematical Reasoning

Mathematical Statement

A mathematical statement is the basic unit of any mathematical reasoning. A sentence is called a mathematical statement if it is either true or false but not both.

Logical Connectives

The words which combine or change simple statements to form new statements or compound statements are called Connectives. The basic connectives (logical) conjunction corresponds to the English word ‘and’, disjunction corresponds to the word ‘or’, and negation corresponds to the word ‘not’.

Algebra of Statements

Idempotent Law

1. $p ∨ p ≡ p$

2. $p ∧ p ≡ p$

$$
\begin{array}{|c|c|c|}
\hline
\;\;\; p \;\;\; & \;\;\; p \vee p \;\;\; & \;\;\; p \wedge p \;\;\; \\
\hline \hline
\mathrm{T} & \mathrm{T} & \mathrm{T} \\
\hline
\mathrm{F} & \mathrm{F} & \mathrm{F} \\
\hline
\end{array}
$$

Associative Law

1. $( p ∨ q ) ∨ r ≡ p ∨ (q ∨ r )$

2. $( p ∧ q ) ∧ r ≡ p ∧ (q ∧ r )$

$$
\begin{array}{|c|c|c|c|c|c|c|}
\hline
{p} & {q} & {r} & {p} \vee {q} & {q} \vee {r} & ({p} \vee {q}) \vee {r} & {p} \vee({q} \vee {r}) \\
\hline
T & T & T & T & T & T & T \\
\hline
T & T & F & T & T & T & T \\
\hline
T & F & T & T & T & T & T \\
\hline
T & F & F & T & F & T & T \\
\hline
F & T & T & T & T & T & T \\
\hline
F & T & F & T & T & T & T \\
\hline
F & F & T & F & T & T & T \\
\hline
F & F & F & F & F & F & F \\
\hline
\end{array}
$$

Distributive Law

1. $p ∧ (q ∨ r ) ≡ ( p ∧ q ) ∨ ( p ∧ r ) $

2. $p ∨ (q ∧ r ) ≡ ( p ∨ q ) ∧ ( p ∨ r ) $

$$
\begin{array}{|c|c|c|c|c|c|c|c|}
\hline
{p} & {q} & {r} & {q} \wedge {r} & {p} \vee({q} \wedge {r}) & {p} \vee {q} & {p} \vee {r} & ({p} \vee {q}) \wedge({p} \vee {r}) \\
\hline
T & T & T & T & T & T & T & T \\
\hline
T & T & F & F & T & T & T & T \\
\hline
T & F & T & F & T & T & T & T \\
\hline
T & F & F & F & T & T & T & T \\
\hline
F & T & T & T & T & T & T & T \\
\hline
F & T & F & F & F & T & F & F \\
\hline
F & F & T & F & F & F & T & F \\
\hline
F & F & F & F & F & F & F & F \\
\hline
\end{array}
$$

Commutative Law

1. $p ∨ q ≡ q ∨ p$

2. $p ∧ q ≡ q ∧ p$

$$
\begin{array}{|c|c|c|c|}
\hline
{p} & {q} & {p} \vee {q} & {q} \vee {p} \\
\hline
T & T & T & T \\
\hline
T & F & T & T \\
\hline
F & T & T & T \\
\hline
F & F & F & F \\
\hline
\end{array}
$$

Identity Law

1. $p ∧ T ≡ p$

2. $p ∧ F ≡ F$

3. $p ∨ T ≡ T$

4. $p ∨ F ≡ p$

$$
\begin{array}{|c|c|c|c|c|}
\hline
{p} & \mathbb{T} & \mathbb{F} & {p} \vee \mathbb{T} & {p} \vee \mathbb{F} \\
\hline
T & T & F & T & T \\
\hline
F & T & F & T & F \\
\hline
\end{array}
$$

Complement Law

5. $p ∨ ～p ≡ T$

6. $p ∧ ～p ≡ F$

7. $\sim (\sim p) ≡ p$

8. $～T ≡ F$

9. $～F ≡ T$

$$
\begin{array}{|c|c|c|c|c|c|c|c|}
\hline
{p} & \neg {p} & \mathbb{T} & \neg \mathbb{T} & \mathbb{F} & \neg \mathbb{F} & {p} \vee \neg {p} & {p} \wedge \neg {p} \\
\hline
T & F & T & F & F & T & T & F \\
\hline
F & T & T & F & F & T & T & F \\
\hline
\end{array}
$$

De-Morgan’s Law

1. $～ ( p ∨ q ) ≡ ～p ∧ ～q$

2. $～ ( p ∧ q ) ≡ ～p ∨ ～q$

Truth table for $～ ( p ∨ q ) and ～p ∧ ～q$

$$
\begin{array}{|c|c|c|c|c|c|c|}
\hline
p & q & \sim p & \sim q & p \vee q & \sim (p \vee q) & \sim p \wedge \sim q \\
\hline \hline
\mathrm{T} & \mathrm{T} & \mathrm{F} & \mathrm{F} & \mathrm{T} & \mathrm{F} & \mathrm{F} \\
\hline
\mathrm{T} & \mathrm{F} & \mathrm{F} & \mathrm{T} & \mathrm{T} & \mathrm{F} & \mathrm{F} \\
\hline
\mathrm{F} & \mathrm{T} & \mathrm{T} & \mathrm{F} & \mathrm{T} & \mathrm{F} & \mathrm{F} \\
\hline
\mathrm{F} & \mathrm{F} & \mathrm{T} & \mathrm{T} & \mathrm{F} & \mathrm{T} & \mathrm{T} \\
\hline
\end{array}
$$

Truth table for $～(p∧q)$ and $～p∨～q$

$$
\begin{array}{|c|c|c|c|c|c|}
\hline p & q & p \wedge q & \sim(p \wedge q) & \sim p \vee \sim q & \sim(p \wedge q) \equiv \sim p \vee \sim q \\
\hline T & T & T & F & F & T \\
T & F & F & T & T & T \\
F & T & F & T & T & T \\
F & F & F & T & T & T \\
\hline
\end{array}
$$

Solved Examples Based on Algebra of Statements:

Example 1: In each question below, a passage followed by several inferences. you have to example each inference separately in the context of the passage and decide upon its degree of truth or falsity.

Corel Draw and Photoshop are the old software used for designing purposes in industries, Adobe illustrator and InDesign are the new ones, but even today Corel and Photoshop are mostly used in industries. These are designing software that are mostly used in Digital Marketing. Digital Marketing is a technique used for advertising on the web, like Facebook marketing, Instagram and E-mail marketing etc. Now these techniques are used by approximately $70 %$ of companies to generate leads for their business. By $2022$ there may be the end of traditional marketing strategies. With the coming of the Internet in India in $1991$ the scene of marketing is dramatically changed. In India, only $35 %$ population is active on the internet.

The government is not doing anything for developing industries in India.

1) if the inference is "" probably true"" though not definitely true in the light of the facts given

2) if the inference is "" definitely true "" it directly follows from the facts given in the passage

3) if you think the data is inadequate, from the facts given you cannot say whether the inference is likely to be true or false

4) if you think the inference is "" probably false "" though not definitely false in the light of the facts given

Solution

Government role is not defined in the passage.

Example 2: Which one is NOT an example of an AND conjunction?

1) $p: x+y=3$ and $x-y=1$
2) $q$: $x^2-4>0$ and $y^2-3<0$
3) $r$: Sam opened the closet and took out clothes
4) $s$ : Delhi in India and Mumbai is in Europe

Solution

'And' Conjunction -

Normally the conjunction 'and' is used between two statements which have some kind of relation but in logic, it can be used even if there is no relation between the statements.

Here , "and " is used in a different sense

Example 3: Which of the following statement is true?

1) $x^2+x+1<0 \forall x \in R$ and $2 x^2+3 x+4>0 \forall x \in R$
2) $x^2-x+3<0 \forall x \in R$ and $x^2+x+1>0 \forall x \in R$
3) $x^2+x+1<0 \forall x \in R$ and $x^2-x+3<0 \forall x \in R$
4) None of these

Solution
Truth value of "And" Conjunction -
The statement $p \wedge q$ has the truth value $T$ whenever both $p$ and $q$ have the truth value $T$.

A compound statement $\mathrm{p} {\wedge} \mathrm{q}$ is the when both p and q are true .

Example 4: The logical statement $[\sim(\sim p \vee q) \vee(p \wedge r)] \wedge(\sim q \wedge r)$ is equivalent to :
1) $(\sim p \wedge \sim q) \wedge r$
2) $(p \wedge \sim q) \vee r$
3) $\sim p \vee r$
4) $(p \wedge r) \wedge \sim q$

Solution

$
\begin{aligned}
& p=\{1,2,5,6\} \quad q=\{2,3,4,5\} \quad r=\{4,5,6,7\} \\
& \sim p=\{3,4,7,8\} \quad \sim q=\{1,6,7,8\} \quad \sim r=\{1,2,3,8\} \\
& {[\sim(\sim p \vee q) \vee(p \wedge r)] \wedge(\sim q \wedge r)} \\
& {[\sim(\{3,4,7,8\} \vee\{2,3,4,5\}) \vee(\{1,2,5,6\} \wedge\{4,5,6,7\})] \wedge(\{1,6,7,8\}} \\
& {[\sim(\{2,3,4,5,7,8\}) \vee\{5,6\}] \wedge\{6,7\}} \\
& {[\{1,6\} \vee\{5,6\}] \wedge\{6,7\}} \\
& \{1,5,6\} \wedge\{6,7\} \\
& \{6\}
\end{aligned}
$

Now check which option gives the same region
Option A

$
\begin{aligned}
& (\sim p \wedge \sim q) \wedge r \\
& \{7,8\} \wedge\{4,5,6,7\}=\{7\}
\end{aligned}
$

Incorrect

Option B

$
\begin{aligned}
& (p \wedge \sim q) \vee r \\
& \{1,6\} \vee\{4,5,6,7\}=\{1,4,5,6,7\}
\end{aligned}
$

Incorrect

Option C

$\begin{aligned}
& \sim p \vee r \\
& \{3,4,7,8\} \vee\{4,5,6,7\}=\{3,4,5,6,7,8\}
\end{aligned}
$

Incorrect

Option D

$
\begin{aligned}
& (p \wedge r) \wedge \sim q \\
& \{5,6\} \wedge\{1,6,7,8\}=\{6\}
\end{aligned}
$

Correct

Example 5: Which of the following is not a disjunction?
1) $3 \times 3=10$ or $7 \times 5=28$
2) $\sin x>1$ and $\cos x<1$
3) $1+2=3$ or $3+5=8$
4) All are disjunctions

Solution

Disjunction 'OR' -

Two statements can be connected by the word "OR" to form a compound statement called the disjunction of original statements.

We use AND is conjunction and OR is a disjunction