Algebra of Statements: Mathematical Reasoning

Algebra of Statements: Mathematical Reasoning

Edited By Komal Miglani | Updated on Oct 15, 2024 12:00 PM IST

In mathematical logic, the algebra of statements (or propositional calculus) is a branch of algebra that deals with the manipulation and transformation of logical statements. This system is based on various logical operators and rules, which help in the formation and simplification of complex logical expressions. This concept in used in various fields like engineering, physics etc.

This article is about the algebra of statements which falls under the category of Discrete Mathematics. This concept is not only important for board exams but also for competitive exams.

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’.

Name of Connective

Connective Word

Symbol

Conjunction

And

Disjunction

Or

Negation

Not

Conditional

‘if-then' or 'implication'

➝ or ⇒

Biconditional

‘If and only if' or 'double implication'

↔️ or ⇔



Algebra of Statements

Idempotent Law

  1. $p ∨ p ≡ p$

  2. $p ∧ p ≡ p$

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

Associative Law

  1. $( p ∨ q ) ∨ r ≡ p ∨ (q ∨ r )$

  2. $( p ∧ q ) ∧ r ≡ p ∧ (q ∧ r )$

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

Distributive Law

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

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

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\begin{array}{|c|c|c|c|c|c|c|c|}
\hline \boldsymbol{p} & \boldsymbol{q} & \boldsymbol{r} & \boldsymbol{q} \wedge \boldsymbol{r} & \boldsymbol{p} \vee(\boldsymbol{q} \wedge \boldsymbol{r}) & \boldsymbol{p} \vee \boldsymbol{q} & \boldsymbol{p} \vee \boldsymbol{r} & (\boldsymbol{p} \vee \boldsymbol{q}) \wedge(\boldsymbol{p} \vee \boldsymbol{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. $p ∧ q ≡ q ∧ p$

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

Identity Law

  1. $p ∧ T ≡ p$

  2. $p ∧ F ≡ F$

  3. $p ∨ T ≡ T$

  4. $p ∨ F ≡ p$

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

Complement Law

  1. $p ∨ ~p ≡ T$

  2. $p ∧ ~p ≡ F$

  3. $\sim (\sim p) ≡ p$

  4. $~T ≡ F$

  5. $~F ≡ T$

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

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 $~ ( p ∧ q ) and ~p ∨ ~q$

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

Recommended Video Based on Algebra of Statements


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.

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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

Summary

The algebra of statements provides a foundational framework for reasoning and manipulating logical expressions. Understanding and applying these laws is crucial for anyone studying mathematics, computer science, philosophy, or any field that requires rigorous logical reasoning.

Frequently Asked Questions (FAQs)

1. What is Algebra of statements?

The algebra of statements (or propositional calculus) is a branch of algebra that deals with the manipulation and transformation of logical statements.

2. State the associative law.
  1. $( p ∨ q ) ∨ r  ≡ p ∨ (q ∨ r )$

  2. $( p ∧ q ) ∧ r  ≡ p ∧ (q ∧ r )$

3. Give the distributive law.
  1. $p ∧ (q  ∨ r ) ≡ ( p ∧ q ) ∨ ( p ∧ r ) $

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

4. Give the commutative law.
  1. $p ∨ q ≡ q ∨ p$

  2. $p ∧ q ≡ q ∧ p$

5. State De'Morgans law.
  1. $~ ( p ∨ q ) ≡ ~p ∧ ~q$

  2. $~ ( p ∧ q ) ≡ ~p ∨ ~q$

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