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  • Which one of the following is a tautology ?  Option: 1 <img alt="\left ( P\wedge \left ( P\rightarrow Q \right ) \right )\rightarrow Q" src="https://learn.careers360.com/latex-image/?%5Cleft%20%28%20P%5Cwedge%20%5Cleft%20%28%20P%5C

Which one of the following is a tautology ?
 
Option: 1 \left ( P\wedge \left ( P\rightarrow Q \right ) \right )\rightarrow Q
Option: 2 P\wedge (P\vee Q)
Option: 3 Q\rightarrow (P\wedge (P\rightarrow Q))  
Option: 4 P\vee (P\wedge Q)

Answers (1)

best_answer

 

 

Tautology And Contradiction -

Tautology

A compound statement is called tautology if it is always true for all possible truth values of its component statement.

For example,    ( p ⇒ q ) ∨ ( q ⇒ p ) 

 

Contradiction (fallacy)

A compound statement is called a contradiction if it is always false for all possible truth values of its component statement.

For example, ∼( p ⇒ q ) ∨ ( q ⇒ p ) 

 

Truth Table

\begin{array}{|c|c|c|c|c|c|c|c|}\hline \mathrm{\;\;\;\;\;}p\mathrm{\;\;\;\;\;}&\mathrm{\;\;\;}q\mathrm{\;\;\;}&\mathrm{\;\;\;\;\;}p\rightarrow q\mathrm{\;\;\;\;\;}&\mathrm{\;\;\;} q\rightarrow p\mathrm{\;\;\;} &\mathrm{\;\;\;}\left ( p\rightarrow q \right )\vee\left ( q\rightarrow p \right )\mathrm{\;\;}&\mathrm{\;\;\;}\sim\left ( \left ( p\rightarrow q \right )\vee\left ( q\rightarrow p \right ) \right ) \mathrm{\;\;} \\\hline \hline \mathrm{T}&\mathrm{T} & \mathrm{T} &\mathrm{T}&\mathrm{T}&\mathrm{F} \\ \hline \mathrm{T}&\mathrm{F} & \mathrm{F} &\mathrm{T}&\mathrm{T}&\mathrm{F} \\ \hline \mathrm{F}&\mathrm{T} & \mathrm{T} &\mathrm{F}&\mathrm{T}&\mathrm{F} \\ \hline \mathrm{F}&\mathrm{F} & \mathrm{T} &\mathrm{T}&\mathrm{T}&\mathrm{F} \\ \hline\end{array}

Quantifiers

Quantifiers are phrases like ‘These exist’ and “for every”. We come across many mathematical statements containing these phrases. 

For example – 

p : For every prime number x, √x is an irrational number.

q : There exists a triangle whose all sides are equal.

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Practise Session - 2 -

Q1. Write the truth table for the following statement pattern:
        \\ \mathrm{1.\;\;\;\sim(p \vee((\sim q \Rightarrow q) \wedge q))}\\\mathrm{2.\;\;\;(p\wedge q)\vee(\sim r)}\\

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\begin{array}{c|c|c}{\mathbf{p}} & {\mathbf{q}} & {(\mathbf{p} \rightarrow \mathbf{q})} \\ \hline F & {F} & {\mathbf{T}} \\ \hline F & {T} & {\mathbf{T}} \\ \hline T & {F} & {F} \\ \hline T & {T} & {\mathbf{T}}\end{array}   \begin{array}{c|c|c}{\mathbf{p}} & {\mathbf{q}} & {(\mathbf{p} \wedge \mathbf{q})} \\ \hline F & {F} & {\mathbf{F}} \\ \hline F & {T} & {F} \\ \hline T & {F} & {F} \\ \hline T & {T} & {T}\end{array}

\begin{array}{c|c|c}{\mathbf{p}} & {\mathbf{q}} & {(p \wedge(p \rightarrow q)) \rightarrow q} \\ \hline F & {F} & {\mathbf{T}} \\ \hline F & {T} & {\mathbf{T}} \\ \hline T & {F} & {T} \\ \hline T & {T} & {\mathbf{T}}\end{array}

Correct Option (1)

Posted by

Kuldeep Maurya

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