Consider Statement -1 : left( {p wedge ∼ q} right) wedge left( { ∼ p wedge q} right) is a

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Mathematical Reasoning

medium

Consider

Statement $-1 :$$\left( {p \wedge \sim q} \right) \wedge \left( { \sim p \wedge q} \right)$ is a fallacy.

Statement $-2 :$$(p \rightarrow q) \leftrightarrow ( \sim q \rightarrow \sim p )$ is a tautology.

Statement $-1$ is false, Statement $-2$ is true

Statement $-1$ is true, Statement $-2$ is false

Statement $-1$ is true, Statement $-2$ is true; Statement $-2$ is a correct explanation for Statement $-1$

Statement $-1$ is true, Statement $-2$ is true; Statement $-2$ is not a correct explanation for Statement $-1$

(AIEEE-2009) (JEE MAIN-2013)

Solution

Statement-ll: $\quad(p \rightarrow q) \leftrightarrow(\sim q \rightarrow-p)$

$\equiv(p \rightarrow q) \leftrightarrow(p \rightarrow q)$

which is always true

so statement- -II is true

$ \text { Statement-l: } (p \wedge \sim q) \wedge(\sim p \wedge q)$

$=p \wedge \sim q \wedge \sim p \wedge q $

$=p \wedge \sim p \wedge \sim q \wedge q $

$=f \wedge f $

$=f $

so statement- – is true

Alternate

Statement-ll: $\quad(p \rightarrow q) \leftrightarrow(\sim q \rightarrow \sim p)$

$\sim \mathrm{q} \rightarrow \sim \mathrm{p}$ is contrapositive

of $p \rightarrow q$ hence $(p \rightarrow q) \leftrightarrow(p \rightarrow q)$

will be a tautology

statement-ll $\quad(p \wedge \sim q) \wedge(\sim p \wedge q)$

Standard 11

Mathematics

The statement $(\mathrm{p} \wedge(\mathrm{p} \rightarrow \mathrm{q}) \wedge(\mathrm{q} \rightarrow \mathrm{r})) \rightarrow \mathrm{r}$ is :

medium

(JEE MAIN-2021)

Negation of statement "If I will go to college, then I will be an engineer" is –

normal

$( S 1)( p \Rightarrow q ) \vee( p \wedge(\sim q ))$ is a tautology $( S 2)((\sim p ) \Rightarrow(\sim q )) \wedge((\sim p ) \vee q )$ is a Contradiction. Then

hard

(JEE MAIN-2023)

If $p , q$ and $r$ are three propositions, then which of the following combination of truth values of $p , q$ and $r$ makes the logical expression $\{(p \vee q) \wedge((\sim p) \vee r)\} \rightarrow((\sim q) \vee r)$ false ?