Mathematical Reasoning * — JEE Main PYQs

Among the statements
(S1): $(p\Rightarrow q)\vee ((\sim p)\wedge q)$ is a tautology
(S2): $(q\Rightarrow p)\Rightarrow ((\sim p)\wedge q)$ is a contradiction

Negation of $(p\Rightarrow q)\Rightarrow (q\Rightarrow p)$ is

Statement $(P\Rightarrow Q)\wedge (R\Rightarrow Q)$ is logically equivalent to

The negation of $(\mathrm{p}\wedge (\sim \mathrm{q}))\vee (\sim \mathrm{p})$ is equivalent to

The statement $\sim [\mathrm{p}\vee (\sim (\mathrm{p}\wedge \mathrm{q}))]$ is equivalent to

The negation of the statement $(p\vee q)\wedge (q\vee (\sim r))$ is

Among the two statements
(S1): $(p\Rightarrow q)\wedge (q\wedge (\sim q))$ is a contradiction and
(S2): $(p\wedge q)\vee ((\sim p)\wedge q)\vee (p\wedge (\sim q))\vee ((\sim p)\wedge (\sim q))$ is a tautology

The statement $(p\wedge (\sim q))\vee ((\sim p)\wedge q)\vee ((\sim p)\wedge (\sim q))$ is equivalent to

The negation of the statement
$((\mathrm{A}\wedge (\mathrm{B}\vee \mathrm{C}))\Rightarrow (\mathrm{A}\vee \mathrm{B}))\Rightarrow \mathrm{A}$ is

Negation of $\mathrm{p}\wedge (\mathrm{q}\wedge \sim (\mathrm{p}\wedge \mathrm{q}))$ is

The converse of the statement $((\sim p)\wedge q)\Rightarrow r$ is

The number of ordered triplets of the truth values of $p,q$ and $r$ such that the truth value of the statement $(p\vee q)\wedge (p\vee r)\Rightarrow (q\vee r)$ is True, is equal to

The compound statement
$(\sim (\mathrm{P}\wedge \mathrm{Q}))\vee ((\sim \mathrm{P})\wedge \mathrm{Q})\Rightarrow ((\sim \mathrm{P})\wedge (\sim \mathrm{Q}))$ is equivalent to

Among the statements:
$(S1):((p\vee q)\Rightarrow r)\iff (p\Rightarrow r)$
$(S2):((p\vee q)\Rightarrow r)\iff ((p\Rightarrow r)\vee (q\Rightarrow r))$

Which of the following statements is a tautology ?

The statement $\mathrm{B}\Rightarrow ((\sim \mathrm{A})\vee \mathrm{B})$ is equivalent to :

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

Let $\Delta ,\nabla \in \{\wedge ,\vee \}$ be such that $(\mathrm{p}\to \mathrm{q})\Delta (\mathrm{p}\nabla \mathrm{q})$ is a tautology. Then

Let p and q be two statements.

Then $\sim (p\wedge (p\Rightarrow \sim q))$ is equivalent to

The number of values of $\mathrm{r}\in \{\mathrm{p},\mathrm{q},\sim \mathrm{p},\sim \mathrm{q}\}$ for which $((\mathrm{p}\wedge \mathrm{q})\Rightarrow (\mathrm{r}\vee \mathrm{q}))\wedge ((\mathrm{p}\wedge \mathrm{r})\Rightarrow \mathrm{q})$ is a tautology, is:

Consider the following statements:

P: I have fever
Q : I will not take medicine
R : I will take rest

The statement "If I have fever, then I will take medicine and I will take rest" is equivalent to:

(S1) $(\mathrm{p}\Rightarrow \mathrm{q})\vee (\mathrm{p}\wedge (\sim \mathrm{q}))$ is a tautology
(S2) $((\sim \mathrm{p})\Rightarrow (\sim \mathrm{q}))\wedge ((\sim \mathrm{p})\vee \mathrm{q})$ is a Contradiction. Then

The negation of the expression $\mathrm{q}\vee ((\sim \mathrm{q})\wedge \mathrm{p})$ is equivalent to

The statement $(\mathrm{p}\wedge (\sim \mathrm{q}))\Rightarrow (\mathrm{p}\Rightarrow (\sim \mathrm{q}))$ is

The statement $(p\Rightarrow q)\vee (p\Rightarrow r)$ is NOT equivalent to:

Let the operations *, ⊙ $\in \{\wedge ,\vee \}$. If $(\mathrm{p}\ast \mathrm{q})\odot (\mathrm{p}\odot \sim \mathrm{q})$ is a tautology, then the ordered pair ( $\ast ,\odot$ ) is :

Let
$\mathbf{p}$ : Ramesh listens to music.
q: Ramesh is out of his village
r : It is Sunday
s : It is Saturday
Then the statement "Ramesh listens to music only if he is in his village and it is Sunday or Saturday" can be expressed as

$(p^{\wedge }r)\iff (p^{\wedge }(\sim q))$ is equivalent to $(\sim p)$ when $r$ is

Which of the following statements is a tautology?

Negation of the Boolean expression $p\iff (q\Rightarrow p)$ is

If the truth value of the statement $(P\wedge (\sim R))\to ((\sim R)\wedge Q)$ is F, then the truth value of which of the following is F ?

Consider the following statements :
P : Ramu is intelligent
Q : Ramu is rich
R : Ramu is not honest
The negation of the statement "Ramu is intelligent and honest if and only if Ramu is not rich" can be expressed as :

The statement $(\sim (p\iff \sim q))\wedge q$ is :

The maximum number of compound propositions, out of $\mathrm{p}\vee \mathrm{r}\vee \mathrm{s},\mathrm{p}\vee \mathrm{r}\vee \sim \mathrm{s},\mathrm{p}\vee \sim \mathrm{q}\vee \mathrm{s}$, …

Which of the following statement is a tautology?

Let $\mathrm{p},\mathrm{q},\mathrm{r}$ be three logical statements. Consider the compound statements
$S_1:((\sim p)\vee q)\vee ((\sim p)\vee r)$ and
${\mathrm{S}}_2:\mathrm{p}\to (\mathrm{q}\vee \mathrm{r})$
Then, which of the following is NOT true ?

Negation of the Boolean statement $(\mathrm{p}\vee \mathrm{q})\Rightarrow ((\sim \mathrm{r})\vee \mathrm{p})$ is equivalent to:

Let $r\in \{p,q,\sim p,\sim q\}$ be such that the logical statement $r\vee (\sim p)\Rightarrow (p\wedge q)\vee r$ is a tautology. Then ' $r$ ' is equal to :

Let $\Delta \in \{\wedge ,\vee ,\Rightarrow ,\iff \}$ be such that $(\mathrm{p}\wedge \mathrm{q})\Delta ((\mathrm{p}\vee \mathrm{q})\Rightarrow \mathrm{q})$ is a tautology. Then $\Delta$ is equal to :

Let $\Delta ,\nabla \in \{\wedge ,\vee \}$ be such that $p\nabla q\Rightarrow ((p\Delta q)\nabla r)$ is a tautology.

Then $(p\nabla q)\Delta r$ is logically equivalent to:

The Boolean expression $(\sim (p\wedge q))\vee q$ is equivalent to :

The statement $(\mathrm{p}\wedge \mathrm{q})\Rightarrow (\mathrm{p}\wedge \mathrm{r})$ is equivalent to :

The number of choices of $\Delta \in \{\wedge ,\vee ,\Rightarrow ,\iff \}$, such that $(\mathrm{p}\Delta \mathrm{q})\Rightarrow ((\mathrm{p}\Delta \sim \mathrm{q})\vee ((\sim \mathrm{p})\Delta \mathrm{q}))$ is a tautology, is

Consider the following statements :
A : Rishi is a judge.
B : Rishi is honest.
C : Rishi is not arrogant.
The negation of the statement "if Rishi is a judge and he is not arrogant, then he is honest" is

The negation of the Boolean expression $((\sim q)\wedge p)\Rightarrow ((\sim p)\vee q)$ is logically equivalent to

Consider the following two propositions:
P1 : $\sim (\mathrm{p}\to \sim \mathrm{q})$
P2: $(\mathrm{p}\wedge \sim \mathrm{q})\wedge ((\sim \mathrm{p})\vee \mathrm{q})$
If the proposition $p\to ((\sim p)\vee q)$ is evaluated as FALSE, then:

If the truth value of the Boolean expression $((p\vee q)\wedge (q\to r)\wedge (\sim r))\to (p\wedge q)$ is false, then the truth values of the statements $p,q,r$ respectively can be:

The statement among the following that is a tautology is:

Consider the two statements:
(S1): $(p\to q)\vee (\sim q\to p)$ is a tautology.
(S2): $(p\wedge \sim q)\wedge (\sim p\vee q)$ is a fallacy.
Then:

The Boolean expression $(p\wedge \sim q)\Rightarrow (q\vee \sim p)$ is equivalent to :