Let p, q, r be three statements such that the truth value of $(p \wedge q) \to (\sim q \vee r)$ is F. Then the truth values of p, q, r are respectively : Options: (A) T, F, T (B) F, T, F (C) T, T, T (D) T, T, F

### Analyzing the Setup We are tasked with determining the truth values of $p$, $q$, and $r$ given that the logical expression $(p \wedge q) \to (\sim q \vee r)$ evaluates to **False**. In formal logic, an implication $A \to B$ is only **False** when the antecedent $A$ is **True** and the consequent $B$ is **False**. Therefore, for the expression $(p \wedge q) \to (\sim q \vee r)$ to be **False**, we must satisfy the following two conditions: 1. The antecedent $(p \wedge q)$ must be **True**. 2. The consequent $(\sim q \vee r)$ must be **False**. ### The Conjunction Gate We begin by analyzing the antecedent: $(p \wedge q) \equiv T$. The conjunction operator $\wedge$ is a strict gatekeeper. It requires both $p$ and $q$ to be **True** for the entire statement to be **True**. Since the conjunction is **True**, we must conclude that: $$p \equiv T$$ $$q \equiv T$$ ### The Disjunction and the Final Reveal Next, we analyze the consequent: $(\sim q \vee r) \equiv F$. The disjunction operator $\vee$ is only **False** if both of its components are **False**. This implies: $$\sim q \equiv F$$ $$r \equiv F$$ We must verify this against our previous findings. Since we established that $q \equiv T$, it follows that $\sim q$ is indeed **False**. This confirms our logic is consistent. We have successfully determined the values for all variables: **$p = T$** **$q = T$** **$r = F$**

JEE Main 2020 - 3 Sep (Evening)

LEVELBoard

Animated Solution for Mathematics - Sets and Relations: Let p, q, r be three statements such that the truth value of $(p \land q) \to (\sim q \lor r)$ is F. Then the truth values of p, q, r are respectively :

Q: Wait, why is the implication False only when the first part is True and the second is False? Can't it be False if both are False?

That is a fantastic question! In logic, the implication $A \to B$ is defined as False only when $A$ is True and $B$ is False. If both are False, the implication is actually True. Think of it like a promise: if I promise to give you a chocolate ($A$) if you study ($B$), the only way I break my promise is if you study ($A$ is True) but I don't give you the chocolate ($B$ is False).

Q: Why does $p \wedge q = T$ mean both $p$ and $q$ must be True? What if only one is True?

Great catch! The conjunction operator $\wedge$ acts like a strict gatekeeper. It requires both inputs to be True for the output to be True. If even one of them is False, the whole statement collapses to False. So, for $p \wedge q$ to be True, there is no other option but for both $p$ and $q$ to be True.

Q: I got confused with the negation $\sim q$. How do I handle that?

Negation is just flipping the switch! If $\sim q$ is False, it means the negation of $q$ is False, which implies $q$ itself must be True. It's like saying 'It is not the case that it is raining' is False, which means it must be raining. Just take it one step at a time!

Select Answer:

Visualized Solution

The Main Implication

Given expression: $(p \land q) \to (\sim q \lor r) \equiv F$
We need to find the truth values of $p$ , $q$ , and $r$ .

The Implication Rule $A \to B \equiv F$

Recall the rule for implication: $A \to B$ is False if and only if $A$ is True and $B$ is False.
This is the only case where an implication fails.

Breaking Down the Implication

Applying the rule to our expression:
The antecedent must be True: $(p \land q) \equiv T$
The consequent must be False: $(\sim q \lor r) \equiv F$

Analyzing $p \land q \equiv T$

We have a conjunction (AND): $(p \land q) \equiv T$
For an AND statement to be True, both components must be True.

Finding $p$ and $q$

Therefore, we can conclude:
$p \equiv T$
$q \equiv T$

Analyzing $\sim q \lor r \equiv F$

Now look at the disjunction (OR): $(\sim q \lor r) \equiv F$
For an OR statement to be False, both components must be False.

Finding $r$

Therefore, we conclude:
$\sim q \equiv F$ (which matches our earlier finding that $q \equiv T$ )
$r \equiv F$

Final Truth Values

Summarizing our results:
$p = T$
$q = T$
$r = F$
The correct sequence is T, T, F.

00:00 / 00:00

The Sigma Insight: Types of Sets and Set Operations

Solution Diagram

Analyzing the Setup

We are tasked with determining the truth values of

p

q

, and

r

given that the logical expression

(p \land q) \to (\sim q \lor r)

evaluates to False.

In formal logic, an implication

A \to B

is only False when the antecedent

A

is True and the consequent

B

is False.

Therefore, for the expression

(p \land q) \to (\sim q \lor r)

to be False, we must satisfy the following two conditions:

1. The antecedent

(p \land q)

must be True.

2. The consequent

(\sim q \lor r)

must be False.

The Conjunction Gate

We begin by analyzing the antecedent:

(p \land q) \equiv T

The conjunction operator

\land

is a strict gatekeeper. It requires both

p

and

q

to be True for the entire statement to be True.

Since the conjunction is True, we must conclude that:

p \equiv T

q \equiv T

The Disjunction and the Final Reveal

Next, we analyze the consequent:

(\sim q \lor r) \equiv F

The disjunction operator

\lor

is only False if both of its components are False. This implies:

\sim q \equiv F

r \equiv F

We must verify this against our previous findings. Since we established that

q \equiv T

, it follows that

\sim q

is indeed False.

This confirms our logic is consistent. We have successfully determined the values for all variables:

$p = T$

$q = T$

$r = F$

Common Traps & FAQs

Wait, why is the implication False only when the first part is True and the second is False? Can't it be False if both are False?

That is a fantastic question! In logic, the implication

A \to B

is defined as False only when

A

is True and

B

is False. If both are False, the implication is actually True. Think of it like a promise: if I promise to give you a chocolate (

A

) if you study (

B

), the only way I break my promise is if you study (

A

is True) but I don't give you the chocolate (

B

is False).

Why does

p \land q = T

mean both

p

and

q

must be True? What if only one is True?

Great catch! The conjunction operator

\land

acts like a strict gatekeeper. It requires both inputs to be True for the output to be True. If even one of them is False, the whole statement collapses to False. So, for

p \land q

to be True, there is no other option but for both

p

and

q

to be True.

I got confused with the negation

\sim q

. How do I handle that?

Negation is just flipping the switch! If

\sim q

is False, it means the negation of

q

is False, which implies

q

itself must be True. It's like saying 'It is not the case that it is raining' is False, which means it must be raining. Just take it one step at a time!

Animated Solution for Mathematics - Sets and Relations: Let p, q, r be three statements such that the truth value of $(p \land q) \to (\sim q \lor r)$ is F. Then the truth values of p, q, r are respectively :

Select Answer:

Visualized Solution

The Main Implication

The Implication Rule $A \to B \equiv F$

Breaking Down the Implication

Analyzing $p \land q \equiv T$

Finding $p$ and $q$

Analyzing $\sim q \lor r \equiv F$

Finding $r$

Final Truth Values

The Sigma Insight: Types of Sets and Set Operations

Analyzing the Setup

The Conjunction Gate

The Disjunction and the Final Reveal

Similar Questions

If $(P \land \sim q) \land (p \land r) \to\sim p \lor q$ is false, then the truth values of $p, q$ and $r$ are respectively

If the truth value of the statement $P \to (\sim p \lor r)$ is false(F), then the truth values of the statements p, q, r are respectively :

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

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 ${(p \lor q) \land ((\sim p) \lor r)} \to ((\sim q) \lor r)$ false?

The statement $(p \land q) \Rightarrow (p \land r)$ is equivalent to :

If $P ⟹ (q \lor r)$ is false, then the truth values of p, q, r are respectively :-

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

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

Among the two statements $(S 1) : (p \Rightarrow q) \land (p \land (\sim q))$ is a contradiction and $(S 2) : (p \land q) \lor ((\sim p) \land q) \lor (p \land (\sim q)) \lor ((\sim p) \land (\sim q))$ is a tautology

The proposition $p \to\sim (p \land \sim q)$ is equivalent to:

Select Answer:

Visualized Solution

The Main Implication

The Implication Rule .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A→B≡F

Breaking Down the Implication

Analyzing .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } p∧q≡T

Finding .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } p and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } q

Analyzing .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ∼q∨r≡F

Finding .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } r

Final Truth Values

The Sigma Insight: Types of Sets and Set Operations

Analyzing the Setup

The Conjunction Gate

The Disjunction and the Final Reveal

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If (P∧∼q)∧(p∧r)→∼p∨q is false, then the truth values of p,q and r are respectively

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the truth value of the statement P→(∼p∨r) is false(F), then the truth values of the statements p, q, r are respectively :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the truth value of the statement (P∧(∼R))→((∼R)∧Q) is F, then the truth value of which of the following is F ?

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The statement (p∧q)⇒(p∧r) is equivalent to :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If P⟹(q∨r) is false, then the truth values of p, q, r are respectively :-

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Statement (P⇒Q)∧(R⇒Q) is logically equivalent to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the truth value of the Boolean expression ((p∨q)∧(q→r)∧(∼r))→(p∧q) is false then the truth values of the statements p,q,r respectively can be:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Among the two statements (S1):(p⇒q)∧(p∧(∼q)) is a contradiction and (S2):(p∧q)∨((∼p)∧q)∨(p∧(∼q))∨((∼p)∧(∼q)) is a tautology

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The proposition p→∼(p∧∼q) is equivalent to:

The Implication Rule $A \to B \equiv F$

Analyzing $p \land q \equiv T$

Finding $p$ and $q$

Analyzing $\sim q \lor r \equiv F$

Finding $r$

If $(P \land \sim q) \land (p \land r) \to\sim p \lor q$ is false, then the truth values of $p, q$ and $r$ are respectively

If the truth value of the statement $P \to (\sim p \lor r)$ is false(F), then the truth values of the statements p, q, r are respectively :

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

The statement $(p \land q) \Rightarrow (p \land r)$ is equivalent to :

If $P ⟹ (q \lor r)$ is false, then the truth values of p, q, r are respectively :-

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

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

Among the two statements $(S 1) : (p \Rightarrow q) \land (p \land (\sim q))$ is a contradiction and $(S 2) : (p \land q) \lor ((\sim p) \land q) \lor (p \land (\sim q)) \lor ((\sim p) \land (\sim q))$ is a tautology

The proposition $p \to\sim (p \land \sim q)$ is equivalent to: