If p, q and r are three propositions, then which of the following combination of truth values of p,

Q: Why is the implication false only when $T \rightarrow F$?

Think of an implication as a promise. If I say 'If it rains, I will carry an umbrella' ($P \rightarrow Q$), the only way I have lied to you is if it actually rains ($P$ is True) but I show up without an umbrella ($Q$ is False). If it doesn't rain, I haven't broken my promise, regardless of whether I have an umbrella or not. That is why $F \rightarrow T$ and $F \rightarrow F$ are still considered True.

Q: Can I just use a truth table?

You absolutely could! A truth table with 8 rows ($2^3$) would work perfectly. However, in the heat of a JEE Advanced exam, time is your most precious currency. The algebraic method we used—working backwards from the False condition—is like taking a shortcut through the woods while everyone else is walking the long road. It is faster, more elegant, and less prone to calculation errors.

Q: Why does $(\sim p \vee F)$ become $\sim p$?

Think of the OR operator as a 'logical addition' where False acts like zero. Just as adding zero to a number doesn't change its value, ORing a proposition with False doesn't change the truth value of that proposition. If $\sim p$ is True, the whole thing is True. If $\sim p$ is False, the whole thing is False. Thus, $(\sim p \vee F) \equiv \sim p$.

Question

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?

Options:

(A) p=T, q=F, r=T

(B) p=T, q=T, r=F

(C) p=F, q=T, r=F

(D) p=T, q=F, r=F

Sigma Percentile · Accepted Answer

### Analyzing the Setup

The expression provided is $\{(p \vee q) \wedge (\sim p \vee r)\} ightarrow (\sim q \vee r)$. In the context of JEE Advanced, we treat this as an implication of the form $X ightarrow Y$.

Here, the antecedent is $X = \{(p \vee q) \wedge (\sim p \vee r)\}$ and the consequent is $Y = (\sim q \vee r)$. We are tasked with finding the truth values of $p$, $q$, and $r$ that make the entire expression **False**.

### The Implication Trap

In propositional logic, an implication $A ightarrow B$ is **False** if and only if the antecedent $A$ is **True** and the consequent $B$ is **False**. This is the only scenario where the implication fails.

$$X ightarrow Y = F \iff (X = T) 	ext{ and } (Y = F)$$

By setting $X = T$ and $Y = F$, we create a clear path to isolate the individual truth values of the variables.

### Deconstructing the Consequent

We begin with the consequent $Y = (\sim q \vee r) = F$. Since the OR operator ($\vee$) only returns **False** when both components are **False**, we must satisfy two conditions simultaneously:

1. $\sim q = F \implies q = T$
2. $r = F$

We have successfully determined two of our variables: $q$ is **True** and $r$ is **False**.

### The Antecedent Dance

Now, we substitute $q = T$ and $r = F$ into the antecedent $X = \{(p \vee q) \wedge (\sim p \vee r)\}$ and set it to **True**:

$$X = \{(p \vee T) \wedge (\sim p \vee F)\} = T$$

We simplify the components using the laws of logic:
*   $(p \vee T) = T$ (Any proposition ORed with True is True).
*   $(\sim p \vee F) = \sim p$ (False acts as the identity element in an OR operation).

Substituting these back into $X$, we get:

$$X = \{T \wedge \sim p\} = T$$

### The Final Revelation

Since $T \wedge \sim p$ simplifies to $\sim p$, the condition for the antecedent becomes:

$$\sim p = T$$

If the negation of $p$ is **True**, then $p$ must be **False**. We have arrived at the final set of values:

**$p = F, q = T, r = F$**

This systematic approach of working backwards from the condition of the implication ensures that you navigate complex logical structures with precision and speed.

Animated Solution for Mathematics - Sets and Relations: 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?

Select Answer:

Visualized Solution

Understanding the Implication Structure

Condition for $X \to Y$ to be False

Analyzing the Consequent $Y$

Finding Truth Values for $q$ and $r$

Substituting $q$ and $r$ into $X$

Simplifying the First Bracket $(p \lor T)$

Simplifying the Second Bracket $(\sim p \lor F)$

Final Simplification of $X$

Conclusion and Final Answer

The Sigma Insight: Types of Sets and Set Operations

Analyzing the Setup

The Implication Trap

Deconstructing the Consequent

The Antecedent Dance

The Final Revelation

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Select Answer:

Visualized Solution

Understanding the Implication Structure

Condition for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } X→Y to be False

Analyzing the Consequent .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Y

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

Simplifying the First Bracket .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } (p∨T)

Simplifying the Second Bracket .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } (∼p∨F)

Final Simplification of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } X