Q.Using the rules of negation, write the negation of the following, with justification: $\sim q \to p$

∼(p↔q) is true whenever p and q have opposite truth values. Checking p↔∼q: when p=T,q=F, ∼q=T, so p↔∼q=T (matches ∼(p↔q)=T); when p=T,q=T, ∼q=F, p↔∼q=F (matches ∼(p↔q)=F); similarly for the remaining two rows. All four rows agree, so p↔∼q ≡ ∼(p↔q).

ANSWER: (D)

1. Let p: "an integer is greater than 4 and less than 5" (itself a conjunction, but treat it as the single hypothesis of the conditional), and q: "it is a multiple of 3" (the conclusion).
2. The original statement has the form p → q ("if p then q").
3. The logical negation of a conditional p → q is NOT another conditional — it is the conjunction p ∧ (~q), i.e., "p is true AND q is false." This is because p → q is false exactly when p holds but q fails, so its negation is precisely that scenario.
4. Applying this: negation = "the integer is greater than 4 and less than 5 (p holds)" AND "it is NOT a multiple of 3 (q fails)."
5. This matches option (C) exactly: "An integer is greater than 4 and less than 5 but it is not a multiple of 3."
6. Option (B) is wrong because it negates p inside a new conditional, which is not how implication negation works. Options (A) and (D) incorrectly split or alter the conjunction "greater than 4 and less than 5" instead of keeping it intact as required.

ANSWER: (C)

For $A\to B$ with $A=p\vee\sim q$, $B=p\wedge\sim q$, the contrapositive is $\sim B\to\sim A$. $\sim B=\sim(p\wedge\sim q)=\sim p\vee q$, and $\sim A=\sim(p\vee\sim q)=\sim p\wedge q$. So the contrapositive is $(\sim p\vee q)\to(\sim p\wedge q)$. Its negation, using $\sim(X\to Y)=X\wedge\sim Y$, is $(\sim p\vee q)\wedge\sim(\sim p\wedge q) = (\sim p\vee q)\wedge(p\vee\sim q)$.

ANSWER: (D)

Let X=(p∧q) and Y=((p∨r)→~q). The negation of X→Y is X∧~Y. First, ~Y = ~((p∨r)→q) = (p∨r)∧(~q) = (p∨r)∧q. So the overall negation is (p∧q)∧[(p∨r)∧q] = (p∧q)∧(p∨r) [since q∧q=q]. Now (p∧q)∧(p∨r) distributes to (p∧q∧p)∨(p∧q∧r) = (p∧q)∨(p∧q∧r), and since p∧q∧r already implies p∧q, this simplifies by absorption to just p∧q.

ANSWER: (A)

This is from the Mathematical Logic topic of the NCERT/CBSE-aligned Class 12 Mathematics syllabus.

Let $p$: 'It is raining' and $q$: 'The weather is pleasant'.

The inner statement 'if it is raining then the weather is not pleasant' is $p \to \sim q$.

The full statement is 'It is not true that ($p \to \sim q$)', i.e. $\sim(p \to \sim q)$.

Using $a \to b \equiv \sim a \vee b$:

Taking the negation and applying De Morgan's law:

So the simplified statement is $p \wedge q$ = 'It is raining and the weather is pleasant.'