Q.The negation of $p \wedge (q \to r)$ is ________.
A) $\sim p \wedge (\sim q \to \sim r)$
B) $p \vee (\sim q \vee r)$
C) $\sim p \wedge (\sim q \to \sim r)$
D) $\sim p \vee (q \wedge \sim r)$

∼(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)= …

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)." …

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) = …

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

$p \to \sim q \equiv \sim p \vee \sim q.$ …