Q.Write the dual of the statement pattern $(p \wedge \sim q) \vee (\sim p \wedge q)$.

$I$ lies completely inside $Q$, so every integer is a rational number — the universal statement "All integers are rational numbers" (option (ii)).

The diagram nests three sets: $I \subseteq Q \subseteq U$, i.e. the circle of integers $I$ is contained entirely within the circle of rational numbers $Q$ (which in turn lies inside the universe of reals $U$).

When one region lies completely inside another, it expresses a universal relationship: every element of the inner set is also an element of the outer set. Here every element of $I$ is also in $Q$, i.e. every integer is a rational number.

Checking the options: …

$p \vee q$ = "He swims or water is warm", and $\sim$ negates the entire disjunction, giving "It is not true that he swims or water is warm" — i.e. neither statement holds.

We are given:

• $p$ : He swims
• $q$ : Water is warm

For the symbolic statement $\sim(p \vee q)$:

• $p \vee q$ is the disjunction, read as "He swims or water is warm".
• The connective $\sim$ placed before the bracket negates the whole disjunction. …

"He is intelligent or strong" is the disjunction $p \vee q$; "It is wrong that ..." applies negation to the entire statement, giving $eg (p \vee q)$.

We are given $p$: He is intelligent, and $q$: He is strong.

The phrase "he is intelligent or strong" connects the two statements with the connective "or", so it is written as the disjunction $p \vee q$.

The opening phrase "It is wrong that" is a negation applied to the whole statement that follows it, so we negate the complete disjunction:

$eg (p \vee q).$

…

To form the dual, swap every $\vee$ with $\wedge$ and every $\wedge$ with $\vee$; the answer is $(p \wedge q) \vee (r \wedge s)$.

The principle of duality states that the dual of a statement pattern is obtained by replacing each disjunction $\vee$ with a conjunction $\wedge$ and each conjunction $\wedge$ with a disjunction $\vee$ (and, where present, $T$ with $F$ and $F$ with $T$). The variables $p, q, r, s$ themselves stay exactly as they are.

Start with the given statement:

$(p \vee q) \wedge (r \vee s)$

Interchange the connectives: …

Rule: $\sim(\exists x,\ P(x)) \equiv \forall x,\ \sim P(x)$. Apply it to the given statement — the $\exists$ becomes $\forall$ and the predicate "is odd" becomes "is not odd".

The given statement is $\exists n \in N,\ (n^2 + 2)$ is an odd number. To negate a statement with an existential quantifier, change $\exists$ to $\forall$ and negate the open sentence:

$\sim\big(\exists n \in N,\ P(n)\big) \equiv \forall n \in N,\ \sim P(n).$

…

"Some continuous functions are differentiable" is an existential ("there exists a continuous function that is differentiable"). Its negation is universal: "every continuous function is not differentiable," i.e. "No continuous function is differentiable."

The statement "Some continuous functions are differentiable" asserts that at least one continuous function is differentiable — an existential claim $\exists x,\ P(x)$ where $P(x)$ is "$x$ is differentiable" over continuous functions $x$.

…

Apply De Morgan's law to the disjunction, then negate each conditional with $\sim(p \to q) \equiv p \wedge \sim q$. The result simplifies to $p \wedge \sim q \wedge \sim r$.

Negate the whole statement:

$\sim\big[(p \to q) \vee (p \to r)\big].$

By De Morgan's law $\sim(A \vee B) \equiv \sim A \wedge \sim B$:

$\sim(p \to q) \wedge \sim(p \to r).$

Use $\sim(p \to q) \equiv p \wedge \sim q$ (since $p \to q \equiv \sim p \vee q$):

$(p \wedge \sim q) \wedge (p \wedge \sim r).$

By associativity, commutativity and idempotence ($p \wedge p \equiv p$), this simplifies to …