Mathematical reasoning practice · Mixed difficulty

Option D ($(\sim p \Rightarrow q) \vee (\sim q \Rightarrow p)$) is not a tautology because its truth value depends on the values of p and q, and it is not always true.

Option A (Assertion is correct, reason is correct; reason is a correct explanation for assertion) is correct because the assertion is a statement, the reason correctly defines a statement, and the reason explains why the assertion is a statement.

Option D ($p \Rightarrow q$) is correct because a false premise implies any conclusion, so $p \Rightarrow q$ is true when $p$ is false and $q$ is true.

Option A ($\sim q$) is correct because the given Boolean expression simplifies to $\sim p \wedge \sim q$, which is equivalent to $\sim (p \vee q)$, and further simplifies to $\sim q$ given the initial conditions.

Option A ($\to ,\\,\to$) is correct because it results in a tautology when substituted into the given Boolean expression.

Option C ($p \wedge \sim q$) is correct because it represents the negation of the given conditional statement.

Option A, $(p \vee q) \rightarrow p$, is not a tautology because its truth value is not always true for all possible values of p and q.

Option B is correct because it is not logically equivalent to the given proposition; the original statement asserts that every real number must be either rational or irrational, while option B implies that if a number is not rational OR not irrational, then it is not real, which is not the same.

Option C ($(p \wedge r) \wedge \sim q$) is correct because the logical statement simplifies to this expression using De Morgan's laws and distributive properties.

Option A (For an integer $n,$ if $n$ is even, then ${n^3} - 1$ is odd) is correct because it correctly represents the contrapositive of the given statement.

Option B (Pairs is not in France or Londan is not in England) is correct because it represents the negation of the conjunction using De Morgan's Law.

Option C (Both ($S_1$) and ($S_2$) are true) is correct because $S_1$ simplifies to a tautology and $S_2$ simplifies to a fallacy.

Option D ($-(p \vee q)$ is true only when both $p$ and $q$ are false) is correct because the negation of a disjunction is true only when both components are false.

The correct answer is (p ∨ q) ∨ ~q because the dual is found by replacing AND with OR.

Option D ($\mathrm{p} \vee \sim \mathrm{p}$) is correct because the given Boolean expression simplifies to a tautology, which is always true.

The correct answer is C ($\left( {\left( { \sim q} \right) \wedge p} \right) \vee \left( {p \vee \left( { \sim p} \right)} \right)$) because it simplifies to a tautology (t) due to the presence of $p \vee \left( { \sim p} \right)$.

Option A (If a number is not divisible by 3, it is not divisible by 9) is the correct contrapositive of the given statement.

Option A (Neither (S1) and (S2) is True) is correct because both statements are neither tautologies nor contradictions after simplification.

Option B ($(p \wedge \sim q) \vee (p \vee q)$) is correct because it is the correct negation of the given statement using De Morgan's laws and logical equivalences.

Option B ($p,q,r$ are all true) is correct because for $(p \wedge q) \wedge (q \wedge r)$ to be true, all three propositions must be true.

Option B ($\sim q \wedge \left( {p \wedge r} \right)$) is correct because it applies De Morgan's Law to negate the disjunction and then simplifies.

Option C (it is not that the Earth is round or it is not that $3+4=7$) is the correct interpretation of $(\sim p) \vee (\sim q)$.

Option D ($\sim \left( {P \Rightarrow Q} \right) \Leftrightarrow P \wedge \sim Q$) is correct because it represents a tautology, meaning the statement is always true regardless of the truth values of P and Q.

Option B (F, T) is correct because given p is True and q is False, then (p -> q) is False and (q -> p) v (~p) is True.

Option B ($\sim p$) is correct because the negation of $p \vee(p \wedge \sim q)$ simplifies to just p, and the negation of p is $\sim p$.

The correct answer is None of the above because the negation of 'p or q' is 'not p and not q', which is not represented by any of the options.

Option A (there exists $M > 0$ , such that $x < M$ for all $x \in S$) is correct because the negation of "for all" is "there exists", and the negation of $x \ge M$ is $x < M$.

Option B ($\Rightarrow$ and $\wedge$ respectively) is correct because $p \Rightarrow (p \wedge q)$ is not a tautology, as it can be false when $p$ is true and $q$ is false.

Option C is correct because it correctly applies De Morgan's law to negate the given proposition.

Option C (Both P1 and P2 are FALSE) is correct because when $p \to (\sim p \vee q)$ is FALSE, it implies $p$ is TRUE and $q$ is FALSE, making both P1 and P2 FALSE.