Mathematical logic Questions in English

(D) To determine the nature of the proposition $p \Rightarrow \sim (p \wedge \sim q)$,we construct its truth table:

$p$	$q$	$\sim q$	$p \wedge \sim q$	$\sim (p \wedge \sim q)$	$p \Rightarrow \sim (p \wedge \sim q)$
$T$	$T$	$F$	$F$	$T$	$T$
$T$	$F$	$T$	$T$	$F$	$F$
$F$	$T$	$F$	$F$	$T$	$T$
$F$	$F$	$T$	$F$	$T$	$T$

Since the final column contains both $T$ and $F$,the proposition is neither a tautology nor a contradiction. Thus,the correct option is $(d)$.

(A) To find the dual of a Boolean expression,we replace $\vee$ with $\wedge$,$\wedge$ with $\vee$,$1$ with $0$,and $0$ with $1$.
Given expression: $(x \vee y) \wedge (x \vee 1) = x \vee (x \wedge y) \vee y$
Applying the rules:
$1$. Replace $\vee$ with $\wedge$ and $\wedge$ with $\vee$.
$2$. Replace $1$ with $0$.
The dual expression becomes: $(x \wedge y) \vee (x \wedge 0) = x \wedge (x \vee y) \wedge y$.
Comparing this with the given options,it matches option $A$.

(C) For Statement-$I$: $\sim (p \leftrightarrow q) \equiv \sim ((p$ $\rightarrow q) \wedge (q$ $\rightarrow p))$
$\equiv \sim (p$ $\rightarrow q) \vee \sim (q$ $\rightarrow p)$
$\equiv (p \wedge \sim q) \vee (q \wedge \sim p)$.
Thus,Statement-$I$ is True.
For Statement-$II$: $p$ $\rightarrow (p$ $\rightarrow q) \equiv \sim p \vee (\sim p \vee q) \equiv (\sim p \vee \sim p) \vee q \equiv \sim p \vee q$.
This is not a tautology (it depends on the truth values of $p$ and $q$).
Thus,Statement-$II$ is False.

(B) Let $p$ be the statement '$96$ is divisible by $2$' and $q$ be the statement '$96$ is divisible by $3$'.
The given statement is $p \land q$.
The negation of a conjunction $p \land q$ is $\sim(p \land q) \equiv \sim p \lor \sim q$.
Here,$\sim p$ is '$96$ is not divisible by $2$' and $\sim q$ is '$96$ is not divisible by $3$'.
Thus,the negation is '$96$ is not divisible by $2$ or $96$ is not divisible by $3$'.

(B) The negation of a conditional statement $p \Rightarrow q$ is given by the logical equivalence $\sim (p \Rightarrow q) \equiv p \wedge (\sim q)$.
Here,$p$ is '$5$ is not greater than $2$' and $q$ is '$\text{Jaipur is the capital of Rajasthan}$'.
Therefore,$\sim q$ is '$\text{Jaipur is not the capital of Rajasthan}$'.
The negation is $p \wedge (\sim q)$,which is '$5$ is not greater than $2$ and $\text{Jaipur is not the capital of Rajasthan}$'.
Thus,option $B$ is correct.

(B) We evaluate each pair for logical equivalence:
$A$. $\sim (\sim p) \equiv p$ (Law of Double Negation). These are equivalent.
$B$. $p \vee (p \wedge q) \equiv p$ (Absorption Law). Since $p \not\equiv q$,these are not logically equivalent.
$C$. $\sim (p \wedge q) \equiv \sim p \vee \sim q$ (De Morgan's Law). These are equivalent.
$D$. $\sim (\sim p \wedge q) \equiv \sim (\sim p) \vee \sim q \equiv p \vee \sim q$ (De Morgan's Law). These are equivalent.
Thus,the pair in option $B$ is not logically equivalent.

(C) The truth table for $p \to (p \leftrightarrow q)$ is as follows:
$p$ | $q$ | $p \leftrightarrow q$ | $p \to (p \leftrightarrow q)$
$T$ | $T$ | $T$ | $T$
$T$ | $F$ | $F$ | $F$
$F$ | $T$ | $F$ | $T$
$F$ | $F$ | $T$ | $T$
Now,evaluating option $C$: $(q \to p) \to (p \to q)$
$p$ | $q$ | $q \to p$ | $p \to q$ | $(q \to p) \to (p \to q)$
$T$ | $T$ | $T$ | $T$ | $T$
$T$ | $F$ | $T$ | $F$ | $F$
$F$ | $T$ | $F$ | $T$ | $T$
$F$ | $F$ | $T$ | $T$ | $T$
Since the truth values match for all combinations of $p$ and $q$,the statement $p \to (p \leftrightarrow q)$ is logically equivalent to $(q \to p) \to (p \to q)$.

(B) The statement $p$ is "$2 \times 4 = 8$",which is true,so $p = T$.
The statement $q$ is "$4$ divides $7$",which is false,so $q = F$.
Now,we evaluate the truth values for each biconditional statement:
$(i)$ $p \leftrightarrow q$ is $T \leftrightarrow F$,which is $F$.
$(ii)$ $\sim p \leftrightarrow q$ is $\sim T \leftrightarrow F$,which is $F \leftrightarrow F$,so it is $T$.
$(iii)$ $\sim q \leftrightarrow p$ is $\sim F \leftrightarrow T$,which is $T \leftrightarrow T$,so it is $T$.
$(iv)$ $\sim p \leftrightarrow \sim q$ is $\sim T \leftrightarrow \sim F$,which is $F \leftrightarrow T$,so it is $F$.
Thus,the truth values are $F, T, T, F$.

(D) The converse of a conditional statement of the form "If $P$,then $Q$" is "If $Q$,then $P$".
Here,the statement is "If $p < q$,then $p - x < q - x$".
Let $P$ be $p < q$ and $Q$ be $p - x < q - x$.
The converse is "If $Q$,then $P$",which is "If $p - x < q - x$,then $p < q$".

(B) We need to find the negation of $p \wedge (\sim q \vee \sim r)$.
Using De Morgan's Law,$\sim (A \wedge B) = \sim A \vee \sim B$.
$\sim (p \wedge (\sim q \vee \sim r)) = \sim p \vee \sim (\sim q \vee \sim r)$.
Applying De Morgan's Law again,$\sim (\sim q \vee \sim r) = (\sim \sim q \wedge \sim \sim r) = (q \wedge r)$.
So,the expression becomes $\sim p \vee (q \wedge r)$.
Using the distributive law,$A \vee (B \wedge C) = (A \vee B) \wedge (A \vee C)$.
Therefore,$\sim p \vee (q \wedge r) = (\sim p \vee q) \wedge (\sim p \vee r)$.

(A) To determine the nature of the statement,we construct a truth table:

$p$	$q$	$\sim p$	$\sim q$	$(p \wedge \sim q)$	$(\sim p \vee q)$	$(p \wedge \sim q) \wedge (\sim p \vee q)$
$T$	$T$	$F$	$F$	$F$	$T$	$F$
$T$	$F$	$F$	$T$	$T$	$F$	$F$
$F$	$T$	$T$	$F$	$F$	$T$	$F$
$F$	$F$	$T$	$T$	$F$	$T$	$F$

Since the final column contains only $F$ (False) for all possible truth values of $p$ and $q$,the statement is a contradiction.

(D) For the implication $(p$ $\rightarrow q)$ $\rightarrow (q$ $\rightarrow r)$ to be false,the antecedent $(p \rightarrow q)$ must be true and the consequent $(q \rightarrow r)$ must be false.
For $(q \rightarrow r)$ to be false,$q$ must be true and $r$ must be false.
Substituting $q = T$ into the antecedent $(p \rightarrow q)$,we get $(p \rightarrow T)$,which is always true regardless of the truth value of $p$.
Thus,$p$ can be either true or false,$q$ must be true,and $r$ must be false.
Comparing this with the options,the combination $(p, q, r) = (F, T, F)$ satisfies the condition.

(B) To determine the truth value,we analyze the logical equivalence of $\sim (p \leftrightarrow \sim q)$.
We know that $(p \leftrightarrow \sim q)$ is true when $p$ and $\sim q$ have the same truth value,which means $p$ and $q$ have opposite truth values.
Thus,$(p \leftrightarrow \sim q)$ is equivalent to $\sim (p \leftrightarrow q)$.
Therefore,$\sim (p \leftrightarrow \sim q) \equiv \sim (\sim (p \leftrightarrow q)) \equiv (p \leftrightarrow q)$.
Hence,the statement in option $B$ is true.

(B) Let $p$ be the statement: "$I$ will go to college".
Let $q$ be the statement: "$I$ will be an engineer".
The given statement is in the form $p \rightarrow q$.
The negation of $p \rightarrow q$ is given by $\sim(p \rightarrow q) \equiv p \wedge \sim q$.
Therefore,the negation is: "$I$ will go to college and $I$ will not be an engineer."

(D) $1$. Analyze $p$: The function $f(x) = x$ is an odd function because $f(-x) = -x = -f(x)$. Thus,$p$ is false ($\sim p$ is true).
$2$. Analyze $q$: The binomial coefficient ${}^nC_m$ represents the number of ways to choose $m$ items from $n$,which is always a non-negative integer (whole number) for $0 \le m \le n$. Thus,$q$ is true.
$3$. Analyze $r$: If $\overrightarrow{c} = \overrightarrow{a} + \lambda \overrightarrow{b}$,then $\overrightarrow{c} - \overrightarrow{a} - \lambda \overrightarrow{b} = 0$. Since there exists a non-zero linear combination equal to the zero vector,the vectors $\overrightarrow{a}, \overrightarrow{b}, \overrightarrow{c}$ are linearly dependent. Thus,$r$ is true.
$4$. Evaluate the options:
- $A: (p \wedge q) = (F \wedge T) = F$
- $B: (p \vee q) \wedge \sim r = (F \vee T) \wedge F = T \wedge F = F$
- $C: \sim (q \wedge r) \vee p = \sim (T \wedge T) \vee F = \sim T \vee F = F \vee F = F$
- $D: \sim p \vee (q \wedge r) = \sim F \vee (T \wedge T) = T \vee T = T$
Therefore,the statement in option $D$ is true.

(D) The implication $p \rightarrow r$ is false only when $p$ is true and $r$ is false.
Here,the statement is $p \rightarrow \sim q$.
For this to be false,$p$ must be true and $\sim q$ must be false.
If $\sim q$ is false,then $q$ must be true.
Therefore,the statement $p \rightarrow \sim q$ is false when $p$ is true and $q$ is true.

(D) Let the given expression be $S = (p \wedge \sim q \wedge \sim r) \vee (\sim p \wedge q \wedge \sim r) \vee (\sim p \wedge \sim q \wedge r)$.
This expression represents the condition where exactly one of the three propositions $p, q, r$ is true.
Consider the expression $(p \vee q \vee r) \wedge \sim ((p \wedge q) \vee (q \wedge r) \vee (r \wedge p))$.
$(p \vee q \vee r)$ is true if at least one of $p, q, r$ is true.
$((p \wedge q) \vee (q \wedge r) \vee (r \wedge p))$ is true if at least two of $p, q, r$ are true.
Therefore,their conjunction is true if and only if exactly one of $p, q, r$ is true.
This matches the given expression $S$.
Thus,the correct option is $D$.

(A) tautology is a statement that is always true for all possible truth values of its components.
Let us simplify the expression $S = p \wedge ((q \vee t) \wedge p)$.
Since $q \vee t$ is a tautology (as $t$ is a tautology),we have $q \vee t \equiv t$.
Thus,$S = p \wedge (t \wedge p) = p \wedge p = p$.
Now,check option $D$: $S \leftrightarrow t$ becomes $p \leftrightarrow t$.
Since $p$ can be false,$p \leftrightarrow t$ is not a tautology.
Check option $A$: The expression is $[(p \to (q \vee r) \wedge (p \vee t))] \leftrightarrow [\sim(q \vee r) \to \sim p]$.
Using $p \to q \equiv \sim p \vee q$,the right side is $\sim(\sim(q \vee r)) \vee \sim p = (q \vee r) \vee \sim p$.
The left side simplifies to $p \to ((q \vee r) \wedge (p \vee t))$.
Since $p \vee t$ is a tautology,this is $p \to (q \vee r) \equiv \sim p \vee (q \vee r)$.
Both sides are identical,hence the statement is a tautology.

(A) First,analyze the given statement: $(p \to \sim p) \to (p \to q)$.
Using the equivalence $A \to B \equiv \sim A \lor B$,we get $\sim (p \to \sim p) \lor (\sim p \lor q)$.
Since $\sim (p \to \sim p) \equiv \sim (\sim p \lor \sim p) \equiv \sim (\sim p) \equiv p$,the expression simplifies to $p \lor \sim p \lor q$,which is a tautology $(T)$.
Now,check the options:
Option $A$: $(p \to p) \to (p \to \sim p) \equiv T \to (\sim p \lor \sim p) \equiv \sim p \lor \sim p \equiv \sim p$. This is not a tautology.
Option $B$: $q \to (p \to q) \equiv \sim q \lor (\sim p \lor q) \equiv (\sim q \lor q) \lor \sim p \equiv T \lor \sim p \equiv T$.
Option $C$: $(q \to \sim p) \to (q \to p) \equiv \sim (\sim q \lor \sim p) \lor (\sim q \lor p) \equiv (q \land p) \lor \sim q \lor p \equiv (q \lor \sim q) \land (p \lor \sim q) \lor p \equiv T \land (p \lor \sim q \lor p) \equiv p \lor \sim q$. This is not a tautology.
Since the original statement is a tautology and options $A$ and $C$ are not,the question implies finding which is not equivalent. Given the structure,$A$ is clearly not equivalent.

(C) Let $p$ be the statement "$3^2 = 10$" and $q$ be the statement "$I$ get second prize".
The given statement is of the form $p \rightarrow q$.
We know that the logical equivalence of an implication is $p \rightarrow q \equiv \sim p \vee q$.
Here,$\sim p$ is "$3^2 \neq 10$" and $q$ is "$I$ get second prize".
Therefore,the statement is equivalent to "$3^2 \neq 10$ or $I$ get second prize".

(C) Given statements are:
$A$: Lotuses are Pink
$B$: The Earth is a planet
The negation $\sim A$ is: Lotuses are not Pink
The logical expression $(\sim A) \vee B$ represents the disjunction: $(\sim A) \text{ or } B$
Substituting the statements:
$(\sim A) \vee B$: Lotuses are not pink or the Earth is a planet.

(C) The given statement $P$ is of the form: "For every $x$,$Q(x)$ or $R(x)$",where $Q(x)$ is $x > 5$ and $R(x)$ is $x < 5$.
To find the negation of a statement involving a universal quantifier ("For every"),we change the quantifier to an existential quantifier ("There exists") and negate the inner statement.
The negation of "For every $x$,$Q(x)$ or $R(x)$" is "There exists $x$ such that $\sim(Q(x) \lor R(x))$".
Using De Morgan's Law,$\sim(Q(x) \lor R(x))$ is equivalent to $\sim Q(x) \land \sim R(x)$.
Here,$\sim(x > 5)$ is $x \leq 5$ and $\sim(x < 5)$ is $x \geq 5$.
Thus,the negation is: "There exists a real number $x$ such that $x \leq 5$ and $x \geq 5$".
This is equivalent to saying: "There exists a real number $x$ such that neither $x > 5$ nor $x < 5$" (which is true for $x = 5$).
Therefore,option $C$ is correct.

(B) To determine the nature of the statement $(p \to q) \leftrightarrow (q \vee \sim p)$,we construct a truth table:

$p$	$q$	$\sim p$	$p \to q$	$q \vee \sim p$	$(p \to q) \leftrightarrow (q \vee \sim p)$
$T$	$T$	$F$	$T$	$T$	$T$
$T$	$F$	$F$	$F$	$F$	$T$
$F$	$T$	$T$	$T$	$T$	$T$
$F$	$F$	$T$	$T$	$T$	$T$

Since the final column contains only $T$ (True) values,the statement is a tautology.

(A) To find the negation of the statement $q \wedge (\sim p \vee \sim r)$,we apply De Morgan's laws and the negation rules for logical connectives.
Let the statement be $S = q \wedge (\sim p \vee \sim r)$.
The negation is $\sim S = \sim (q \wedge (\sim p \vee \sim r))$.
Using De Morgan's law $\sim (A \wedge B) = \sim A \vee \sim B$,we get:
$\sim S = \sim q \vee \sim (\sim p \vee \sim r)$.
Using De Morgan's law $\sim (A \vee B) = \sim A \wedge \sim B$,we get:
$\sim S = \sim q \vee (\sim (\sim p) \wedge \sim (\sim r))$.
Since $\sim (\sim p) = p$ and $\sim (\sim r) = r$,the expression simplifies to:
$\sim S = \sim q \vee (p \wedge r)$.
Thus,the correct option is $A$.

(A) The inverse of the conditional statement $p \to (\sim q \wedge \sim r)$ is given by $\sim p \to \sim(\sim q \wedge \sim r)$.
Using De Morgan's Law,this simplifies to $\sim p \to (q \vee r)$.
$A$ conditional statement is false only when the antecedent is true and the consequent is false.
Therefore,$\sim p \equiv T$ and $(q \vee r) \equiv F$.
From $\sim p \equiv T$,we get $p \equiv F$.
From $(q \vee r) \equiv F$,both $q$ and $r$ must be false,so $q \equiv F$ and $r \equiv F$.
Thus,the truth values are $p \equiv F, q \equiv F, r \equiv F$.

(B) Let $p$ be the statement 'Jaipur is the capital of Rajasthan' and $q$ be the statement 'Jaipur is in India'.
The given statement is of the form $p \to q$.
The contrapositive of $p \to q$ is defined as $(\sim q) \to (\sim p)$.
Here,$\sim q$ is 'Jaipur is not in India' and $\sim p$ is 'Jaipur is not the capital of Rajasthan'.
Therefore,the contrapositive is 'If Jaipur is not in India,then Jaipur is not the capital of Rajasthan'.

(B) To determine the nature of the statement,we construct a truth table:

$p$	$q$	$p \wedge q$	$(p \wedge q) \rightarrow p$	$\sim q$	$q \wedge \sim q$	$[(p \wedge q)$ $\rightarrow p]$ $\rightarrow (q \wedge \sim q)$
$T$	$T$	$T$	$T$	$F$	$F$	$F$
$T$	$F$	$F$	$T$	$T$	$F$	$F$
$F$	$T$	$F$	$T$	$F$	$F$	$F$
$F$	$F$	$F$	$T$	$T$	$F$	$F$

Since the final column contains only $F$ (False) for all possible truth values of $p$ and $q$,the statement is a contradiction.

(B) Let $p$ be the statement: "India wins the match".
Let $q$ be the statement: "India will reach the final".
The given statement is in the form of an implication: $p \implies q$.
The negation of an implication $p \implies q$ is given by $\sim(p \implies q) \equiv p \land \sim q$.
Here,$p$ is "India wins the match" and $\sim q$ is "India will not reach the final".
Therefore,the negation is: "India wins the match and India will not reach the final".

(A) The given logical statement is $(p \wedge \sim q) \wedge r \to \sim r = F$.
An implication $A \to B$ is false $(F)$ only when $A$ is true $(T)$ and $B$ is false $(F)$.
Therefore,$(p \wedge \sim q) \wedge r = T$ and $\sim r = F$.
From $\sim r = F$,we get $r = T$.
Checking the first part: if $r = T$,then $(p \wedge \sim q) \wedge T = T$,which implies $(p \wedge \sim q) = T$.
This is possible if $p = T$ and $q = F$.
Since the condition holds for $r = T$,the truth value of $r$ is $T$.

(A) To determine the nature of the statement $(p \wedge q) \Rightarrow p$,we construct a truth table:

$p$	$q$	$p \wedge q$	$(p \wedge q) \Rightarrow p$
$T$	$T$	$T$	$T$
$T$	$F$	$F$	$T$
$F$	$T$	$F$	$T$
$F$	$F$	$F$	$T$

Since the truth value of the statement $(p \wedge q) \Rightarrow p$ is $T$ for all possible truth values of $p$ and $q$,it is a tautology.

(B) By the definition of logical equivalence,the biconditional statement $p \Leftrightarrow q$ is defined as the conjunction of the two conditional statements $p \Rightarrow q$ and $q \Rightarrow p$.
Therefore,$p \Leftrightarrow q \equiv (p$ $\Rightarrow q) \wedge (q$ $\Rightarrow p)$.

(A) The negation of the statement $\sim p \vee (p \vee (\sim q))$ is given by $\sim (\sim p \vee (p \vee \sim q))$.
Using De Morgan's Law,$\sim (A \vee B) \equiv \sim A \wedge \sim B$,we get:
$p \wedge \sim (p \vee \sim q)$.
Applying De Morgan's Law again,$\sim (p \vee \sim q) \equiv \sim p \wedge \sim (\sim q) \equiv \sim p \wedge q$.
Thus,the expression becomes $p \wedge (\sim p \wedge q)$.
By the associative property,this is equivalent to $(p \wedge \sim p) \wedge q$,which is $F \wedge q = F$.
However,evaluating the options provided,the expression $(\sim p \wedge q) \wedge p$ is equivalent to $(\sim p \wedge p) \wedge q = F \wedge q = F$. Both the original negation and option $A$ evaluate to a contradiction.

(B) To determine the nature of the statement $(p \wedge q) \to (p \vee q)$,we construct a truth table:

$p$	$q$	$p \wedge q$	$p \vee q$	$(p \wedge q) \to (p \vee q)$
$T$	$T$	$T$	$T$	$T$
$T$	$F$	$F$	$T$	$T$
$F$	$T$	$F$	$T$	$T$
$F$	$F$	$F$	$F$	$T$

Since all the truth values in the final column are $T$ (True),the statement is a tautology.

(B) The contrapositive of a conditional statement $p \Rightarrow q$ is $\neg q \Rightarrow \neg p.$ The truth value of a contrapositive is the same as the truth value of the original statement.
For statement $P: p \Rightarrow q$ where $p$ is '$7$ is an odd number' (True) and $q$ is '$7$ is divisible by $2$' (False).
Since $T \Rightarrow F$ is $F,$ the truth value $V_1 = F.$
For statement $Q: p \Rightarrow q$ where $p$ is '$7$ is a prime number' (True) and $q$ is '$7$ is an odd number' (True).
Since $T \Rightarrow T$ is $T,$ the truth value $V_2 = T.$
Thus,the ordered pair $(V_1, V_2) = (F, T).$

(C) The contrapositive of a conditional statement $p \to q$ is defined as $\sim q \to \sim p$.
Let $p$ be the statement: "The side of a square doubles."
Let $q$ be the statement: "Its area increases four times."
The contrapositive $\sim q \to \sim p$ is: "If the area of a square does not increase four times,then its side is not doubled."
Thus,option $C$ is the correct answer.

(D) The contrapositive of a conditional statement "If $P$,then $Q$" is defined as "If not $Q$,then not $P$".
Given the statement: "If it is raining $(P)$,then $I$ will not come $(Q)$".
Here,$P$ is "it is raining" and $Q$ is "$I$ will not come".
Therefore,"not $Q$" is "$I$ will come" and "not $P$" is "it is not raining".
Thus,the contrapositive is "If $I$ will come,then it is not raining".

(D) Let the statements be $P$,$Q$,and $R$.
"Suman is brilliant and dishonest" is represented as $P \wedge \sim R$.
"Suman is rich" is represented as $Q$.
The statement "Suman is brilliant and dishonest if and only if Suman is rich" is represented as $(P \wedge \sim R) \leftrightarrow Q$.
We know that the negation of $A \leftrightarrow B$ is $\sim A \leftrightarrow B$ or $A \leftrightarrow \sim B$.
Therefore,the negation of $(P \wedge \sim R) \leftrightarrow Q$ is $(P \wedge \sim R) \leftrightarrow \sim Q$,which is equivalent to $\sim Q \leftrightarrow (P \wedge \sim R)$.

(B) The given statement is "If it does not rain,then $I$ go to school".
Let $p$ be the statement "It does not rain" and $q$ be the statement "$I$ go to school".
The given statement is in the form $p \Rightarrow q$.
The contrapositive of $p \Rightarrow q$ is $\sim q \Rightarrow \sim p$.
Here,$\sim q$ is "$I$ do not go to school" and $\sim p$ is "It rains".
Therefore,the contrapositive is "If $I$ do not go to school,then it rains".

(C) Given the expression $\sim (p \vee \sim q) \vee \sim (p \vee q)$.
Using De Morgan's Law,$\sim (A \vee B) \equiv \sim A \wedge \sim B$.
So,$\sim (p \vee \sim q) \equiv \sim p \wedge q$ and $\sim (p \vee q) \equiv \sim p \wedge \sim q$.
The expression becomes $(\sim p \wedge q) \vee (\sim p \wedge \sim q)$.
Using the Distributive Law,we factor out $\sim p$:
$\sim p \wedge (q \vee \sim q)$.
Since $(q \vee \sim q)$ is a tautology $(T)$,
$\sim p \wedge T \equiv \sim p$.

(B) The given statement is $p \Rightarrow (q \vee r)$.
Using the logical equivalence $A$ $\Rightarrow (B \vee C) \equiv (A$ $\Rightarrow B) \vee (A$ $\Rightarrow C)$,we can rewrite the expression.
Thus,$p \Rightarrow (q \vee r)$ is equivalent to $(p$ $\Rightarrow q) \vee (p$ $\Rightarrow r)$.

(C) The given statement is in the form $p \Rightarrow q$,where $p$ is "$I$ am not feeling well" and $q$ is "$I$ will go to the doctor".
The contrapositive of $p \Rightarrow q$ is defined as $\neg q \Rightarrow \neg p$.
Here,$\neg q$ is "$I$ will not go to the doctor" and $\neg p$ is "$I$ am feeling well".
Therefore,the contrapositive is "If $I$ will not go to the doctor,then $I$ am feeling well".

(C) For Statement $-1$:
$A \to (B \to A) \equiv \sim A \vee (\sim B \vee A) \equiv (\sim A \vee A) \vee \sim B \equiv T \vee \sim B \equiv T$.
$A \to (A \vee B) \equiv \sim A \vee (A \vee B) \equiv (\sim A \vee A) \vee B \equiv T \vee B \equiv T$.
Since both are equivalent to $T$ (Tautology),Statement $-1$ is true.
For Statement $-2$:
Let $P = (A \wedge B) \to (\sim A \vee B)$.
If $A=T, B=T$,then $P = (T \wedge T) \to (F \vee T) = T \to T = T$.
Then $\sim P = \sim T = F$.
Since the statement is not true for all truth values,it is not a tautology. Thus,Statement $-2$ is false.

(B) We construct the truth table for the given statement and the options. The statement $p \to (q \to p)$ is equivalent to $\neg p \vee (\neg q \vee p)$,which simplifies to $(\neg p \vee p) \vee \neg q$,which is $T \vee \neg q = T$ (a tautology).
Checking option $B$: $p \to (p \vee q)$ is equivalent to $\neg p \vee (p \vee q)$,which simplifies to $(\neg p \vee p) \vee q$,which is $T \vee q = T$ (a tautology).
Since both statements are tautologies,they are logically equivalent.

$p$	$q$	$q \to p$	$p \to (q \to p)$	$p \vee q$	$p \to (p \vee q)$
$T$	$T$	$T$	$T$	$T$	$T$
$T$	$F$	$T$	$T$	$T$	$T$
$F$	$T$	$F$	$T$	$T$	$T$
$F$	$F$	$T$	$T$	$F$	$T$

(A) We evaluate the truth value of each implication by testing counterexamples:
For option $(D)$,$Q \to P$: Let $m = 8$ and $n = 4$. Here $n^2 = 16$. Since $8$ divides $16$,$Q$ is true. However,$8$ does not divide $4$,so $P$ is false. Thus,$Q \to P$ is false.
For option $(C)$,$Q \to R$: Let $m = 12$ and $n = 6$. Here $n^2 = 36$. Since $12$ divides $36$,$Q$ is true. However,$12$ is not prime,so $R$ is false. Thus,$Q \to R$ is false.
For option $(B)$,$P \wedge Q \to R$: Let $m = 4$ and $n = 8$. $P$ is true ($4$ divides $8$) and $Q$ is true ($4$ divides $64$). However,$m = 4$ is not prime,so $R$ is false. Thus,$P \wedge Q \to R$ is false.
For option $(A)$,$Q \wedge R \to P$: If $m$ is prime $(R)$ and $m$ divides $n^2$ $(Q)$,then by Euclid's Lemma,$m$ must divide $n$ $(P)$. This is a standard mathematical theorem. Therefore,$Q \wedge R \to P$ is true.

(C) The logical statement $r: p \to (\sim p \vee q)$ has a truth value $F$.
An implication $A \to B$ is false if and only if $A$ is true and $B$ is false.
Therefore,$p$ must be $T$ and $(\sim p \vee q)$ must be $F$.
For the disjunction $(\sim p \vee q)$ to be $F$,both $\sim p$ and $q$ must be $F$.
Since $\sim p$ is $F$,it implies $p$ is $T$.
Thus,$p$ is $T$ and $q$ is $F$.

(A) For option $(A)$: The contrapositive of $p \Rightarrow q$ is $\sim q \Rightarrow \sim p$. Here $p$ is $3x + 2 = 8$ and $q$ is $x = 2$. Thus,$\sim q \Rightarrow \sim p$ is $x \neq 2 \Rightarrow 3x + 2 \neq 8$. This is true.
For option $(B)$: The converse of $p \Rightarrow q$ is $q \Rightarrow p$. The converse of $\tan x = 0 \Rightarrow x = 0$ is $x = 0 \Rightarrow \tan x = 0$. Thus,$(B)$ is false.
For option $(C)$: $p \Rightarrow q$ is equivalent to $\sim p \vee q$,not $p \vee \sim q$. Thus,$(C)$ is false.
For option $(D)$: $p \vee q$ (disjunction) and $p \wedge q$ (conjunction) have different truth tables. Thus,$(D)$ is false.

(A) By definition,the biconditional statement $p \Leftrightarrow q$ is logically equivalent to the conjunction of the two conditional statements $p \Rightarrow q$ and $q \Rightarrow p$.
Therefore,$p \Leftrightarrow q \equiv (p$ $\Rightarrow q) \wedge (q$ $\Rightarrow p)$.

(B) The conditional statement $p \to q$ is defined by the truth table where it is false only when $p$ is true and $q$ is false.
This is logically equivalent to the disjunction of the negation of $p$ and $q$.
Therefore,$p \to q \equiv \sim p \vee q$.

(C) Let $p$: The sun is shining.
Let $q$: $I$ shall play tennis in the afternoon.
The given statement is of the form $p \to q$.
The negation of a conditional statement $p \to q$ is given by $\sim(p \to q) \equiv p \wedge \sim q$.
Therefore,the negation is "The sun is shining and $I$ shall not play tennis in the afternoon" which corresponds to $p \wedge \sim q$.

(C) We test the possible combinations for $(\oplus, \Theta)$ where $\oplus, \Theta \in \{\wedge, \vee\}$.
Case $1$: $(\oplus, \Theta) = (\wedge, \vee)$
$(p \wedge q) \wedge (\sim p \vee q) \equiv (p \wedge q \wedge \sim p) \vee (p \wedge q \wedge q)$
$\equiv (F \wedge q) \vee (p \wedge q) \equiv F \vee (p \wedge q) \equiv p \wedge q$.
This matches the given expression.
Case $2$: $(\oplus, \Theta) = (\wedge, \wedge)$
$(p \wedge q) \wedge (\sim p \wedge q) \equiv (p \wedge \sim p) \wedge q \equiv F \wedge q \equiv F$.
Case $3$: $(\oplus, \Theta) = (\vee, \vee)$
$(p \vee q) \wedge (\sim p \vee q) \equiv (p \wedge \sim p) \vee q \equiv F \vee q \equiv q$.
Case $4$: $(\oplus, \Theta) = (\vee, \wedge)$
$(p \vee q) \wedge (\sim p \wedge q) \equiv (p \wedge \sim p \wedge q) \vee (q \wedge \sim p \wedge q) \equiv F \vee (q \wedge \sim p) \equiv q \wedge \sim p$.
Thus,the correct ordered pair is $(\wedge, \vee)$.