Mathematical logic Questions in English

(C) Let us construct the truth table for the statement $\sim (p \rightarrow q) \Leftrightarrow (\sim p \vee \sim q)$.
$1$. For $p = T, q = T$: $\sim (T$ $\rightarrow T) \Leftrightarrow (\sim T \vee \sim T) \implies \sim (T) \Leftrightarrow (F \vee F) \implies F \Leftrightarrow F$,which is $T$.
$2$. For $p = T, q = F$: $\sim (T$ $\rightarrow F) \Leftrightarrow (\sim T \vee \sim F) \implies \sim (F) \Leftrightarrow (F \vee T) \implies T \Leftrightarrow T$,which is $T$.
$3$. For $p = F, q = T$: $\sim (F$ $\rightarrow T) \Leftrightarrow (\sim F \vee \sim T) \implies \sim (T) \Leftrightarrow (T \vee F) \implies F \Leftrightarrow T$,which is $F$.
$4$. For $p = F, q = F$: $\sim (F$ $\rightarrow F) \Leftrightarrow (\sim F \vee \sim F) \implies \sim (T) \Leftrightarrow (T \vee T) \implies F \Leftrightarrow T$,which is $F$.
Since the truth values of the statement are not all $T$ (not a tautology) and not all $F$ (not a contradiction),it is neither a tautology nor a contradiction.

(B) Let $p$ be the statement "$I$ become a teacher" and $q$ be the statement "$I$ will open a school".
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 "$I$ become a teacher" and $\sim q$ is "$I$ will not open a school".
Therefore,the negation is "$I$ will become a teacher and $I$ will not open a school".

(B) Using De Morgan's Law,$\sim (p \vee q) \equiv (\sim p \wedge \sim q)$.
So,the expression becomes $(\sim p \wedge \sim q) \vee (\sim p \wedge q)$.
By the Distributive Law,this is equivalent to $\sim p \wedge (\sim q \vee q)$.
Since $(\sim q \vee q) \equiv t$ (Tautology),the expression becomes $\sim p \wedge t$.
By the Identity Law,$\sim p \wedge t \equiv \sim p$.

(B) We know that $p \rightarrow q \equiv \sim p \vee q$.
Applying this to the given expression:
$(p$ $\rightarrow \sim p) \wedge (\sim p$ $\rightarrow p) \equiv (\sim p \vee \sim p) \wedge (p \vee p)$.
Using the idempotent law $(p \vee p \equiv p)$:
$\equiv (\sim p) \wedge (p)$.
Since $p \wedge \sim p \equiv c$ (where $c$ is a contradiction),the statement is a contradiction.

(A) Let $S = (p \wedge \sim q) \wedge (\sim p \vee q)$.
Using the distributive law,we expand the expression:
$S = [(p \wedge \sim q) \wedge \sim p] \vee [(p \wedge \sim q) \wedge q]$
Using the associative and commutative laws:
$S = [(p \wedge \sim p) \wedge \sim q] \vee [p \wedge (\sim q \wedge q)]$
Since $(p \wedge \sim p) = c$ (contradiction) and $(\sim q \wedge q) = c$:
$S = (c \wedge \sim q) \vee (p \wedge c)$
$S = c \vee c = c$
Since the result is a contradiction,the statement is a contradiction.

(D) We know that the implication $p \rightarrow q$ is equivalent to $\sim p \vee q$.
Applying this to the expression $\sim (p \rightarrow \sim q)$:
$\sim (p \rightarrow \sim q) \equiv \sim (\sim p \vee \sim q)$.
Using De Morgan's Law,$\sim (\sim p \vee \sim q) \equiv (\sim \sim p \wedge \sim \sim q) \equiv (p \wedge q)$.
Therefore,the expression $\sim (p \rightarrow \sim q)$ is logically equivalent to $(p \wedge q)$.

(D) statement in logic is a declarative sentence that is either true or false,but not both.
$(a)$ 'Open the door' is an imperative sentence (command).
$(b)$ 'Do your homework' is an imperative sentence (command).
$(c)$ 'Hurrah! We won the match' is an exclamatory sentence.
$(d)$ 'The sum of two and two is five' is a declarative sentence that is false. Since it has a definite truth value (false),it is a mathematical statement.

(A) conditional statement $p \rightarrow S$ is false if and only if $p$ is $T$ and $S$ is $F$.
Here,$p \rightarrow (q \vee r)$ is false.
This implies $p$ is $T$ and $(q \vee r)$ is $F$.
For the disjunction $(q \vee r)$ to be $F$,both $q$ and $r$ must be $F$.
Therefore,the truth values are $p = T, q = F, r = F$.

(A) Statement-$1$: $\sim (p \Leftrightarrow \sim q)$ is equivalent to $p \Leftrightarrow q$.
Statement-$2$: $\sim (p \Leftrightarrow \sim q)$ is a tautology.

$p$	$q$	$\sim q$	$(p \Leftrightarrow q)$	$(p \Leftrightarrow \sim q)$	$\sim (p \Leftrightarrow \sim q)$
$T$	$T$	$F$	$T$	$F$	$T$
$T$	$F$	$T$	$F$	$T$	$F$
$F$	$T$	$F$	$F$	$T$	$F$
$F$	$F$	$T$	$T$	$F$	$T$

From the truth table,we can see that the column for $\sim (p \Leftrightarrow \sim q)$ is identical to the column for $(p \Leftrightarrow q)$,so Statement-$1$ is true.
However,the column for $\sim (p \Leftrightarrow \sim q)$ contains both $T$ and $F$ values,so it is not a tautology. Thus,Statement-$2$ is false.

(D) We know the following logical identities:
$p \vee (\sim p) = t$ (Tautology)
$p \wedge (\sim p) = c$ (Contradiction)
$p \wedge p = p$
$p \vee p = p$
Comparing these with the given options,$p \vee p = p$ is true.

(A) Let $p$ be the statement: "$12$ is a multiple of $3$".
Let $q$ be the statement: "$12$ is a multiple of $4$".
The given statement is $p \wedge q$.
The negation of $p \wedge q$ is $\sim (p \wedge q) = \sim p \vee \sim q$ (by De Morgan's Law).
Therefore,the negation is: "$12$ is not a multiple of $3$ or $12$ is not a multiple of $4$".

(A) We know that the implication $p \rightarrow (q \vee r)$ is false only when the antecedent $p$ is true and the consequent $(q \vee r)$ is false.
For the disjunction $(q \vee r)$ to be false,both $q$ and $r$ must be false.
Therefore,the truth values are $p = T, q = F, r = F$.

(D) To check if $[p \wedge (p$ $\rightarrow q)]$ $\rightarrow q$ is a tautology:
$1$. Use the implication law $p \rightarrow q \equiv \sim p \vee q$:
$[p \wedge (\sim p \vee q)] \rightarrow q$
$2$. Apply the distributive law:
$[(p \wedge \sim p) \vee (p \wedge q)] \rightarrow q$
$3$. Since $p \wedge \sim p \equiv F$ (contradiction):
$[F \vee (p \wedge q)] \rightarrow q$
$4$. Simplify using the identity law $F \vee r \equiv r$:
$(p \wedge q) \rightarrow q$
$5$. Apply the implication law again:
$\sim (p \wedge q) \vee q$
$6$. Apply De Morgan's law:
$(\sim p \vee \sim q) \vee q$
$7$. Use the associative law:
$\sim p \vee (\sim q \vee q)$
$8$. Since $\sim q \vee q \equiv T$ (tautology):
$\sim p \vee T \equiv T$
Since the result is $T$,the statement is a tautology.

(B) We evaluate the truth value of the statement $p$ $\Rightarrow (q$ $\Rightarrow p)$:
$p$ $\Rightarrow (q$ $\Rightarrow p) \equiv \sim p \vee (q$ $\Rightarrow p)$
$\equiv \sim p \vee (\sim q \vee p)$
$\equiv (\sim p \vee p) \vee \sim q$
$\equiv T \vee \sim q \equiv T$
Now,we evaluate the options:
For option $B$: $p \Rightarrow (p \vee q) \equiv \sim p \vee (p \vee q)$
$\equiv (\sim p \vee p) \vee q$
$\equiv T \vee q \equiv T$
Since both the given statement and option $B$ are tautologies (always true),they are logically equivalent.

(A) statement is a contradiction if its truth value is always false for all possible truth values of its components.
Let us evaluate the expression $(p \wedge q) \wedge (\sim (p \vee q))$.
Using De Morgan's Law,$\sim (p \vee q) \equiv (\sim p \wedge \sim q)$.
So,the expression becomes $(p \wedge q) \wedge (\sim p \wedge \sim q)$.
By associative and commutative laws,this is equivalent to $(p \wedge \sim p) \wedge (q \wedge \sim q)$.
Since $(p \wedge \sim p)$ is always false $(F)$ and $(q \wedge \sim q)$ is always false $(F)$,the entire expression is $F \wedge F$,which is always $F$.
Therefore,$(p \wedge q) \wedge (\sim (p \vee q))$ is a contradiction.

(A) The given statement is $(p \wedge q) \rightarrow p$.
Using the logical equivalence $(A \rightarrow B) \equiv (\neg A \vee B)$,we get:
$\neg (p \wedge q) \vee p$
Applying De Morgan's Law:
$(\neg p \vee \neg q) \vee p$
By the Associative and Commutative laws:
$(\neg p \vee p) \vee \neg q$
Since $(\neg p \vee p) \equiv T$ (Tautology):
$T \vee \neg q \equiv T$
Therefore,the statement is a tautology.

(C) Let $p$ and $q$ be the following statements:
$p$: $A$ quadrilateral is a square.
$q$: $A$ quadrilateral is a rhombus.
The given statement is $p \rightarrow q$.
The negation of a conditional statement $p \rightarrow q$ is given by $\sim(p \rightarrow q) \equiv p \wedge \sim q$.
Therefore,the negation is: "$A$ quadrilateral is a square and it is not a rhombus."

(A) The statement $p$ is defined as: $x \in S$ is a rational number such that $x > 0$.
The negation of a statement involving a condition $x > 0$ is $x \leq 0$.
Therefore,the negation $\sim p$ is: $x \in S$ is a rational number such that $x \leq 0$.

(C) Given that the implication $(p \vee \sim r) \rightarrow (q \wedge r)$ is false.
An implication $A \rightarrow B$ is false only when $A$ is true and $B$ is false.
Therefore,$(p \vee \sim r)$ is true and $(q \wedge r)$ is false.
Since $(q \wedge r)$ is false and $q$ is true,$r$ must be false.
Now,substitute $r = \text{False}$ into $(p \vee \sim r) = \text{True}$.
$(p \vee \sim \text{False}) = \text{True} \Rightarrow (p \vee \text{True}) = \text{True}$.
Since $(p \vee \text{True})$ is always true regardless of the value of $p$,$p$ can be either true or false.

(D) The dual of a statement is obtained by replacing $\wedge$ with $\vee$,$\vee$ with $\wedge$,$c$ with $t$,and $t$ with $c$ simultaneously.
Given the statement $p \wedge (\sim p) = c$,replacing the operators and constants gives the dual statement $p \vee (\sim p) = t$.

(C) The dual of a compound statement is obtained by replacing 'and' $(land)$ with 'or' $(lor)$ and vice versa.
Given statement: $p \land q$,where $p$ is "Rina is healthy" and $q$ is "Mina is beautiful".
The dual statement is $p \lor q$.
Therefore,the dual statement is "Rina is healthy or Mina is beautiful".

(B) The converse of $p \Rightarrow q$ is $q \Rightarrow p$.
The contrapositive of $q \Rightarrow p$ is $\sim p \Rightarrow \sim q$.
Therefore,the contrapositive of the converse of $p \Rightarrow q$ is $\sim p \Rightarrow \sim q$.

(D) statement in logic is a declarative sentence that is either true or false,but not both.
$(a)$ 'Every set is an infinite set' is a false statement.
$(b)$ 'Every square is a rectangle' is a true statement.
$(c)$ 'The Sun is a star' is a true statement.
$(d)$ 'Close the window' is an imperative sentence (a command),which cannot be classified as true or false.
Therefore,it is not a statement.

(A) Consider the expression $(\sim p \vee \sim q) \vee (p \vee \sim q)$.
Using the associative and commutative laws of logic:
$\equiv (\sim p \vee p) \vee (\sim q \vee \sim q)$
$\equiv t \vee \sim q$
Since $t \vee \text{anything} \equiv t$,where $t$ is a tautology:
$\equiv t$
Therefore,$(\sim p \vee \sim q) \vee (p \vee \sim q)$ is a tautology.

(D) Let the given statements be $p$,$q$,and $r$.
The statement "Suman is brilliant and dishonest" is represented as $(p \wedge \sim r)$.
The statement "Suman is rich if and only if Suman is brilliant and dishonest" is represented as $q \Leftrightarrow (p \wedge \sim r)$.
The negation of a statement $S$ is denoted by $\sim S$.
Therefore,the negation of the given statement is $\sim (q \Leftrightarrow (p \wedge \sim r))$.

(B) Given the statement: $(p \wedge \sim q) \wedge (\sim p \vee q)$
Using De Morgan's Law,we know that $\sim p \vee q \equiv \sim (p \wedge \sim q)$.
Substituting this into the expression,we get: $(p \wedge \sim q) \wedge \sim (p \wedge \sim q)$.
Let $X = (p \wedge \sim q)$. Then the expression becomes $X \wedge \sim X$.
By the law of contradiction (or complement law),$X \wedge \sim X \equiv c$,where $c$ is a contradiction.

(D) statement is a tautology if its truth value is always $T$ (True) for all possible truth values of its components.
$1$. For $(p$ $\Rightarrow q) \wedge p$ $\Rightarrow q$:
$(\sim p \vee q) \wedge p$ $\Rightarrow q = ((\sim p \wedge p) \vee (q \wedge p))$ $\Rightarrow q = (F \vee (p \wedge q))$ $\Rightarrow q = (p \wedge q)$ $\Rightarrow q = \sim (p \wedge q) \vee q = \sim p \vee \sim q \vee q = \sim p \vee T = T$.
$2$. For $(p \vee q) \wedge (\sim p) \Rightarrow q$:
$((p \wedge \sim p) \vee (q \wedge \sim p))$ $\Rightarrow q = (F \vee (q \wedge \sim p))$ $\Rightarrow q = (q \wedge \sim p)$ $\Rightarrow q = \sim (q \wedge \sim p) \vee q = \sim q \vee p \vee q = T \vee p = T$.
$3$. For $(p$ $\Rightarrow q) \wedge (q$ $\Rightarrow r)$ $\Rightarrow (p$ $\Rightarrow r)$:
This is the standard law of hypothetical syllogism,which is a tautology.
Since all the given statements are tautologies,the correct answer is 'None of these'.

(C) We know that $p \Leftrightarrow q$ is equivalent to $(p$ $\Rightarrow q) \wedge (q$ $\Rightarrow p)$.
Therefore,$\sim (p \Leftrightarrow q) = \sim ((p$ $\Rightarrow q) \wedge (q$ $\Rightarrow p))$.
Using De Morgan's Law,$\sim (A \wedge B) = \sim A \vee \sim B$,we get:
$\sim (p$ $\Rightarrow q) \vee \sim (q$ $\Rightarrow p)$.
Since $\sim (p \Rightarrow q) \equiv p \wedge \sim q$,the expression becomes:
$(p \wedge \sim q) \vee (q \wedge \sim p)$.

(A) In Boolean algebra,the disjunction (sum) operation is denoted by $\vee$.
For any logical statement $p$,the identity element $c$ for disjunction must satisfy $p \vee c = p$.
Since $p \vee F = p$ (where $F$ is a contradiction/false statement),the identity element for disjunction is the contradiction $F$.
In the context of Boolean algebra,the contradiction is denoted as $c$ or $\sim t$ (where $t$ is a tautology).
Therefore,the identity element is $\sim t$.

(C) Let $p: x \in A$,$q: x \in B$,and $r: x \in A \cup B$.
The given statement is $p \vee q \Rightarrow r$.
The contrapositive of an implication $P \Rightarrow Q$ is $\sim Q \Rightarrow \sim P$.
Here,the contrapositive is $\sim r \Rightarrow \sim (p \vee q)$.
Using De Morgan's Law,$\sim (p \vee q) \equiv (\sim p) \wedge (\sim q)$.
Thus,the contrapositive is: If $x \notin A \cup B$,then $x \notin A$ and $x \notin B$.

(A) In the Boolean algebra of logical statements,the conjunction operation is denoted by $\wedge$.
We know that for any statement $p$,$p \wedge t = p = t \wedge p$,where $t$ represents a tautology.
Therefore,$t$ is the identity element for the conjunction operation.
Since $t = \sim c$ (where $c$ is a contradiction),the identity element is $\sim c$.

(C) mathematical statement is a declarative sentence that is either true or false,but not both.
$1$. "$I$ am a lion" is not a statement because its truth value depends on the speaker.
$2$. "Logic is an interesting subject" is not a statement because "interesting" is subjective.
$3$. "$A$ triangle is a circle and $10$ is a prime number" is a compound statement formed by two components connected by "and". Since both components are false (a triangle is not a circle,and $10$ is not a prime number),the compound statement is false. Therefore,it is a valid mathematical statement.

(D) $1$. $p \wedge q$ is true only when both $p$ and $q$ are true. Thus,option $A$ is incorrect.
$2$. $p \rightarrow q$ is false only when $p$ is true and $q$ is false. Thus,option $B$ is incorrect.
$3$. $p \Leftrightarrow q$ is true when both $p$ and $q$ have the same truth value (both true or both false). Thus,option $C$ is incorrect.
$4$. $\sim (p \vee q)$ is true only when $(p \vee q)$ is false,which occurs only when both $p$ and $q$ are false. Thus,option $D$ is correct.

(D) The statement is $(p \wedge q) \vee (q \wedge r)$.
By distributive law,this is equivalent to $q \wedge (p \vee r)$.
For this statement to be true,both $q$ must be true $AND$ $(p \vee r)$ must be true.
This means $q$ is true,and at least one of $p$ or $r$ is true.
Looking at the options,if $q$ and $r$ are true and $p$ is false,then $(p \vee r)$ is true,so $q \wedge (p \vee r)$ is true.
Thus,option $D$ is a valid condition for the statement to be true.

(D) The negation of a conditional statement $p \rightarrow q$ is given by $\sim (p \rightarrow q) \equiv p \wedge \sim q$.
Applying this to the given statement $p \rightarrow (q \wedge r)$:
$\sim (p \rightarrow (q \wedge r)) \equiv p \wedge \sim (q \wedge r)$.
Using De Morgan's Law,$\sim (q \wedge r) \equiv \sim q \vee \sim r$.
Therefore,the negation is $p \wedge (\sim q \vee \sim r)$.

(B) The given statement is $p$ $\rightarrow (q$ $\rightarrow p)$.
Using the logical equivalence $A \rightarrow B \equiv \sim A \vee B$,we have:
$p$ $\rightarrow (q$ $\rightarrow p) \equiv \sim p \vee (\sim q \vee p)$.
By the associative and commutative laws,this is equivalent to $(\sim p \vee p) \vee \sim q$,which simplifies to $T \vee \sim q = T$ (a tautology).
Now,let us check option $B$: $p \rightarrow (q \vee p) \equiv \sim p \vee (q \vee p) \equiv (\sim p \vee p) \vee q \equiv T \vee q = T$.
Since both expressions are tautologies,the statement $p$ $\rightarrow (q$ $\rightarrow p)$ is equivalent to $p \rightarrow (q \vee p)$.

(D) To evaluate Statement $-1$,we analyze the truth table for $\sim (p \leftrightarrow \sim q)$.
Recall 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,$\sim (p \leftrightarrow \sim q)$ is equivalent to $\sim (\sim (p \leftrightarrow q))$,which simplifies to $(p \leftrightarrow q)$.
So,Statement $-1$ is true.
To evaluate Statement $-2$,we check if $\sim (p \leftrightarrow \sim q)$ is a tautology.
Since $\sim (p \leftrightarrow \sim q) \equiv (p \leftrightarrow q)$,and $(p \leftrightarrow q)$ is not true for all values of $p$ and $q$ (it is false when $p$ and $q$ have different truth values),the expression is not a tautology.
Thus,Statement $-2$ is false.

(A) Let the given statements be:
$P :$ Suman is brilliant
$Q :$ Suman is rich
$R :$ Suman is honest
The statement "Suman is brilliant and dishonest" is represented as $(P \wedge \sim R)$.
The statement "Suman is brilliant and dishonest if and only if Suman is rich" is represented as $(P \wedge \sim R) \leftrightarrow Q$.
Since the biconditional operator is commutative,this is equivalent to $Q \leftrightarrow (P \wedge \sim R)$.
The negation of a statement $S$ is denoted by $\sim S$.
Therefore,the negation of the given statement is $\sim (Q \leftrightarrow (P \wedge \sim R))$.

(A) Let $p$ be the statement "$I$ become a teacher" and $q$ be the statement "$I$ will open a school".
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 "$I$ become a teacher" and $\sim q$ is "$I$ will not open a school".
Therefore,the negation is "$I$ will become a teacher and $I$ will not open a school".

(D) Statement-$2$: $(p \rightarrow q) \leftrightarrow (\sim q \rightarrow \sim p)$
Since $(\sim q \rightarrow \sim p)$ is the contrapositive of $(p \rightarrow q)$,they have the same truth values.
Thus,$(p \rightarrow q) \leftrightarrow (p \rightarrow q)$ is always true,which means it is a tautology.
So,Statement-$2$ is true.
Statement-$1$: $(p \wedge \sim q) \wedge (\sim p \wedge q)$
Using the associative and commutative laws:
$= (p \wedge \sim p) \wedge (\sim q \wedge q)$
$= F \wedge F = F$
Since the result is always false,it is a fallacy.
So,Statement-$1$ is true.
Both statements are true,and Statement-$2$ is not an explanation for Statement-$1$.

(C) To determine the nature of the statement $\sim(p \leftrightarrow \sim q)$,we construct a truth table:
$1$. The biconditional $p \leftrightarrow \sim q$ is true when $p$ and $\sim q$ have the same truth value.
$2$. The negation $\sim(p \leftrightarrow \sim q)$ is true when $p \leftrightarrow \sim q$ is false.
Truth Table:
$p$ | $q$ | $\sim q$ | $p \leftrightarrow \sim q$ | $\sim(p \leftrightarrow \sim q)$
$T$ | $T$ | $F$ | $F$ | $T$
$T$ | $F$ | $T$ | $T$ | $F$
$F$ | $T$ | $F$ | $T$ | $F$
$F$ | $F$ | $T$ | $F$ | $T$
Comparing this to $p \leftrightarrow q$:
$p$ | $q$ | $p \leftrightarrow q$
$T$ | $T$ | $T$
$T$ | $F$ | $F$
$F$ | $T$ | $F$
$F$ | $F$ | $T$
Since the truth values of $\sim(p \leftrightarrow \sim q)$ and $(p \leftrightarrow q)$ are identical for all combinations of $p$ and $q$,the statement is equivalent to $(p \leftrightarrow q)$.

(A) Let the given expression be $P = \sim s \vee (\sim r \wedge s)$.
We want to find the negation $\sim P = \sim (\sim s \vee (\sim r \wedge s))$.
Using De Morgan's Law,$\sim (A \vee B) = \sim A \wedge \sim B$,we get:
$\sim P = \sim (\sim s) \wedge \sim (\sim r \wedge s)$.
This simplifies to $s \wedge (\sim (\sim r) \vee \sim s)$ by applying De Morgan's Law again.
So,$\sim P = s \wedge (r \vee \sim s)$.
Using the Distributive Law,$A \wedge (B \vee C) = (A \wedge B) \vee (A \wedge C)$,we get:
$\sim P = (s \wedge r) \vee (s \wedge \sim s)$.
Since $(s \wedge \sim s)$ is a contradiction (False),we have:
$\sim P = (s \wedge r) \vee F = s \wedge r$.

(A) Let the expression be $E = (p \wedge \sim q) \vee q \vee (\sim p \wedge q)$.
Using the distributive law,$(p \wedge \sim q) \vee q \equiv (p \vee q) \wedge (\sim q \vee q)$.
Since $(\sim q \vee q) \equiv T$ (Tautology),we have $(p \vee q) \wedge T \equiv p \vee q$.
Now,substitute this back into the expression:
$E \equiv (p \vee q) \vee (\sim p \wedge q)$.
Using the distributive law again,$(p \vee q) \vee (\sim p \wedge q) \equiv (p \vee q \vee \sim p) \wedge (p \vee q \vee q)$.
Since $(p \vee \sim p) \equiv T$,we have $(T \vee q) \wedge (p \vee q)$.
Since $(T \vee q) \equiv T$,we have $T \wedge (p \vee q) \equiv p \vee q$.
Thus,the expression is equivalent to $p \vee q$.

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

$p$	$q$	$\sim p$	$p \to q$	$\sim p \to q$	$(\sim p \to q) \to q$	$(p \to q) \to [(\sim p \to q) \to q]$
$T$	$T$	$F$	$T$	$T$	$T$	$T$
$T$	$F$	$F$	$F$	$T$	$F$	$T$
$F$	$T$	$T$	$T$	$T$	$T$	$T$
$F$	$F$	$T$	$T$	$F$	$T$	$T$

Since the final column contains only $T$ (True) values for all possible truth values of $p$ and $q$,the statement is a tautology.

(D) Using De Morgan's Law,$\sim(p \vee q) \equiv (\sim p \wedge \sim q)$.
Substituting this into the expression:
$(\sim p \wedge \sim q) \vee (\sim p \wedge q)$
Applying the Distributive Law:
$\sim p \wedge (\sim q \vee q)$
Since $(\sim q \vee q) \equiv T$ (Tautology):
$\sim p \wedge T \equiv \sim p$.

(B) Let $p$ be the statement: "The examination is difficult".
Let $q$ be the statement: "$I$ shall pass".
Let $r$ be the statement: "$I$ study hard".
The given proposition is: $p \Rightarrow (r \Rightarrow q)$.
We know that the negation of an implication $A \Rightarrow B$ is $A \wedge \sim B$.
Therefore,$\sim(p \Rightarrow (r \Rightarrow q)) \equiv p \wedge \sim(r \Rightarrow q)$.
Since $\sim(r \Rightarrow q) \equiv r \wedge \sim q$,the negation is $p \wedge (r \wedge \sim q)$.
This translates to: "The examination is difficult and $I$ study hard but $I$ shall not pass."

(C) The implication $p \Rightarrow q$ is a logical connective that is false only in one specific case.
According to the truth table for implication:
- If $p$ is true and $q$ is true,$p \Rightarrow q$ is true.
- If $p$ is true and $q$ is false,$p \Rightarrow q$ is false.
- If $p$ is false and $q$ is true,$p \Rightarrow q$ is true.
- If $p$ is false and $q$ is false,$p \Rightarrow q$ is true.
Therefore,$p \Rightarrow q$ is false only when $p$ is true and $q$ is false.

(C) The biconditional statement $p \Leftrightarrow \sim q$ is true if and only if both $p$ and $\sim q$ have the same truth value.
Case $1$: $p$ is true and $\sim q$ is true. This implies $p$ is true and $q$ is false.
Case $2$: $p$ is false and $\sim q$ is false. This implies $p$ is false and $q$ is true.
Comparing this with the given options,option $C$ ($p$ is false and $q$ is true) makes the statement true.