Mathematical logic Questions in English

(A) Let $p$ : The surface area increases.
Let $q$ : The pressure decreases.
The given statement is $p \rightarrow q$.
The inverse of a conditional statement $p \rightarrow q$ is defined as $\sim p \rightarrow \sim q$.
Here,$\sim p$ is "The surface area does not increase" and $\sim q$ is "The pressure does not decrease".
Therefore,the inverse statement is "If the surface area does not increase,then the pressure does not decrease.".
Thus,Option $(A)$ is correct.

(D) Let $p$: $A$ quadrilateral is a square.
Let $q$: All sides of the quadrilateral are equal.
Statement $1$ is $p \rightarrow q$.
Statement $2$ is $q \rightarrow p$.
By definition,the converse of a conditional statement $p \rightarrow q$ is $q \rightarrow p$.
Therefore,Statement $2$ is the converse of Statement $1$.

(B) Let $S$ be the statement $(p \wedge q) \rightarrow (p \vee \sim q)$.
The inverse of $S$ is $\sim(p \wedge q) \rightarrow \sim(p \vee \sim q)$.
Using the logical equivalence $A \rightarrow B \equiv \sim A \vee B$,the inverse is:
$\sim[\sim(p \wedge q)] \vee \sim(p \vee \sim q)$
$\equiv (p \wedge q) \vee (\sim p \wedge q)$ (By De Morgan's Law)
$\equiv (q \wedge p) \vee (q \wedge \sim p)$ (By Commutative Law)
$\equiv q \wedge (p \vee \sim p)$ (By Distributive Law)
$\equiv q \wedge T$ (By Complement Law,where $T$ is a tautology)
$\equiv q$ (By Identity Law).
Now,the negation of the inverse is $\sim(q) = \sim q$.

(C) We simplify the expression using logical laws:
$(p \wedge \sim q) \vee q \vee (\sim p \wedge q)$
Using the distributive law on the first two terms:
$\equiv ((p \vee q) \wedge (\sim q \vee q)) \vee (\sim p \wedge q)$
Since $(\sim q \vee q) \equiv T$ (Complement law):
$\equiv ((p \vee q) \wedge T) \vee (\sim p \wedge q)$
$\equiv (p \vee q) \vee (\sim p \wedge q)$
Applying the distributive law again:
$\equiv (p \vee q \vee \sim p) \wedge (p \vee q \vee q)$
Since $(p \vee \sim p) \equiv T$ and $(q \vee q) \equiv q$ (Idempotent law):
$\equiv (T \vee q) \wedge (p \vee q)$
Since $(T \vee q) \equiv T$:
$\equiv T \wedge (p \vee q)$
$\equiv p \vee q$
Therefore,the expression is equivalent to $p \vee q$.

(B) Given that $q$ is false and $p \wedge q \leftrightarrow r$ is true.
Since $q \equiv F$,then $p \wedge q \equiv F$.
For the biconditional $p \wedge q \leftrightarrow r$ to be true,$r$ must have the same truth value as $p \wedge q$.
Therefore,$r \equiv F$.
Now,let us evaluate the options:
$(A)$ $p \vee r \equiv p \vee F \equiv p$,which is not a tautology.
$(B)$ $(p \wedge r)$ $\rightarrow (p \vee r) \equiv (p \wedge F)$ $\rightarrow (p \vee F) \equiv F$ $\rightarrow p$. Since $F \rightarrow p$ is always true for any truth value of $p$,this is a tautology.
$(C)$ $(p \vee r)$ $\rightarrow (p \wedge r) \equiv (p \vee F)$ $\rightarrow (p \wedge F) \equiv p$ $\rightarrow F$,which is not a tautology.
$(D)$ $p \wedge r \equiv p \wedge F \equiv F$,which is a contradiction.

(B) The contrapositive of the statement $(p \vee \sim q) \rightarrow (p \wedge \sim q)$ is given by $\sim (p \wedge \sim q) \rightarrow \sim (p \vee \sim q)$.
Using the equivalence $A \rightarrow B \equiv \sim A \vee B$,we get:
$\sim (p \wedge \sim q) \vee \sim (p \vee \sim q)$
Applying De Morgan's law:
$(\sim p \vee q) \vee (\sim p \wedge q)$
Now,we find the negation of this contrapositive:
$\sim [(\sim p \vee q) \vee (\sim p \wedge q)]$
Using De Morgan's law again:
$\sim (\sim p \vee q) \wedge \sim (\sim p \wedge q)$
$(p \wedge \sim q) \wedge (p \vee \sim q)$
Wait,let us re-evaluate the negation of the contrapositive directly:
The contrapositive is $C = \sim (p \wedge \sim q) \rightarrow \sim (p \vee \sim q)$.
The negation of $A \rightarrow B$ is $A \wedge \sim B$.
Thus,$\sim C = \sim (p \wedge \sim q) \wedge \sim [\sim (p \vee \sim q)]$
$\sim C = (\sim p \vee q) \wedge (p \vee \sim q)$.

(D) To determine when the statement $[(p$ $\rightarrow q) \wedge \sim q]$ $\rightarrow r$ is a tautology,we analyze the expression $[(p \rightarrow q) \wedge \sim q]$.
Note that $(p \rightarrow q) \equiv (\sim p \vee q)$.
Thus,$[(p \rightarrow q) \wedge \sim q] \equiv [(\sim p \vee q) \wedge \sim q]$.
Using the distributive law,this becomes $(\sim p \wedge \sim q) \vee (q \wedge \sim q)$.
Since $(q \wedge \sim q) \equiv F$ (a contradiction),the expression simplifies to $(\sim p \wedge \sim q) \vee F \equiv \sim p \wedge \sim q$.
Now,the original statement is $(\sim p \wedge \sim q) \rightarrow r$.
For this to be a tautology,the implication must be true for all truth values of $p$ and $q$.
If we set $r \equiv \sim p \wedge \sim q$,the statement becomes $(\sim p \wedge \sim q) \rightarrow (\sim p \wedge \sim q)$,which is always true.
However,looking at the options,we check if $r \equiv \sim q$ works.
If $r \equiv \sim q$,the statement is $(\sim p \wedge \sim q) \rightarrow \sim q$.
Since $(\sim p \wedge \sim q)$ is a subset of $\sim q$ (i.e.,if $\sim p \wedge \sim q$ is true,then $\sim q$ must be true),the implication is always true.
Therefore,the statement is a tautology when $r \equiv \sim q$.

(A) Let $p$: The number is an odd number.
Let $q$: The number is divisible by $3$.
The given statement is $p \leftrightarrow q$.
The negation of a biconditional statement is $\sim(p \leftrightarrow q) \equiv (p \wedge \sim q) \vee (q \wedge \sim p)$.
Thus,the negation is: "The number is an odd number but not divisible by $3$ $OR$ the number is divisible by $3$ but not an odd number."

(B) Given expression: $[p \wedge (q \vee r)] \vee [\sim r \wedge \sim q \wedge p]$
Using the commutative law on the second part: $[p \wedge (q \vee r)] \vee [p \wedge \sim q \wedge \sim r]$
Applying De Morgan's law to the second part: $[p \wedge (q \vee r)] \vee [p \wedge \sim (q \vee r)]$
Using the distributive law: $p \wedge [(q \vee r) \vee \sim (q \vee r)]$
Using the complement law: $p \wedge T$ (where $T$ is a tautology)
Using the identity law: $p$
Therefore,the statement is equivalent to $p$.

(B) The statement $p \leftrightarrow (q \rightarrow p)$ is false.
This biconditional is false if the truth values of $p$ and $(q \rightarrow p)$ are different.
If $p$ is $T$,then $(q \rightarrow T)$ is $T$,so $T \leftrightarrow T$ is $T$.
If $p$ is $F$,then $(q \rightarrow F)$ must be $T$ for the biconditional to be false.
For $(q \rightarrow F)$ to be $T$,$q$ must be $F$.
Thus,$p \equiv F$ and $q \equiv F$.
Now,check option $(B)$: $p$ $\rightarrow (p \vee \sim q) \equiv F$ $\rightarrow (F \vee \sim F) \equiv F$ $\rightarrow (F \vee T) \equiv F$ $\rightarrow T \equiv T$.
Since the result is $T$,option $(B)$ is the correct statement pattern.

(D) Let the given statement be $S = (p \wedge \sim q) \vee q \vee (\sim p \wedge q)$.
Using the associative and commutative laws,we can rearrange the terms:
$S = \{(p \wedge \sim q) \vee (\sim p \wedge q)\} \vee q$
We know that $(p \wedge \sim q) \vee (\sim p \wedge q)$ is the logical expression for the exclusive $OR$,denoted as $p \oplus q$ or $\sim(p \Leftrightarrow q)$.
So,$S = (p \oplus q) \vee q$.
Using the distributive law: $(p \oplus q) \vee q \equiv (p \vee q) \wedge (\sim q \vee q) \equiv (p \vee q) \wedge T \equiv p \vee q$.
Alternatively,using a truth table:
| $p$ | $q$ | $p \wedge \sim q$ | $\sim p \wedge q$ | $(p \wedge \sim q) \vee (\sim p \wedge q)$ | $S = \{(p \wedge \sim q) \vee (\sim p \wedge q)\} \vee q$ | $p \vee q$ |
|---|---|---|---|---|---|---|
| $T$ | $T$ | $F$ | $F$ | $F$ | $T$ | $T$ |
| $T$ | $F$ | $T$ | $F$ | $T$ | $T$ | $T$ |
| $F$ | $T$ | $F$ | $T$ | $T$ | $T$ | $T$ |
| $F$ | $F$ | $F$ | $F$ | $F$ | $F$ | $F$ |
Since the columns for $S$ and $p \vee q$ are identical,the statement is equivalent to $p \vee q$.

(C) Using De Morgan's law,$\sim(p \vee q) \equiv (\sim p \wedge \sim q)$.
So,the expression becomes $(\sim p \wedge \sim q) \vee (\sim p \wedge q)$.
Applying the distributive law,we factor out $\sim p$:
$\sim p \wedge (\sim q \vee q)$.
Since $(\sim q \vee q) \equiv t$ (a tautology),
the expression simplifies to $\sim p \wedge t \equiv \sim p$.

(A) The given statement is of the form 'If $p$,then $q$',where $p$ is '$\forall x, x$ is a complex number' and $q$ is '$x^2 < 0$'.
The negation of 'If $p$,then $q$' is '$p$ and (not $q$)' where the negation of '$\forall x, P(x)$' is '$\exists x, \neg P(x)$'.
However,for the conditional statement $p \implies q$,the negation is $p \land \neg q$.
Here,$p$ is '$\forall x, x$ is a complex number' and $\neg q$ is '$x^2 \geq 0$'.
Thus,the negation is '$\forall x, x$ is a complex number and $x^2 \geq 0$'.

(D) The contrapositive of a conditional statement $p \rightarrow q$ is defined as $\sim q \rightarrow \sim p$.
Given the statement $(\sim p \wedge q) \rightarrow (q \wedge \sim r)$,we identify $p$ as $(\sim p \wedge q)$ and $q$ as $(q \wedge \sim r)$.
The contrapositive is $\sim (q \wedge \sim r) \rightarrow \sim (\sim p \wedge q)$.
Applying De Morgan's Laws:
$\sim (q \wedge \sim r) \equiv \sim q \vee \sim (\sim r) \equiv \sim q \vee r$.
$\sim (\sim p \wedge q) \equiv \sim (\sim p) \vee \sim q \equiv p \vee \sim q$.
Thus,the contrapositive is $(\sim q \vee r) \rightarrow (p \vee \sim q)$.

(B) To determine the equivalent statement,we evaluate the truth table for the given pattern $p$ $\rightarrow (q$ $\rightarrow p)$ and compare it with the options.
| $p$ | $q$ | $q \rightarrow p$ | $p$ $\rightarrow (q$ $\rightarrow p)$ |
|---|---|---|---|
| $T$ | $T$ | $T$ | $T$ |
| $T$ | $F$ | $T$ | $T$ |
| $F$ | $T$ | $F$ | $T$ |
| $F$ | $F$ | $T$ | $T$ |
The statement $p$ $\rightarrow (q$ $\rightarrow p)$ is a tautology (always true).
Now,check option $B$: $p \rightarrow (p \vee q)$.
| $p$ | $q$ | $p \vee q$ | $p \rightarrow (p \vee q)$ |
|---|---|---|---|
| $T$ | $T$ | $T$ | $T$ |
| $T$ | $F$ | $T$ | $T$ |
| $F$ | $T$ | $T$ | $T$ |
| $F$ | $F$ | $F$ | $T$ |
Since both $p$ $\rightarrow (q$ $\rightarrow p)$ and $p \rightarrow (p \vee q)$ result in a tautology,they are logically equivalent.
Thus,the correct option is $B$.

(D) To find the negation of the statement $p \vee (q \rightarrow \sim r)$,we apply De Morgan's Law: $\sim(p \vee (q$ $\rightarrow \sim r)) \equiv \sim p \wedge \sim(q$ $\rightarrow \sim r)$.
Using the logical equivalence $\sim(A \rightarrow B) \equiv A \wedge \sim B$,we get $\sim(q \rightarrow \sim r) \equiv q \wedge \sim(\sim r)$.
Since $\sim(\sim r) \equiv r$,the expression simplifies to $q \wedge r$.
Therefore,the final negation is $\sim p \wedge (q \wedge r)$.

(C) The expression $n^2+n = n(n+1)$ is the product of two consecutive natural numbers,which is always even. Thus,$p$ is true.
The expression $n^2-n = n(n-1)$ is also the product of two consecutive natural numbers,which is always even. Thus,$q$ is false.
Now,evaluating the truth values:
$p \wedge q = T \wedge F = F$
$p \vee q = T \vee F = T$
$p$ $\rightarrow q = T$ $\rightarrow F = F$
Therefore,the truth values are $F, T, F$.

(B) Let $p$ be the statement "The payment is made" and $q$ be the statement "The work is finished in time".
The given statement is $p \leftrightarrow q$.
We know that the negation of a biconditional statement is $\sim(p \leftrightarrow q) \equiv (p \wedge \sim q) \vee (\sim p \wedge q)$.
This translates to: "The payment is made and the work is not finished in time,$OR$ the payment is not made and the work is finished in time".
Comparing this with the given options,option $B$ is correct.

(A) statement pattern is a contradiction if its truth value is always false $(c)$ for all possible truth values of its components.
For $S_3 \equiv (\sim p \wedge q) \wedge (\sim q)$:
$S_3 \equiv \sim p \wedge (q \wedge \sim q)$ [Associative law]
$S_3 \equiv \sim p \wedge c$ [Since $q \wedge \sim q \equiv c$]
$S_3 \equiv c$ [Since any statement $\wedge c \equiv c$]
Since $S_3$ results in a contradiction,option $A$ is correct.

(B) Given the expression $p \rightarrow (q \vee r)$.
Using the logical equivalence $p \rightarrow q \equiv \sim p \vee q$,we get:
$p \rightarrow (q \vee r) \equiv \sim p \vee (q \vee r)$
By the associative law of disjunction:
$\equiv (\sim p \vee q) \vee (\sim p \vee r)$
Applying the definition of implication $p \rightarrow q \equiv \sim p \vee q$ again to both parts:
$\equiv (p$ $\rightarrow q) \vee (p$ $\rightarrow r)$

(C) We simplify the given expression using the laws of logic:
$(p \wedge q) \vee (\sim p \wedge q) \vee (r \wedge \sim q)$
Using the distributive law on the first two terms:
$\equiv \{(p \vee \sim p) \wedge q\} \vee (r \wedge \sim q)$
Since $(p \vee \sim p) \equiv t$ (tautology):
$\equiv (t \wedge q) \vee (r \wedge \sim q)$
$\equiv q \vee (r \wedge \sim q)$
Applying the distributive law $A \vee (B \wedge C) \equiv (A \vee B) \wedge (A \vee C)$:
$\equiv (q \vee r) \wedge (q \vee \sim q)$
Since $(q \vee \sim q) \equiv t$:
$\equiv (q \vee r) \wedge t$
$\equiv q \vee r$

(A) The contrapositive of a statement of the form $p \rightarrow q$ is $\sim q \rightarrow \sim p$.
Here,$p$ is '$A \subseteq B$ and $B \subseteq D$' and $q$ is '$A \subseteq C$'.
The negation $\sim q$ is '$A \nsubseteq C$'.
The negation $\sim p$ is '$\sim(A \subseteq B \text{ and } B \subseteq D)$',which by De Morgan's law is '$A \nsubseteq B$ or $B \nsubseteq D$'.
Thus,the contrapositive is: If $A \nsubseteq C$,then $A \nsubseteq B$ or $B \nsubseteq D$.

(C) Given that the truth value of the implication $(p \wedge \sim r) \rightarrow (\sim p \vee q)$ is $F$.
An implication $A \rightarrow B$ is $F$ only when $A$ is $T$ and $B$ is $F$.
Therefore,$(p \wedge \sim r)$ is $T$ and $(\sim p \vee q)$ is $F$.
For $(p \wedge \sim r)$ to be $T$,both $p$ must be $T$ and $\sim r$ must be $T$.
If $p$ is $T$,then $\sim p$ is $F$.
For $(\sim p \vee q)$ to be $F$ given $\sim p$ is $F$,$q$ must also be $F$.
Since $\sim r$ is $T$,$r$ must be $F$.
Thus,the truth values are $p = T, q = F, r = F$.

(D) The statement '$\text{A man is neither happy nor rich}$' means the man is not happy $AND$ the man is not rich.
Symbolically,this is represented as $(\sim p \wedge \sim q)$.
By De Morgan's Law,$(\sim p \wedge \sim q) \equiv \sim(p \vee q)$.
Thus,the correct symbolic form is $\sim(p \vee q)$.

(A) We need to find the negation of the expression $\sim s \vee (\sim r \wedge s)$.
Let $P = \sim s \vee (\sim r \wedge s)$.
The negation is $\sim P = \sim (\sim s \vee (\sim r \wedge s))$.
Using De Morgan's Law,$\sim (A \vee B) \equiv \sim A \wedge \sim B$:
$\sim P \equiv \sim (\sim s) \wedge \sim (\sim r \wedge s)$.
Using the double negation law $\sim (\sim s) \equiv s$ and De Morgan's Law $\sim (A \wedge B) \equiv \sim A \vee \sim B$:
$\sim P \equiv s \wedge (\sim (\sim r) \vee \sim s)$.
Simplifying further:
$\sim P \equiv s \wedge (r \vee \sim s)$.
Using the distributive law $A \wedge (B \vee C) \equiv (A \wedge B) \vee (A \wedge C)$:
$\sim P \equiv (s \wedge r) \vee (s \wedge \sim s)$.
Since $(s \wedge \sim s) \equiv F$ (a contradiction):
$\sim P \equiv (s \wedge r) \vee F \equiv s \wedge r$.

(D) To determine the nature of the statement pattern $\sim(p \leftrightarrow \sim q)$,we construct a truth table:
| $p$ | $q$ | $\sim p$ | $\sim q$ | $p \leftrightarrow \sim q$ | $\sim(p \leftrightarrow \sim q)$ | $p \leftrightarrow q$ |
|---|---|---|---|---|---|---|
| $T$ | $T$ | $F$ | $F$ | $F$ | $T$ | $T$ |
| $T$ | $F$ | $F$ | $T$ | $T$ | $F$ | $F$ |
| $F$ | $T$ | $T$ | $F$ | $T$ | $F$ | $F$ |
| $F$ | $F$ | $T$ | $T$ | $F$ | $T$ | $T$ |
From the truth table,we observe that the column for $\sim(p \leftrightarrow \sim q)$ is identical to the column for $(p \leftrightarrow q)$.
Therefore,the statement pattern $\sim(p \leftrightarrow \sim q)$ is equivalent to $(p \leftrightarrow q)$.

(A) First,determine the truth values of the statements:
$P: 11$ is a prime number,which is $T$ (True).
$Q: 7$ is a factor of $176$. Since $176 \div 7 = 25.14$,$7$ is not a factor,so $Q$ is $F$ (False).
$R$: $LCM$ of $3$ and $7$ is $21$,which is $T$ (True).
Now,evaluate the options:
Option $A$: $P \vee (\sim Q \wedge R) \equiv T \vee (\sim F \wedge T) \equiv T \vee (T \wedge T) \equiv T \vee T \equiv T$.
Option $B$: $(\sim P) \wedge (\sim Q \wedge R) \equiv F \wedge (T \wedge T) \equiv F \wedge T \equiv F$.
Option $C$: $(P \wedge Q) \vee (\sim R) \equiv (T \wedge F) \vee F \equiv F \vee F \equiv F$.
Option $D$: $(\sim P) \vee (Q \wedge R) \equiv F \vee (F \wedge T) \equiv F \vee F \equiv F$.
Therefore,the statement in option $A$ is true.

(D) Let $p$ be the statement '$I$ study' and $q$ be the statement '$I$ fail'.
The given statement is $p \vee q$.
The negation of a disjunction is given by De Morgan's Law: $\sim(p \vee q) \equiv \sim p \wedge \sim q$.
Here,$\sim p$ is '$I$ do not study' and $\sim q$ is '$I$ do not fail'.
Therefore,the negation is '$I$ do not study and $I$ do not fail'.

(B) We analyze each statement:
$S_1: \sim(p \rightarrow \sim q) \equiv p \wedge \sim(\sim q) \equiv p \wedge q$. Thus,$S_1$ is a tautology (correct).
$S_2: (p \wedge q) \wedge (\sim p \vee \sim q) \equiv (p \wedge q) \wedge \sim(p \wedge q)$,which is a contradiction,not a tautology.
$S_3: [p \wedge (p$ $\rightarrow \sim q)]$ $\rightarrow q \equiv [p \wedge (\sim p \vee \sim q)]$ $\rightarrow q \equiv [(p \wedge \sim p) \vee (p \wedge \sim q)]$ $\rightarrow q \equiv (F \vee (p \wedge \sim q))$ $\rightarrow q \equiv (p \wedge \sim q)$ $\rightarrow q \equiv \sim(p \wedge \sim q) \vee q \equiv (\sim p \vee q) \vee q \equiv \sim p \vee q$. This is a contingency,not a contradiction.
$S_4: p$ $\rightarrow (q$ $\rightarrow p) \equiv \sim p \vee (\sim q \vee p) \equiv (\sim p \vee p) \vee \sim q \equiv T \vee \sim q \equiv T$. This is a tautology,not a contingency.
Therefore,only $S_1$ is correct.

(B) The implication $p \rightarrow (\sim p \vee q)$ is false only when the antecedent $p$ is true and the consequent $(\sim p \vee q)$ is false.
Since $p$ is true,$\sim p$ is false.
For the disjunction $(\sim p \vee q)$ to be false,both $\sim p$ and $q$ must be false.
Since $\sim p$ is already false,$q$ must be false.
Therefore,$p$ is true and $q$ is false.

(B) Given that $p = F$ and $q = F$.
Step $1$: Evaluate $(\sim p \vee q) \leftrightarrow \sim(p \wedge q)$
$\sim p = \sim F = T$
$\sim p \vee q = T \vee F = T$
$p \wedge q = F \wedge F = F$
$\sim(p \wedge q) = \sim F = T$
Therefore,$(\sim p \vee q) \leftrightarrow \sim(p \wedge q) = T \leftrightarrow T = T$.
Step $2$: Evaluate $\sim p \leftrightarrow (p \rightarrow \sim q)$
$\sim p = T$
$\sim q = \sim F = T$
$p$ $\rightarrow \sim q = F$ $\rightarrow T = T$
Therefore,$\sim p \leftrightarrow (p \rightarrow \sim q) = T \leftrightarrow T = T$.
Thus,the truth values are $T, T$.

(C) The converse of a conditional statement $p \rightarrow q$ is defined as $q \rightarrow p$.
In this case,let $p$ be 'quadrilateral $ABCD$ is a square' and $q$ be 'all sides of $ABCD$ are equal'.
Statement $I$ is $p \rightarrow q$.
Statement $II$ is $q \rightarrow p$.
Therefore,statement $II$ is the converse of statement $I$.

(D) First,we evaluate the truth values of the given statements:
$p: 25$ is an odd prime number. Since $25 = 5 \times 5$,it is a composite number. Thus,$p \equiv F$.
$q: 14$ is a composite number. Since $14 = 2 \times 7$,it is a composite number. Thus,$q \equiv T$.
$r: 64$ is a perfect square number. Since $64 = 8^2$,it is a perfect square. Thus,$r \equiv T$.
Now,we evaluate the options:
$A: \sim(q \wedge r) \vee p \equiv \sim(T \wedge T) \vee F \equiv \sim T \vee F \equiv F \vee F \equiv F$.
$B: (p \wedge q) \vee r \equiv (F \wedge T) \vee T \equiv F \vee T \equiv T$.
$C: (p \vee q) \wedge (\sim r) \equiv (F \vee T) \wedge (\sim T) \equiv T \wedge F \equiv F$.
$D: \sim p \vee (q \wedge r) \equiv \sim F \vee (T \wedge T) \equiv T \vee T \equiv T$.
Note: Both $B$ and $D$ result in $T$. Based on the provided solution,$D$ is the intended answer.

(A) The inverse of a conditional statement $p \rightarrow q$ is defined as $\sim p \rightarrow \sim q$.
Given the statement pattern $(p \vee q) \rightarrow (p \wedge q)$,we identify $p$ as $(p \vee q)$ and $q$ as $(p \wedge q)$.
Applying the definition,the inverse is $\sim(p \vee q) \rightarrow \sim(p \wedge q)$.
Using De Morgan's laws,$\sim(p \vee q) \equiv (\sim p \wedge \sim q)$ and $\sim(p \wedge q) \equiv (\sim p \vee \sim q)$.
Thus,the inverse is $(\sim p \wedge \sim q) \rightarrow (\sim p \vee \sim q)$.

(A) First,consider the statement $p$ $\rightarrow (q$ $\rightarrow p)$. This is equivalent to $\neg p \vee (\neg q \vee p)$,which simplifies to $(\neg p \vee p) \vee \neg q = T \vee \neg q = T$. Thus,it is a tautology.
Next,consider $p \rightarrow (p \vee q)$. This is equivalent to $\neg p \vee (p \vee q)$,which simplifies to $(\neg p \vee p) \vee q = T \vee q = T$. Thus,it is also a tautology.
Finally,the expression $[p$ $\rightarrow (q$ $\rightarrow p)]$ $\rightarrow [p$ $\rightarrow (p \vee q)]$ becomes $T \rightarrow T$,which is $T$. Therefore,the statement pattern is a tautology.

(C) Let $p$ be the statement: "$A$ quadrilateral $ABCD$ is a rhombus".
Let $q$ be the statement: "Its opposite sides are parallel".
The given statement is $p \rightarrow q$.
The contrapositive of $p \rightarrow q$ is $\sim q \rightarrow \sim p$,which translates to: "If the opposite sides of a quadrilateral $ABCD$ are not parallel,then the quadrilateral $ABCD$ is not a rhombus".
The converse of $p \rightarrow q$ is $q \rightarrow p$,which translates to: "If the opposite sides of a quadrilateral $ABCD$ are parallel,then the quadrilateral $ABCD$ is a rhombus".
Comparing these with the given options,option $C$ is correct.

(D) To simplify the logical expression $p \wedge (\sim p \vee \sim q)$,we apply the distributive law of logic:
$p \wedge (\sim p \vee \sim q) \equiv (p \wedge \sim p) \vee (p \wedge \sim q)$.
Since $(p \wedge \sim p)$ is a contradiction,it is equivalent to $F$ (False).
Therefore,the expression becomes:
$F \vee (p \wedge \sim q) \equiv p \wedge \sim q$.
Thus,the correct expression is $p \wedge \sim q$.

(B) Let $p$ be the statement '$x \in A \cap B$' and $q$ be the statement '$x \in A \text{ and } x \in B$'.
The given statement is in the form $p \rightarrow q$.
The negation of an implication $p \rightarrow q$ is given by $\sim(p \rightarrow q) \equiv p \wedge \sim q$.
Here,$p$ is '$x \in A \cap B$' and $\sim q$ is the negation of '$x \in A \text{ and } x \in B$',which by De Morgan's Law is '$x \notin A \text{ or } x \notin B$'.
Therefore,the negation is '$x \in A \cap B \text{ and } (x \notin A \text{ or } x \notin B)$'.

(C) Given statements:
$p$: It rains today
$q$: $I$ am going to school
$r$: $I$ will meet my friend
$s$: $I$ will go to watch a movie
The statement is: "If (it does not rain today $OR$ $I$ will not go to school),then ($I$ will meet my friend $AND$ $I$ will go to watch a movie)."
Symbolically:
"It does not rain today" is $\sim p$.
"$I$ will not go to school" is $\sim q$.
"$I$ will meet my friend" is $r$.
"$I$ will go to watch a movie" is $s$.
Combining these using logical connectives:
$(\sim p \vee \sim q) \rightarrow (r \wedge s)$
Using De Morgan's Law,$\sim p \vee \sim q \equiv \sim(p \wedge q)$.
Therefore,the symbolic form is $\sim(p \wedge q) \rightarrow (r \wedge s)$.

(D) To find the negation of the implication $(p \wedge q) \rightarrow (\sim p \vee r)$,we use the logical equivalence $\sim(A \rightarrow B) \equiv A \wedge \sim B$.
Let $A = (p \wedge q)$ and $B = (\sim p \vee r)$.
The negation is $\sim(A \rightarrow B) \equiv A \wedge \sim B$.
Substituting the values:
$(p \wedge q) \wedge \sim(\sim p \vee r)$
Using De Morgan's Law,$\sim(\sim p \vee r) \equiv \sim(\sim p) \wedge \sim r \equiv p \wedge \sim r$.
Therefore,the expression becomes:
$(p \wedge q) \wedge (p \wedge \sim r)$
By the associative and idempotent laws,this simplifies to:
$p \wedge q \wedge (\sim r)$.

(C) Given: $p = T, q = T, r = F$.
Evaluating each option:
$(A)$ $(p \vee q) \vee r \equiv (T \vee T) \vee F \equiv T \vee F \equiv T$.
$(B)$ $(p \wedge q)$ $\rightarrow r \equiv (T \wedge T)$ $\rightarrow F \equiv T$ $\rightarrow F \equiv F$.
$(C)$ $(p$ $\rightarrow r)$ $\rightarrow q \equiv (T$ $\rightarrow F)$ $\rightarrow T \equiv F$ $\rightarrow T \equiv T$.
$(D)$ $(p \leftrightarrow q)$ $\rightarrow r \equiv (T \leftrightarrow T)$ $\rightarrow F \equiv T$ $\rightarrow F \equiv F$.
Since the question asks which statement is correct,and option $(C)$ evaluates to $T$,option $(C)$ is the correct statement.

(C) The negation of a universal quantifier statement $\forall x \in S, P(x)$ is given by $\exists x \in S$ such that $\neg P(x)$.
Here,the statement is $\forall x \in \mathbb{R}, x^2+1=0$.
Applying the rule,the negation is $\exists x \in \mathbb{R}$ such that $x^2+1 \neq 0$.

(A) Given the logical statement: $(p$ $\rightarrow q) \wedge (p$ $\rightarrow \sim p)$
Using the implication law $a \rightarrow b \equiv \sim a \vee b$:
$(p \rightarrow q) \equiv \sim p \vee q$
$(p \rightarrow \sim p) \equiv \sim p \vee \sim p \equiv \sim p$
Now,the expression becomes: $(\sim p \vee q) \wedge (\sim p)$
Using the absorption law: $A \wedge (A \vee B) \equiv A$
Here,let $A = \sim p$ and $B = q$.
So,$(\sim p) \wedge (\sim p \vee q) \equiv \sim p$
Therefore,the statement is equivalent to $\sim p$.

(B) Given: $p = T, q = T, r = F, s = F$.
For the first expression $\sim(p \rightarrow q) \leftrightarrow (p \wedge s)$:
$\sim(T \rightarrow T) \leftrightarrow (T \wedge F)$
$= \sim(T) \leftrightarrow (F)$
$= F \leftrightarrow F = T$.
For the second expression $(\sim p \rightarrow q) \wedge (r \leftrightarrow s)$:
$(\sim T \rightarrow T) \wedge (F \leftrightarrow F)$
$= (F \rightarrow T) \wedge (T)$
$= T \wedge T = T$.
Thus,the truth values are $T, T$.

(C) Given $p = T, q = T, r = F, s = F$.
For $a : \sim (p \wedge \sim r) \vee (\sim q \vee s)$:
$a \equiv \sim (T \wedge \sim F) \vee (\sim T \vee F)$
$a \equiv \sim (T \wedge T) \vee (F \vee F)$
$a \equiv \sim T \vee F$
$a \equiv F \vee F$
$a \equiv F$.
For $b : (p \vee s) \leftrightarrow (q \wedge r)$:
$b \equiv (T \vee F) \leftrightarrow (T \wedge F)$
$b \equiv T \leftrightarrow F$
$b \equiv F$.
Thus,the truth values of $a$ and $b$ are $F$ and $F$ respectively.

(B) Given expression: $(p \wedge q) \wedge [(p \wedge q) \vee (\sim p \wedge q)]$
Using the distributive law on the term inside the square brackets:
$(p \wedge q) \vee (\sim p \wedge q) \equiv (p \vee \sim p) \wedge q$
Since $(p \vee \sim p) \equiv T$ (Tautology),we have:
$T \wedge q \equiv q$
Substituting this back into the original expression:
$(p \wedge q) \wedge q$
Using the associative and idempotent laws:
$p \wedge (q \wedge q) \equiv p \wedge q$
Thus,the expression is equivalent to $p \wedge q$.

(D) We represent the given statements in logical form:
$S_1 = p \rightarrow q$
$S_2 = p \rightarrow \sim q$
$S_3 = \sim p \vee q$
$S_4 = p \wedge q$
We know the logical equivalence $p \rightarrow q \equiv \sim p \vee q$.
Comparing this with $S_1$,we get $S_1 = p \rightarrow q \equiv \sim p \vee q$.
Since $S_3 = \sim p \vee q$,it follows that $S_1 \equiv S_3$.