Mathematical logic Questions in English

(C) Given statements are:
$p$: It is raining
$q$: It is pleasant
The statement "It is neither raining nor pleasant" means "It is not raining $AND$ it is not pleasant".
This can be written as $(\sim p) \wedge (\sim q)$.
Therefore,the correct option is $C$.

(B) To find the negation of $p \wedge (q \rightarrow r)$,we apply De Morgan's Law and the properties of implication:
$\sim [p \wedge (q \rightarrow r)]$
Since $q \rightarrow r \equiv \sim q \vee r$,we have:
$\equiv \sim [p \wedge (\sim q \vee r)]$
Applying De Morgan's Law $\sim (A \wedge B) \equiv \sim A \vee \sim B$:
$\equiv \sim p \vee \sim (\sim q \vee r)$
Applying De Morgan's Law again $\sim (A \vee B) \equiv \sim A \wedge \sim B$:
$\equiv \sim p \vee (q \wedge \sim r)$

(C) Let $p: -7$ is an integer.
Let $q: \sqrt{-7}$ is a complex number.
The logical form of $S_1$ is $p \rightarrow q$.
The logical form of $S_2$ is $\sim p \lor q$.
We know that the implication $p \rightarrow q$ is logically equivalent to $\sim p \lor q$.
Therefore,$S_1$ and $S_2$ are equivalent statements.

(C) Let $p: 3+6 > 8$ and $q: 2+3 < 6$.
The logical form of the given statement is $p \wedge q$.
The negation of a conjunction is given by De Morgan's Law: $\sim(p \wedge q) \equiv \sim p \vee \sim q$.
Here,$\sim p$ is $3+6 \leq 8$ and $\sim q$ is $2+3 \geq 6$.
Therefore,the negation is $3+6 \leq 8 \text{ or } 2+3 \geq 6$.

(C) Let $p$: Two triangles are congruent.
$q$: Their areas are equal.
The logical form of the given statement is $p \rightarrow q$.
The inverse of the given statement is $\sim p \rightarrow \sim q$.
The contrapositive of the inverse is the contrapositive of $(\sim p \rightarrow \sim q)$,which is $\sim(\sim q) \rightarrow \sim(\sim p)$.
Using the law of double negation,this simplifies to $q \rightarrow p$.
Therefore,the statement is: If areas of two triangles are equal,then they are congruent.

(D) Given statement: $\sim p \rightarrow q$.
The inverse of a statement $A \rightarrow B$ is $\sim A \rightarrow \sim B$.
Therefore,the inverse of $\sim p \rightarrow q$ is $\sim(\sim p) \rightarrow \sim q$,which simplifies to $p \rightarrow \sim q$.
The negation of an implication $A \rightarrow B$ is $A \wedge (\sim B)$.
Thus,the negation of $p \rightarrow \sim q$ is $p \wedge \sim(\sim q)$,which simplifies to $p \wedge q$.

(D) The given statement is of the form $\forall x \in N, P(x)$,where $P(x)$ is the statement '$x^2+x$ is an even number'.
To find the negation of a universal quantifier statement $\forall x, P(x)$,we use the rule $\sim(\forall x, P(x)) \equiv \exists x, \sim P(x)$.
Here,the negation of '$x^2+x$ is an even number' is '$x^2+x$ is not an even number'.
Therefore,the negation of the statement is $\exists x \in N$ such that $x^2+x$ is not an even number.

(D) Let $p$: It is raining and $q$: The weather is pleasant.
The given statement is "It is not true that,if it is raining then the weather is not pleasant".
Symbolically,this is written as $\sim(p \rightarrow \sim q)$.
Using the logical equivalence $\sim(A \rightarrow B) \equiv A \wedge \sim B$,we get:
$\sim(p \rightarrow \sim q) \equiv p \wedge \sim(\sim q)$.
Since $\sim(\sim q) \equiv q$,the expression simplifies to $p \wedge q$.
Thus,the statement is "It is raining and the weather is pleasant".

(D) statement pattern is a tautology if its truth value is $T$ for all possible combinations of truth values of its components.
Based on the provided truth table:
Column $9$ represents $S_{1} \equiv \sim p \rightarrow (q \leftrightarrow p)$. The values are $T, T, F, T$. Not a tautology.
Column $10$ represents $S_{2} \equiv \sim p \vee \sim q$. The values are $F, T, T, T$. Not a tautology.
Column $11$ represents $S_{3} \equiv (p$ $\rightarrow q) \wedge (q$ $\rightarrow p)$. The values are $T, F, F, T$. Not a tautology.
Column $12$ represents $S_{4} \equiv (q \rightarrow p) \vee (\sim p \leftrightarrow q)$. The values are $T, T, T, T$.
Since all entries in column $12$ are $T$,$S_{4}$ is a tautology.

(A) To determine the nature of the statement pattern,we construct a truth table:
| $p$ | $q$ | $\sim p$ | $\sim q$ | $p \vee q$ | $(p \vee q) \wedge \sim p$ | $[(p \vee q) \wedge \sim p] \wedge \sim q$ |
|---|---|---|---|---|---|---|
| $T$ | $T$ | $F$ | $F$ | $T$ | $F$ | $F$ |
| $T$ | $F$ | $F$ | $T$ | $T$ | $F$ | $F$ |
| $F$ | $T$ | $T$ | $F$ | $T$ | $T$ | $F$ |
| $F$ | $F$ | $T$ | $T$ | $F$ | $F$ | $F$ |
Since all entries in the final column are $F$ (False),the statement pattern is a contradiction.

(A) statement pattern is a contradiction if its truth value is $F$ for all possible truth values of its components.
We analyze $S_{1} \equiv (p \rightarrow q) \wedge (p \wedge \sim q)$.
Using logical laws:
$S_{1} \equiv (\sim p \vee q) \wedge (p \wedge \sim q)$
$S_{1} \equiv [(\sim p \vee q) \wedge p] \wedge \sim q$
$S_{1} \equiv [(\sim p \wedge p) \vee (q \wedge p)] \wedge \sim q$
$S_{1} \equiv [F \vee (p \wedge q)] \wedge \sim q$
$S_{1} \equiv (p \wedge q) \wedge \sim q$
$S_{1} \equiv p \wedge (q \wedge \sim q)$
$S_{1} \equiv p \wedge F$
$S_{1} \equiv F$
Since the truth value is always $F$,$S_{1}$ is a contradiction.

(B) To find the negation of a statement involving a quantifier,we follow these rules:
$1$. Replace the existential quantifier $\exists$ (there exists) with the universal quantifier $\forall$ (for all).
$2$. Replace the inequality symbol $>$ with its negation $\leq$.
Therefore,the negation of $\exists x \in A$ such that $x+5 > 8$ is $\forall x \in A, \quad x+5 \leq 8$.

(D) To determine which statement is a tautology,we construct a truth table for each statement pattern.
$A$ statement is a tautology if its truth value is $T$ for all possible combinations of truth values of its component statements.
For $S_2 \equiv [p \wedge (p$ $\rightarrow q)]$ $\rightarrow q$:
| $p$ | $q$ | $p \rightarrow q$ | $p \wedge (p \rightarrow q)$ | $[p \wedge (p$ $\rightarrow q)]$ $\rightarrow q$ |
|---|---|---|---|---|
| $T$ | $T$ | $T$ | $T$ | $T$ |
| $T$ | $F$ | $F$ | $F$ | $T$ |
| $F$ | $T$ | $T$ | $F$ | $T$ |
| $F$ | $F$ | $T$ | $F$ | $T$ |
Since all entries in the final column for $S_2$ are $T$,$S_2$ is a tautology.

(C) To find the dual of a statement pattern,we replace $\wedge$ with $\vee$,$\vee$ with $\wedge$,$t$ with $c$,and $c$ with $t$.
Given the statement pattern: $\sim p \wedge (q \vee t)$.
Replacing $\wedge$ with $\vee$,$\vee$ with $\wedge$,and $t$ with $c$:
The dual is $\sim p \vee (q \wedge c)$.

(A) Given expression: $\sim(p \vee q) \vee(\sim p \wedge q)$
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 $(\sim q \vee q) \equiv T$ (Tautology),the expression simplifies to:
$\sim p \wedge T \equiv \sim p$.

(A) The circuit consists of two main parallel branches.
Branch $1$ contains switches $s_{1}$ and $s_{2}$ in series,represented by $(p \wedge q)$.
Branch $2$ contains switch $s_{1}'$ (which is $\sim p$) in series with a parallel combination of $s_{2}'$ $(\sim q)$,$s_{1}$ $(p)$,and $s_{3}$ $(r)$.
This parallel combination is $(\sim q \vee p \vee r)$.
Thus,the symbolic form is $(p \wedge q) \vee [\sim p \wedge (\sim q \vee p \vee r)] \equiv \ell$.

(B) Let $p$: Raju is courageous,and $q$: Raju will join Indian Army.
The contrapositive of the implication $p \rightarrow q$ is defined as $\sim q \rightarrow \sim p$.
Here,$\sim q$ is 'Raju does not join Indian Army' and $\sim p$ is 'Raju is not courageous'.
Therefore,the contrapositive is: 'If Raju does not join Indian Army,then he is not courageous'.

(D) Given expression: $[p \wedge (q \vee r)] \vee [\sim r \wedge \sim q \wedge p]$
Using the commutative and associative laws,we can rewrite the second part:
$[p \wedge (q \vee r)] \vee [p \wedge (\sim q \wedge \sim r)]$
By De Morgan's law,$\sim q \wedge \sim r \equiv \sim (q \vee r)$:
$[p \wedge (q \vee r)] \vee [p \wedge \sim (q \vee r)]$
Apply the distributive law $p \wedge (A \vee \sim A) = p \wedge T$:
$p \wedge [(q \vee r) \vee \sim (q \vee r)]$
Since $(q \vee r) \vee \sim (q \vee r) \equiv T$ (Tautology):
$p \wedge T = p$

(A) The given expression is $[\sim p \vee (p \wedge \sim q)] \vee q$.
Applying the distributive law:
$= [(\sim p \vee p) \wedge (\sim p \vee \sim q)] \vee q$
$= [T \wedge (\sim p \vee \sim q)] \vee q$
$= (\sim p \vee \sim q) \vee q$
$= \sim p \vee (\sim q \vee q)$
$= \sim p \vee T$
$= T$
Since the result is a tautology $(T)$,the current flows irrespective of the status of the switches.

(A) The implication $A \rightarrow B$ is false only when $A$ is true and $B$ is false.
Given $(\sim p \wedge q) \rightarrow r$ is false,we must have $(\sim p \wedge q) = T$ and $r = F$.
The conjunction $A \wedge B$ is true only when both $A$ and $B$ are true.
Therefore,$\sim p = T$ and $q = T$.
If $\sim p = T$,then $p = F$.
Thus,the truth values are $p = F, q = T, r = F$.

(B) The statement 'Sandeep neither likes tea nor coffee' means 'Sandeep does not like tea $AND$ Sandeep does not like coffee',which is represented as $(\sim p \wedge \sim q)$.
The phrase 'but enjoys a soft drink' adds the condition '$AND$ Sandeep enjoys a soft drink',which is represented as $\wedge r$.
Combining these,the symbolic form is $(\sim p \wedge \sim q) \wedge r$.

(B) Let $p:$ The grass is green.
Let $q:$ It rains in July.
The logical form of the given statement is $p \rightarrow q$.
We know that the implication $p \rightarrow q$ is logically equivalent to $\sim p \vee q$.
Therefore,the statement is equivalent to 'The grass is not green or it rains in July'.

(C) To find the truth values for $\sim(p \wedge q)$,we construct the truth table as follows:

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

The entries in the last column are $F, T, T, T$.

(D) tautology is a statement that is true for all possible truth values of its components.
We construct the truth table for the given options:
| $p$ | $q$ | $\sim p$ | $\sim q$ | $\sim p \vee \sim q$ | $p \vee \sim q$ | $(\sim p \vee \sim q) \vee (p \vee \sim q)$ |
|---|---|---|---|---|---|---|
| $T$ | $T$ | $F$ | $F$ | $F$ | $T$ | $T$ |
| $T$ | $F$ | $F$ | $T$ | $T$ | $T$ | $T$ |
| $F$ | $T$ | $T$ | $F$ | $T$ | $F$ | $T$ |
| $F$ | $F$ | $T$ | $T$ | $T$ | $T$ | $T$ |
As seen in the final column,the statement $(\sim p \vee \sim q) \vee (p \vee \sim q)$ is true for all combinations of $p$ and $q$.
Therefore,it is a tautology.

(A) Let $p$ be the statement that switch $S_{1}$ is closed.
Let $q$ be the statement that switch $S_{2}$ is closed.
Then $\sim p$ represents switch $S_{1}'$ being closed,and $\sim q$ represents switch $S_{2}'$ being closed.
In the given circuit:
$1$. The top branch has switches $S_{1}$ and $S_{2}$ in parallel,which is represented by $(p \vee q)$.
$2$. The bottom branch has switches $S_{1}'$ and $S_{2}'$ in series,which is represented by $(\sim p \wedge \sim q)$.
$3$. These two branches are connected in parallel to each other.
Therefore,the symbolic form of the circuit is $(p \vee q) \vee (\sim p \wedge \sim q) \equiv l$.

(B) Let $p$ be the statement 'He is poor' and $q$ be the statement 'He is happy'.
The given statement 'He is poor but happy' can be written in logical form as $p \wedge q$.
We know that the negation of a conjunction is given by De Morgan's Law: $\sim(p \wedge q) \equiv \sim p \vee \sim q$.
Here,$\sim p$ is 'He is not poor' and $\sim q$ is 'He is not happy'.
Thus,the negation is 'He is not poor or he is not happy'.
Comparing this with the given options,option $B$ represents this logical equivalence.

(B) Given: $p \equiv T, q \equiv T, r \equiv F$.
We evaluate the truth value of each option:
$(A) (p \wedge q)$ $\rightarrow r \equiv (T \wedge T)$ $\rightarrow F \equiv T$ $\rightarrow F \equiv F$.
$(B) (p$ $\rightarrow r)$ $\rightarrow q \equiv (T$ $\rightarrow F)$ $\rightarrow T \equiv F$ $\rightarrow T \equiv T$.
$(C) (p \vee q) \vee r \equiv (T \vee T) \vee F \equiv T \vee F \equiv T$.
$(D) (p \leftrightarrow q) \leftrightarrow r \equiv (T \leftrightarrow T) \leftrightarrow F \equiv T \leftrightarrow F \equiv F$.
Since the question asks for a true statement,and both $(B)$ and $(C)$ result in $T$,we re-examine the options provided in the original prompt. Based on standard logic,$(B)$ and $(C)$ are both true. Given the structure,$(B)$ is the standard intended answer.

(A) Let $p: 5 < 7$,$q: 7 > 2$,and $r: 5 > 2$.
The given statement is of the form $(p \wedge q) \rightarrow r$.
The negation of an implication $A \rightarrow B$ is $A \wedge \sim B$.
Here,$A = (p \wedge q)$ and $B = r$.
So,the negation is $(p \wedge q) \wedge \sim r$.
Substituting the values: $(5 < 7 \wedge 7 > 2) \wedge \sim(5 > 2)$.
Since the negation of $5 > 2$ is $5 \leq 2$,the final statement is $(5 < 7 \text{ and } 7 > 2) \text{ and } 5 \leq 2$.

(A) The given statement is '$p \land q$',where '$p$: Mangoes are delicious' and '$q$: Mangoes are expensive'.
In logic,the word 'but' acts as a conjunction,which is equivalent to 'and' $(\land)$.
The dual of a statement is obtained by replacing 'and' $(\land)$ with 'or' $(\lor)$ and vice versa.
Therefore,the dual of '$p \land q$' is '$p \lor q$'.
Thus,the dual statement is 'Mangoes are delicious or Mangoes are expensive'.

(A) Using the distributive law,we expand the expression:
$p \wedge (q \vee \sim p) \equiv (p \wedge q) \vee (p \wedge \sim p)$
Since $(p \wedge \sim p) \equiv F$ (contradiction),the expression becomes:
$(p \wedge q) \vee F$
By the identity law,$(p \wedge q) \vee F \equiv p \wedge q$.
Thus,the correct option is $A$.

(B) The negation of an implication $A \rightarrow B$ is $A \wedge \sim B$.
Here,$A = (p \vee \sim q)$ and $B = (p \wedge \sim q)$.
So,the negation is $(p \vee \sim q) \wedge \sim(p \wedge \sim q)$.
Applying De Morgan's Law,$\sim(p \wedge \sim q) \equiv \sim p \vee \sim(\sim q) \equiv \sim p \vee q$.
Therefore,the negation is $(p \vee \sim q) \wedge (\sim p \vee q)$.

(B) The given statement is $p \rightarrow q$.
We know that the logical equivalence of the implication is $p \rightarrow q \equiv \sim p \vee q$.
Here,$\sim p$ is 'Seema is not fat' and $q$ is 'She is happy'.
Therefore,the required equivalent statement is 'Seema is not fat or she is happy'.

(D) Given expression: $[p \wedge (q \vee r)] \vee [(\sim p \wedge q) \vee (\sim p \wedge r)]$
Using the distributive law on the second part:
$[p \wedge (q \vee r)] \vee [\sim p \wedge (q \vee r)]$
Using the distributive law again:
$(q \vee r) \wedge (p \vee \sim p)$
Since $(p \vee \sim p) = T$ (Tautology):
$(q \vee r) \wedge T$
$= q \vee r$

(B) tautology is a statement pattern that is always true for all possible truth values of its components.
Check option $(b): p \rightarrow (q \vee p)$
$= \sim p \vee (q \vee p)$
$= (\sim p \vee p) \vee q$
$= T \vee q$
$= T$
Since the result is always true $(T)$,the statement pattern $p \rightarrow (q \vee p)$ is a tautology.

(D) Let $p$ be the statement "Two triangles are congruent" and $q$ be the statement "Their areas are equal". The given statement is $p \rightarrow q$.
The inverse of $p \rightarrow q$ is $\sim p \rightarrow \sim q$,which is: "If two triangles are not congruent,then their areas are not equal."
The contrapositive of the inverse $(\sim p \rightarrow \sim q)$ is $\sim(\sim q) \rightarrow \sim(\sim p)$,which simplifies to $q \rightarrow p$. This is: "If the areas of two triangles are equal,then they are congruent."

(A) Given: $p = T, q = T, r = F, s = F$.
For $a: \sim(p \wedge \sim r) \vee(\sim q \vee s)$
Substitute the values: $\sim(T \wedge \sim F) \vee(\sim T \vee F)$
$= \sim(T \wedge T) \vee(F \vee F)$
$= \sim(T) \vee(F)$
$= F \vee F = F$.
For $b: (p \vee s) \leftrightarrow(q \wedge r)$
Substitute the values: $(T \vee F) \leftrightarrow(T \wedge F)$
$= (T) \leftrightarrow(F)$
Since the truth values are different,the biconditional statement is $F$.
Therefore,the truth values of $a$ and $b$ are $F, F$.

(C) Key Idea: $A$ statement that is neither a tautology nor a contradiction is a contingency.
Option $A$: $(p \vee q) \vee \sim q \equiv p \vee (q \vee \sim q) \equiv p \vee T \equiv T$. This is a tautology.
Option $B$: $(p \vee q) \vee \sim p \equiv (p \vee \sim p) \vee q \equiv T \vee q \equiv T$. This is a tautology.
Option $C$: $(p \vee q) \wedge \sim q \equiv (p \wedge \sim q) \vee (q \wedge \sim q) \equiv (p \wedge \sim q) \vee F \equiv p \wedge \sim q$.
If $p=T, q=T$,then $p \wedge \sim q = F$.
If $p=T, q=F$,then $p \wedge \sim q = T$.
Since the truth value depends on $p$ and $q$,it is a contingency.
Option $D$: $p \rightarrow (p \vee q) \equiv \sim p \vee (p \vee q) \equiv (\sim p \vee p) \vee q \equiv T \vee q \equiv T$. This is a tautology.
Therefore,the correct statement is $(p \vee q) \wedge \sim q$.

(A) Key Idea: The negation of the universal quantifier $\forall$ (for all) is the existential quantifier $\exists$ (there exists),and the negation of the inequality $>$ is $\leq$.
Given the statement: $\forall n \in N, n+7 > 6$.
Applying the rules of negation:
$1$. Replace $\forall$ with $\exists$.
$2$. Negate the condition $n+7 > 6$,which becomes $n+7 \leq 6$.
Therefore,the negation of the given statement is $\exists n \in N$,such that $n+7 \leq 6$.

(C) Given: $p = T, q = T, r = F, s = F$.
Check Option $(A): (q \wedge r) \vee (\sim p \wedge s) \equiv (T \wedge F) \vee (F \wedge F) \equiv F \vee F \equiv F$.
Check Option $(B): (\sim p$ $\rightarrow q)$ $\rightarrow (r \wedge s) \equiv (\sim T$ $\rightarrow T)$ $\rightarrow (F \wedge F) \equiv (F$ $\rightarrow T)$ $\rightarrow F \equiv T$ $\rightarrow F \equiv F$.
Check Option $(C): (p$ $\rightarrow q) \vee (r \leftrightarrow s) \equiv (T$ $\rightarrow T) \vee (F \leftrightarrow F) \equiv T \vee T \equiv T$.
Check Option $(D): (p \wedge \sim r) \wedge (\sim q \vee s) \equiv (T \wedge \sim F) \wedge (\sim T \vee F) \equiv (T \wedge T) \wedge (F \vee F) \equiv T \wedge F \equiv F$.
Thus,option $(C)$ is true.

(A) We know that the negation of an implication is given by $\sim(p \rightarrow q) \equiv p \wedge \sim q$.
Applying this rule to the given expression:
$\sim(p \rightarrow \sim q) \equiv p \wedge \sim(\sim q)$.
Using the law of double negation,$\sim(\sim q) \equiv q$.
Therefore,the expression simplifies to $p \wedge q$.

(C) The logical implication $p \rightarrow q$ can be expressed in several equivalent ways:
$1.$ If $p$ then $q$.
$2.$ $p$ only if $q$.
$3.$ $q$ is necessary for $p$.
$4.$ $p$ is sufficient for $q$.
Comparing these with the given options,option $C$ ($q$ only if $p$) is equivalent to $q \rightarrow p$,which is the converse of $p \rightarrow q$. Therefore,it is not equivalent to $p \rightarrow q$.

(B) Let the given statement be $S = (p \wedge q) \wedge [\sim r \vee (p \wedge q)] \vee (\sim p \wedge q)$.
Using the absorption law,$(p \wedge q) \wedge [\sim r \vee (p \wedge q)]$ simplifies to $(p \wedge q)$.
Thus,the expression becomes $S = (p \wedge q) \vee (\sim p \wedge q)$.
Using the distributive law,we can factor out $q$:
$S = (p \vee \sim p) \wedge q$.
Since $(p \vee \sim p)$ is a tautology $(T)$,
$S = T \wedge q = q$.
Therefore,the statement pattern is equivalent to $q$.

(B) Let $P$ be the statement "Hema gets above $95 \%$ marks" and $Q$ be the statement "Hema gets admission in a good college".
The statement "Getting above $95 \%$ marks is a necessary condition for Hema to get admission in a good college" is equivalent to the implication $Q \implies P$.
The negation of an implication $Q \implies P$ is given by $\sim(Q \implies P) \equiv Q \land \sim P$.
Here,$Q$ is "Hema gets admission in a good college" and $\sim P$ is "Hema does not get above $95 \%$ marks".
Therefore,the negation is: "Hema gets admission in a good college and she does not get above $95 \%$ marks."

(B) Given compound statement is $p \wedge (\sim p \wedge q)$.
Using the associative law,we can rewrite this as $(p \wedge \sim p) \wedge q$.
Since $p \wedge \sim p$ is always false $(F)$,the expression becomes $F \wedge q$.
Since $F \wedge q$ is always false,the statement is a contradiction.

(B) Let $p$ be the statement: "The weather is fine".
Let $q$ be the statement: "My friends will come".
Let $r$ be the statement: "We go for a picnic".
The given statement is $p \rightarrow (q \wedge r)$.
The contrapositive of $p \rightarrow (q \wedge r)$ is $\sim(q \wedge r) \rightarrow \sim p$.
Using De Morgan's Law,$\sim(q \wedge r) \equiv (\sim q \vee \sim r)$.
So,the contrapositive is $(\sim q \vee \sim r) \rightarrow \sim p$.
Translating this back to words: "If my friends do not come or we do not go for a picnic,then the weather will not be fine."

(A) To find the dual of a compound statement,we replace $\wedge$ with $\vee$,$\vee$ with $\wedge$,$t$ (tautology) with $c$ (contradiction),and $c$ (contradiction) with $t$ (tautology).
Given the statement: $\sim p \wedge (q \vee c)$.
Replacing $\wedge$ with $\vee$,$\vee$ with $\wedge$,and $c$ with $t$:
The dual is $\sim p \vee (q \wedge t)$.

(C) tautology is a statement pattern that is always true for all possible truth values of its components.
Let us evaluate each option:
$(A) \ p \vee (q \rightarrow p) \equiv p \vee (\sim q \vee p) \equiv (p \vee p) \vee \sim q \equiv p \vee \sim q$. This is not a tautology as it is false when $p$ is $F$ and $q$ is $T$.
$(B) \ \sim q \rightarrow \sim p \equiv q \vee \sim p$. This is not a tautology as it is false when $q$ is $F$ and $p$ is $T$.
$(D) \ p \wedge \sim p \equiv F$. This is a contradiction.
$(C) \ (q \rightarrow p) \vee (\sim p \leftrightarrow q)$. Let us construct the truth table:

$p, q$	$(q \rightarrow p) \vee (\sim p \leftrightarrow q)$
$T, T$	$T \vee (F \leftrightarrow T) = T \vee F = T$
$T, F$	$T \vee (F \leftrightarrow F) = T \vee T = T$
$F, T$	$F \vee (T \leftrightarrow T) = F \vee T = T$
$F, F$	$T \vee (T \leftrightarrow F) = T \vee F = T$

Since the result is always $T$,option $(C)$ is a tautology.

(B) We evaluate the given options to find the one equivalent to $(\sim p \wedge q)$.
Consider option $B$: $(p \vee q) \wedge \sim p$.
By the Distributive Law,this is equivalent to $(p \wedge \sim p) \vee (q \wedge \sim p)$.
Since $(p \wedge \sim p) = F$ (Complementary Law),we have $F \vee (q \wedge \sim p)$.
By the Identity Law,$F \vee (q \wedge \sim p) = (q \wedge \sim p)$.
By the Commutative Law,$(q \wedge \sim p) = (\sim p \wedge q)$.
Thus,the statement pattern $(\sim p \wedge q)$ is logically equivalent to $(p \vee q) \wedge \sim p$.

(D) Let $S_1 \equiv p$ and $S_2 \equiv q$. Then $S_1' \equiv \sim p$ and $S_2' \equiv \sim q$.
The circuit consists of two parallel branches.
The first branch has switches $S_1$ and $S_2'$ in series,which corresponds to $(p \wedge \sim q)$.
The second branch has switches $S_1'$ and $S_2$ in series,which corresponds to $(\sim p \wedge q)$.
Since the branches are in parallel,the total expression is $(p \wedge \sim q) \vee (\sim p \wedge q)$.
This expression is equivalent to the exclusive $OR$ operation,which is $\sim(p \leftrightarrow q)$.

(C) For any real number $x$,the square $x^2$ is always greater than or equal to $0$ $(x^2 \geq 0)$.
Option $A$ is false because for $x=0$,$x^2=0$,which is not positive.
Option $B$ is false because the square of a real number cannot be negative.
Option $C$ is true because there exists a real number $x=0$ such that $x^2=0$,which is not positive.
Option $D$ is false because there exist irrational numbers like $\sqrt{2}$.