Mathematical logic Questions in English

(C) To determine which combination results in a tautology,we evaluate the truth table for each case:
$1$. For $\Delta=\wedge, \nabla=\vee$: The expression is $(p \rightarrow q) \wedge (p \vee q)$. If $p=T, q=F$,then $(T \rightarrow F) \wedge (T \vee F) = F \wedge T = F$. Not a tautology.
$2$. For $\Delta=\vee, \nabla=\wedge$: The expression is $(p \rightarrow q) \vee (p \wedge q)$. If $p=T, q=F$,then $(T \rightarrow F) \vee (T \wedge F) = F \vee F = F$. Not a tautology.
$3$. For $\Delta=\vee, \nabla=\vee$: The expression is $(p \rightarrow q) \vee (p \vee q)$.
- If $p=T, q=T$: $T \vee T = T$
- If $p=T, q=F$: $F \vee T = T$
- If $p=F, q=T$: $T \vee T = T$
- If $p=F, q=F$: $T \vee F = T$
Since all values are $T$,this is a tautology.
$4$. For $\Delta=\wedge, \nabla=\wedge$: The expression is $(p \rightarrow q) \wedge (p \wedge q)$. If $p=F, q=F$,then $T \wedge F = F$. Not a tautology.
Thus,the correct option is $\Delta=\vee, \nabla=\vee$.

(C) conditional statement $A \rightarrow B$ is false only when $A$ is True and $B$ is False.
Let $A = (p \vee q) \wedge ((\sim p) \vee r)$ and $B = ((\sim q) \vee r)$.
We need to find the case where $A = T$ and $B = F$.
For $B = ((\sim q) \vee r)$ to be False,both $(\sim q)$ and $r$ must be False.
This implies $q = T$ and $r = F$.
Now,substitute $q = T$ and $r = F$ into $A = (p \vee q) \wedge ((\sim p) \vee r)$:
$A = (p \vee T) \wedge ((\sim p) \vee F)$
$A = T \wedge (\sim p)$
For $A$ to be True,$(\sim p)$ must be True,which means $p = F$.
Thus,the combination $p = F, q = T, r = F$ makes the expression False.

(C) The given statement is $B \Rightarrow ((\sim A) \vee B)$.
Using the logical equivalence $P \Rightarrow Q \equiv (\sim P) \vee Q$,we have:
$B \Rightarrow ((\sim A) \vee B) \equiv (\sim B) \vee ((\sim A) \vee B)$
By the commutative and associative laws:
$(\sim B) \vee B \vee (\sim A) \equiv T \vee (\sim A) \equiv T$
Since the statement is a tautology,we check the options for tautologies.
Option $C$ is $A$ $\Rightarrow ((\sim A)$ $\Rightarrow B) \equiv (\sim A) \vee ((\sim A)$ $\Rightarrow B) \equiv (\sim A) \vee (A \vee B) \equiv (\sim A \vee A) \vee B \equiv T \vee B \equiv T$.
Thus,the statement is equivalent to $A$ $\Rightarrow ((\sim A)$ $\Rightarrow B)$.

(B) To determine if a statement is a tautology,we construct truth tables.
For $(S1): ((p \vee q)$ $\Rightarrow r) \Leftrightarrow (p$ $\Rightarrow r)$
This is not a tautology because if $p=F, q=T, r=F$,then $(p \vee q) \Rightarrow r$ is $F$,while $p \Rightarrow r$ is $T$. Thus,the biconditional is $F$.
For $(S2): ((p \vee q)$ $\Rightarrow r) \Leftrightarrow ((p$ $\Rightarrow r) \vee (q$ $\Rightarrow r))$
Using logical equivalence: $(p \vee q)$ $\Rightarrow r \equiv \neg(p \vee q) \vee r \equiv (\neg p \wedge \neg q) \vee r \equiv (\neg p \vee r) \wedge (\neg q \vee r) \equiv (p$ $\Rightarrow r) \wedge (q$ $\Rightarrow r)$.
Since $(p$ $\Rightarrow r) \wedge (q$ $\Rightarrow r)$ is not equivalent to $(p$ $\Rightarrow r) \vee (q$ $\Rightarrow r)$,$(S2)$ is not a tautology.
Therefore,neither $(S1)$ nor $(S2)$ is a tautology.

(A) Let the statements be:
$P$: $I$ have fever
$Q$: $I$ will take medicine
$R$: $I$ will take rest
The given statement is: "If $I$ have fever,then $I$ will take medicine and $I$ will take rest".
This can be written as: $P \rightarrow (Q \wedge R)$.
Using the logical equivalence $A \rightarrow B \equiv \sim A \vee B$:
$P \rightarrow (Q \wedge R) \equiv \sim P \vee (Q \wedge R)$.
Using the distributive law $A \vee (B \wedge C) \equiv (A \vee B) \wedge (A \vee C)$:
$\sim P \vee (Q \wedge R) \equiv (\sim P \vee Q) \wedge (\sim P \vee R)$.
Wait,the original question statement is "$I$ will not take medicine" for $Q$. Let's re-evaluate.
If $Q$ is "$I$ will take medicine",then "$I$ will not take medicine" is $\sim Q$.
The statement is $P \rightarrow (\sim Q \wedge R)$.
Equivalent to $\sim P \vee (\sim Q \wedge R)$.
Applying distributive law: $(\sim P \vee \sim Q) \wedge (\sim P \vee R)$.

(D) To check if $(S1)$ is a tautology,we construct the truth table for $(p \Rightarrow q) \vee (p \wedge (\sim q))$.
Since all values in the final column are $T$,$(S1)$ is a tautology.
To check if $(S2)$ is a contradiction,we construct the truth table for $((\sim p) \Rightarrow (\sim q)) \wedge ((\sim p) \vee q)$.
Since the final column contains both $T$ and $F$,it is a contingency,not a contradiction.
Therefore,only $(S1)$ is correct.

(B) Let the given expression be $S = ((p \wedge q)$ $\Rightarrow (r \vee q)) \wedge ((p \wedge r)$ $\Rightarrow q)$.
Using the identity $A \Rightarrow B \equiv \sim A \vee B$,we get:
$S = (\sim(p \wedge q) \vee (r \vee q)) \wedge (\sim(p \wedge r) \vee q)$
$S = (\sim p \vee \sim q \vee r \vee q) \wedge (\sim p \vee \sim r \vee q)$
Since $\sim q \vee q = T$ (tautology),the first part becomes $\sim p \vee r \vee T = T$.
Thus,$S = T \wedge (\sim p \vee \sim r \vee q) = \sim p \vee \sim r \vee q$.
For $S$ to be a tautology,$\sim p \vee \sim r \vee q$ must be true for all truth values of $p$ and $q$.
Case $1$: $r = p$. Then $\sim p \vee \sim p \vee q = \sim p \vee q$,which is not a tautology.
Case $2$: $r = q$. Then $\sim p \vee \sim q \vee q = \sim p \vee T = T$. (Valid)
Case $3$: $r = \sim p$. Then $\sim p \vee \sim(\sim p) \vee q = \sim p \vee p \vee q = T \vee q = T$. (Valid)
Case $4$: $r = \sim q$. Then $\sim p \vee \sim(\sim q) \vee q = \sim p \vee q \vee q = \sim p \vee q$,which is not a tautology.
Thus,there are $2$ values of $r$ for which the expression is a tautology.

(A) We need to find the negation of the expression $q \vee ((\sim q) \wedge p)$.
Let $E = q \vee ((\sim q) \wedge p)$.
Using De Morgan's Law,$\sim(A \vee B) = (\sim A) \wedge (\sim B)$:
$\sim E = \sim q \wedge \sim((\sim q) \wedge p)$
Using De Morgan's Law again,$\sim(A \wedge B) = (\sim A) \vee (\sim B)$:
$\sim E = \sim q \wedge (q \vee \sim p)$
Using the Distributive Law,$A \wedge (B \vee C) = (A \wedge B) \vee (A \wedge C)$:
$\sim E = (\sim q \wedge q) \vee (\sim q \wedge \sim p)$
Since $(\sim q \wedge q)$ is a contradiction $(F)$:
$\sim E = F \vee (\sim q \wedge \sim p)$
Since $F \vee X = X$:
$\sim E = (\sim q \wedge \sim p)$.

(B) statement is a tautology if its truth value is always true for all possible truth values of its components.
$(A)$ $p$ $\rightarrow (p \land (p$ $\rightarrow q)) \equiv \sim p \lor (p \land (\sim p \lor q)) \equiv \sim p \lor ((p \land \sim p) \lor (p \land q)) \equiv \sim p \lor (F \lor (p \land q)) \equiv \sim p \lor p \land q \equiv (\sim p \lor p) \land (\sim p \lor q) \equiv T \land (\sim p \lor q) \equiv \sim p \lor q$. This is not a tautology.
$(B)$ $(p \land q)$ $\rightarrow (\sim p$ $\rightarrow q) \equiv \sim(p \land q) \lor (\sim(\sim p) \lor q) \equiv (\sim p \lor \sim q) \lor (p \lor q) \equiv (\sim p \lor p) \lor (\sim q \lor q) \equiv T \lor T \equiv T$. This is a tautology.
$(C)$ $(p \land (p$ $\rightarrow q))$ $\rightarrow \sim q \equiv \sim(p \land (\sim p \lor q)) \lor \sim q \equiv \sim((p \land \sim p) \lor (p \land q)) \lor \sim q \equiv \sim(F \lor (p \land q)) \lor \sim q \equiv \sim(p \land q) \lor \sim q \equiv \sim p \lor \sim q \lor \sim q \equiv \sim p \lor \sim q$. This is not a tautology.
$(D)$ $p \lor (p \land q) \equiv p$. This is not a tautology.

(A) Given the expression: $(P$ $\Rightarrow Q) \wedge (R$ $\Rightarrow Q)$
We know that the implication $P \Rightarrow Q$ is logically equivalent to $\sim P \vee Q$.
Substituting this into the expression:
$(\sim P \vee Q) \wedge (\sim R \vee Q)$
Using the distributive law,we can factor out $Q$:
$(\sim P \wedge \sim R) \vee Q$
By De Morgan's Law,$\sim P \wedge \sim R$ is equivalent to $\sim(P \vee R)$:
$\sim(P \vee R) \vee Q$
Using the implication rule $\sim A \vee B \equiv A \Rightarrow B$,we get:
$(P \vee R) \Rightarrow Q$
Therefore,the correct option is $A$.

(A) To determine if the statements are true,we construct truth tables for each.
For $(S1): (p \Rightarrow q) \vee ((\sim p) \wedge q)$
| $p$ | $q$ | $p \Rightarrow q$ | $\sim p \wedge q$ | $(p \Rightarrow q) \vee (\sim p \wedge q)$ |
|---|---|---|---|---|
| $T$ | $T$ | $T$ | $F$ | $T$ |
| $T$ | $F$ | $F$ | $F$ | $F$ |
| $F$ | $T$ | $T$ | $T$ | $T$ |
| $F$ | $F$ | $T$ | $F$ | $T$ |
Since the last column contains an $F$,$(S1)$ is not a tautology. Thus,$(S1)$ is False.
For $(S2): (q \Rightarrow p) \Rightarrow ((\sim p) \wedge q)$
| $p$ | $q$ | $q \Rightarrow p$ | $\sim p \wedge q$ | $(q \Rightarrow p) \Rightarrow (\sim p \wedge q)$ |
|---|---|---|---|---|
| $T$ | $T$ | $T$ | $F$ | $F$ |
| $T$ | $F$ | $T$ | $F$ | $F$ |
| $F$ | $T$ | $F$ | $T$ | $T$ |
| $F$ | $F$ | $T$ | $F$ | $F$ |
Since the last column contains both $T$ and $F$,$(S2)$ is neither a tautology nor a contradiction. Thus,$(S2)$ is False.
Therefore,neither $(S1)$ nor $(S2)$ is True.

(C) Let $S = (p$ $\Rightarrow q)$ $\Rightarrow (q$ $\Rightarrow p)$.
Using the identity $A \Rightarrow B \equiv \sim A \vee B$,we have:
$S \equiv \sim (p$ $\Rightarrow q) \vee (q$ $\Rightarrow p)$
$S \equiv \sim (\sim p \vee q) \vee (\sim q \vee p)$
$S \equiv (p \wedge \sim q) \vee (\sim q \vee p)$
Using the associative and commutative laws:
$S \equiv (p \vee \sim q \vee p) \wedge (\sim q \vee \sim q \vee p)$
$S \equiv (p \vee \sim q) \wedge (\sim q \vee p) \equiv p \vee \sim q$.
Now,the negation of $S$ is $\sim (p \vee \sim q)$.
By De Morgan's Law,$\sim (p \vee \sim q) \equiv \sim p \wedge \sim (\sim q) \equiv \sim p \wedge q$ or $q \wedge (\sim p)$.

(A) Let the given expression be $S = (p \wedge (\sim q)) \vee (\sim p)$.
Using the distributive law,we have:
$S = (p \vee (\sim p)) \wedge ((\sim q) \vee (\sim p))$
Since $(p \vee (\sim p))$ is a tautology $(T)$,we get:
$S = T \wedge ((\sim q) \vee (\sim p)) = (\sim q) \vee (\sim p)$
Now,we need to find the negation of $S$:
$\sim S = \sim ((\sim q) \vee (\sim p))$
Using De Morgan's law,$\sim (A \vee B) = (\sim A) \wedge (\sim B)$:
$\sim S = (\sim (\sim q)) \wedge (\sim (\sim p))$
$\sim S = q \wedge p = p \wedge q$
Thus,the negation is equivalent to $p \wedge q$.

(A) Let the statement be $S = (p \vee q) \wedge (q \vee (\sim r))$.
The negation of $S$ is $\sim S = \sim [(p \vee q) \wedge (q \vee (\sim r))]$.
Using De Morgan's Law,$\sim (A \wedge B) = (\sim A) \vee (\sim B)$:
$\sim S = \sim (p \vee q) \vee \sim (q \vee (\sim r))$.
Applying De Morgan's Law again,$\sim (p \vee q) = (\sim p \wedge \sim q)$ and $\sim (q \vee (\sim r)) = (\sim q \wedge r)$:
$\sim S = (\sim p \wedge \sim q) \vee (\sim q \wedge r)$.

(D) Given statement: $\sim[p \vee (\sim(p \wedge q))]$
Applying De Morgan's Law: $\sim p \wedge \sim(\sim(p \wedge q))$
Using the Law of Double Negation: $\sim p \wedge (p \wedge q)$
Using the Associative Law: $(\sim p \wedge p) \wedge q$
Since $(\sim p \wedge p)$ is a contradiction $(F)$,we have: $F \wedge q = F$
Checking the options,option $D$ is $(p \wedge q) \wedge (\sim p)$,which is equivalent to $(p \wedge \sim p) \wedge q = F \wedge q = F$.
Thus,the statement is equivalent to $(p \wedge q) \wedge (\sim p)$.

(B) To find the number of ordered triplets $(p, q, r)$ for which the statement $(p \vee q) \wedge (p \vee r) \Rightarrow (q \vee r)$ is True,we construct the truth table:
| $p$ | $q$ | $r$ | $p \vee q$ | $p \vee r$ | $(p \vee q) \wedge (p \vee r)$ | $q \vee r$ | $(p \vee q) \wedge (p \vee r) \Rightarrow (q \vee r)$ |
|---|---|---|---|---|---|---|---|
| $T$ | $T$ | $T$ | $T$ | $T$ | $T$ | $T$ | $T$ |
| $T$ | $T$ | $F$ | $T$ | $T$ | $T$ | $T$ | $T$ |
| $T$ | $F$ | $T$ | $T$ | $T$ | $T$ | $T$ | $T$ |
| $T$ | $F$ | $F$ | $T$ | $T$ | $T$ | $F$ | $F$ |
| $F$ | $T$ | $T$ | $T$ | $T$ | $T$ | $T$ | $T$ |
| $F$ | $T$ | $F$ | $T$ | $F$ | $F$ | $T$ | $T$ |
| $F$ | $F$ | $T$ | $F$ | $T$ | $F$ | $T$ | $T$ |
| $F$ | $F$ | $F$ | $F$ | $F$ | $F$ | $F$ | $T$ |
Counting the rows where the final column is $T$,we find there are $7$ such cases.
Thus,the number of ordered triplets is $7$.

(B) The converse of a conditional statement $P \Rightarrow Q$ is defined as $Q \Rightarrow P$.
Given the statement $((\sim p) \wedge q) \Rightarrow r$,where $P = ((\sim p) \wedge q)$ and $Q = r$.
Therefore,the converse is $Q \Rightarrow P$,which is $r \Rightarrow ((\sim p) \wedge q)$.

(D) For $(S1): (p \Rightarrow q) \wedge (q \wedge (\sim q))$
Since $(q \wedge (\sim q))$ is always false $(F)$,the entire expression $(p \Rightarrow q) \wedge F$ is always false. Thus,$(S1)$ is a contradiction.
For $(S2): (p \wedge q) \vee ((\sim p) \wedge q) \vee (p \wedge (\sim q)) \vee ((\sim p) \wedge (\sim q))$
We can simplify this using distributive laws:
$= [q \wedge (p \vee (\sim p))] \vee [(\sim q) \wedge (p \vee (\sim p))]$
$= [q \wedge T] \vee [(\sim q) \wedge T]$
$= q \vee (\sim q) = T$
Since the result is always true $(T)$,$(S2)$ is a tautology.
Therefore,both statements are true.

(A) Let $p = (( A \wedge ( B \vee C ))$ $\Rightarrow ( A \vee B ))$ $\Rightarrow A$.
Using the implication rule $X \Rightarrow Y \equiv \sim X \vee Y$,we have:
$p \equiv \sim (( A \wedge ( B \vee C )) \Rightarrow ( A \vee B )) \vee A$.
Applying the negation rule $\sim (X \Rightarrow Y) \equiv X \wedge \sim Y$:
$p \equiv (( A \wedge ( B \vee C )) \wedge \sim ( A \vee B )) \vee A$.
Using De Morgan's Law $\sim ( A \vee B ) \equiv \sim A \wedge \sim B$:
$p \equiv (( A \wedge ( B \vee C )) \wedge ( \sim A \wedge \sim B )) \vee A$.
Since $( A \wedge \sim A ) \equiv F$ (False),the expression simplifies to:
$p \equiv ( F \wedge ( B \vee C ) \wedge \sim B ) \vee A \equiv F \vee A \equiv A$.
Therefore,the negation of the statement is $\sim p \equiv \sim A$.

(A) Given expression: $(p \wedge (\sim q)) \vee ((\sim p) \wedge q) \vee ((\sim p) \wedge (\sim q))$
Group the terms containing $(\sim p)$:
$(p \wedge (\sim q)) \vee ((\sim p) \wedge (q \vee (\sim q)))$
Since $(q \vee (\sim q)) = t$ (tautology):
$(p \wedge (\sim q)) \vee ((\sim p) \wedge t)$
$(p \wedge (\sim q)) \vee (\sim p)$
Using the distributive law $(A \vee (B \wedge C) = (A \vee B) \wedge (A \vee C))$:
$((\sim p) \vee p) \wedge ((\sim p) \vee (\sim q))$
Since $((\sim p) \vee p) = t$:
$t \wedge ((\sim p) \vee (\sim q))$
$= (\sim p) \vee (\sim q)$

(D) We need to find the negation of the statement $S = p \wedge (q \wedge \sim(p \wedge q))$.
Using De Morgan's Law,$\sim(A \wedge B) = \sim A \vee \sim B$ and $\sim(A \vee B) = \sim A \wedge \sim B$.
$\sim S = \sim [p \wedge (q \wedge \sim(p \wedge q))]$
$= \sim p \vee \sim (q \wedge \sim(p \wedge q))$
$= \sim p \vee (\sim q \vee \sim(\sim(p \wedge q)))$
$= \sim p \vee (\sim q \vee (p \wedge q))$
Using the distributive law,$\sim q \vee (p \wedge q) = (\sim q \vee p) \wedge (\sim q \vee q)$.
Since $(\sim q \vee q) = T$ (Tautology),the expression becomes $(\sim q \vee p) \wedge T = \sim q \vee p$.
Thus,$\sim S = \sim p \vee (\sim q \vee p)$.
Rearranging the terms,we get $(\sim p \vee p) \vee \sim q = T \vee \sim q = T$.
Wait,let us re-evaluate the original expression $p \wedge (q \wedge \sim(p \wedge q))$.
$p \wedge (q \wedge (\sim p \vee \sim q)) = p \wedge ((q \wedge \sim p) \vee (q \wedge \sim q)) = p \wedge ((q \wedge \sim p) \vee F) = p \wedge (q \wedge \sim p) = (p \wedge \sim p) \wedge q = F \wedge q = F$.
The negation of a contradiction $F$ is a tautology $T$. However,looking at the options,let us re-check the simplification.
$\sim [p \wedge (q \wedge (\sim p \vee \sim q))] = \sim [ (p \wedge q) \wedge (\sim p \vee \sim q) ] = \sim [ (p \wedge q \wedge \sim p) \vee (p \wedge q \wedge \sim q) ] = \sim [ F \vee F ] = \sim F = T$.
Given the options provided,there might be a typo in the question or options. If the question intended to ask for the negation of $p \wedge q$,the answer would be $\sim p \vee \sim q$. Given the structure,option $D$ is the closest logical form.

(B) We are given the condition $x_{1} = 1, x_{2} = 0, x_{3} = 0$.
We evaluate the given options:
For option $B$: $x_{1} \cdot x_{2}' = 1 \cdot (0)' = 1 \cdot 1 = 1$.
Thus,the function $f(x_{1}, x_{2}, x_{3}) = x_{1} \cdot x_{2}'$ satisfies the condition.

(B) Given expression: $(x \cdot y)+[(x+y') \cdot y]'$
Using De Morgan's Law $(a \cdot b)' = a' + b'$:
$= (x \cdot y) + [(x+y')' + y']$
Using De Morgan's Law $(a+b)' = a' \cdot b'$ and involution law $(y')' = y$:
$= (x \cdot y) + [x' \cdot (y')' + y']$
$= (x \cdot y) + [x' \cdot y + y']$
$= x \cdot y + x' \cdot y + y'$
Factor out $y$ from the first two terms:
$= y \cdot (x + x') + y'$
Since $x + x' = 1$:
$= y \cdot 1 + y'$
$= y + y'$
Since $y + y' = 1$:
$= 1$

(A) Let $p$ be the switch $S_1$ and $q$ be the switch $S_2$. The symbolic form of the given circuit is $(p \wedge \sim q) \vee (\sim p \wedge q) \vee (\sim p \wedge \sim q)$.
Simplifying the expression:
$= (p \wedge \sim q) \vee [\sim p \wedge (q \vee \sim q)]$
$= (p \wedge \sim q) \vee [\sim p \wedge t]$
$= (p \wedge \sim q) \vee \sim p$
$= \sim p \vee (p \wedge \sim q)$
$= (\sim p \vee p) \wedge (\sim p \vee \sim q)$
$= t \wedge (\sim p \vee \sim q)$
$= \sim p \vee \sim q$
The expression $\sim p \vee \sim q$ represents two switches $S_1'$ and $S_2'$ connected in parallel.

(C) $1$. Analyze statement $p$: For $n=1$,$10(1)-3 = 7$ (prime). For $n=2$,$10(2)-3 = 17$ (prime). For $n=4$,$10(4)-3 = 37$ (prime). For $n=5$,$10(5)-3 = 47$ (prime). For $n=7$,$10(7)-3 = 67$ (prime). For $n=8$,$10(8)-3 = 77 = 7 \times 11$ (not prime). Thus,$p$ is False $(F)$.
$2$. Analyze statement $q$: The sum of squares of direction cosines must be $1$. Here,$(\frac{2}{\sqrt{3}})^2 + (\frac{-2}{\sqrt{3}})^2 + (\frac{-1}{\sqrt{3}})^2 = \frac{4}{3} + \frac{4}{3} + \frac{1}{3} = \frac{9}{3} = 3 \neq 1$. Thus,$q$ is False $(F)$.
$3$. Analyze statement $r$: $\sin x$ is strictly increasing on $[-\frac{\pi}{2}, \frac{\pi}{2}]$. Thus,$r$ is True $(T)$.
$4$. Evaluate options:
$A: (F \wedge F) \leftrightarrow T$ $\Rightarrow F \leftrightarrow T$ $\Rightarrow F$
$B: (F$ $\rightarrow F)$ $\rightarrow \sim T$ $\Rightarrow T$ $\rightarrow F$ $\Rightarrow F$
$C: (\sim F \vee F) \wedge T$ $\Rightarrow (T \vee F) \wedge T$ $\Rightarrow T \wedge T$ $\Rightarrow T$
$D: (\sim F \wedge \sim F) \leftrightarrow \sim T$ $\Rightarrow (T \wedge T) \leftrightarrow F$ $\Rightarrow T \leftrightarrow F$ $\Rightarrow F$
Therefore,the correct option is $C$.

(D) To determine when the statement pattern $[(p$ $\rightarrow q) \wedge \sim q]$ $\rightarrow r$ is a tautology,we first simplify the antecedent $[(p \rightarrow q) \wedge \sim q]$.
Using the implication law $p \rightarrow q \equiv \sim p \vee q$,we get:
$[(\sim p \vee q) \wedge \sim q] \rightarrow r$
By the distributive law,this becomes $[(\sim p \wedge \sim q) \vee (q \wedge \sim q)] \rightarrow r$.
Since $q \wedge \sim q \equiv F$ (a contradiction),we have:
$[(\sim p \wedge \sim q) \vee F]$ $\rightarrow r \equiv (\sim p \wedge \sim q)$ $\rightarrow r$.
For this to be a tautology,the expression must be true for all truth values of $p$ and $q$.
If $r = \sim p$,the expression becomes $(\sim p \wedge \sim q) \rightarrow \sim p$.
Since $(\sim p \wedge \sim q)$ implies $\sim p$,the implication is always true.
Thus,the statement is a tautology when $r = \sim p$.

(D) We represent the circuits using Boolean algebra,where a closed switch is $1$ and an open switch is $0$. The lamp $L$ glows if the circuit is closed.
$(A)$ The circuit has two parallel branches: $(S_1 \land S_2)$ and $(S_1 \land S_3)$. The expression is $(S_1 \land S_2) \lor (S_1 \land S_3) = S_1 \land (S_2 \lor S_3)$.
$(B)$ The circuit has two parallel branches: $S_1$ and $(S_2 \land S_3)$. The expression is $S_1 \lor (S_2 \land S_3)$.
$(C)$ The circuit has $S_1$ in series with a parallel combination of $S_2$ and $S_3$. The expression is $S_1 \land (S_2 \lor S_3)$.
$(D)$ The circuit has $S_1, S_2, S_3$ in series. The expression is $S_1 \land S_2 \land S_3$.
$(E)$ The circuit has two parallel branches: $(S_1 \land S_2)$ and $S_3$. The expression is $(S_1 \land S_2) \lor S_3$.
Comparing the expressions,circuit $(A)$ and circuit $(C)$ have the same Boolean expression: $S_1 \land (S_2 \lor S_3)$.
Therefore,$(A)$ and $(C)$ are equivalent circuits.

(C) Let us construct the truth table for the given statement pattern:
$1$. $p$ | $q$ | $\sim p$ | $\sim q$ | $(q \wedge \sim p)$ | $p \rightarrow (q \wedge \sim p)$ | $(p \vee \sim q)$ | $(p \vee \sim q) \wedge p$ | $[p \rightarrow (q \wedge \sim p)] \vee [(p \vee \sim q) \wedge p]$
$2$. $T$ | $T$ | $F$ | $F$ | $F$ | $F$ | $T$ | $T$ | $T$
$3$. $T$ | $F$ | $F$ | $T$ | $F$ | $F$ | $T$ | $T$ | $T$
$4$. $F$ | $T$ | $T$ | $F$ | $T$ | $T$ | $F$ | $F$ | $T$
$5$. $F$ | $F$ | $T$ | $T$ | $F$ | $T$ | $T$ | $F$ | $T$
Thus,the last column values are $T, T, T, T$.

(B) Given expression: $(\sim p \wedge q) \vee (\sim p \wedge \sim q) \vee (p \wedge \sim q)$
Using the distributive law on the first two terms: $(\sim p \wedge (q \vee \sim q)) \vee (p \wedge \sim q)$
Since $(q \vee \sim q) \equiv T$ (Tautology),the expression becomes: $(\sim p \wedge T) \vee (p \wedge \sim q)$
This simplifies to: $(\sim p) \vee (p \wedge \sim q)$
Applying the distributive law again: $(\sim p \vee p) \wedge (\sim p \vee \sim q)$
Since $(\sim p \vee p) \equiv T$,the expression becomes: $T \wedge (\sim p \vee \sim q)$
Therefore,the final equivalent statement is: $(\sim p) \vee (\sim q)$

(D) Given that the truth value of $q$ is $False$ and $(p \wedge q) \leftrightarrow r$ is $True$.
Since $q$ is $False$,the conjunction $(p \wedge q)$ is always $False$ regardless of the truth value of $p$.
Substituting this into the biconditional statement: $False \leftrightarrow r$ is $True$.
For a biconditional statement to be $True$,both sides must have the same truth value. Therefore,$r$ must be $False$.
Now,let us evaluate the options:
$A) p \wedge q = p \wedge False = False$.
$B) p \vee r = p \vee False = p$. This depends on $p$,so it is not necessarily $True$.
$C) p \wedge r = p \wedge False = False$.
$D) (p \wedge r)$ $\rightarrow (p \vee r) = (p \wedge False)$ $\rightarrow (p \vee False) = False$ $\rightarrow p$.
Since $False \rightarrow p$ is always $True$ for any truth value of $p$,option $D$ is $True$.

(D) Let $P$ be the statement: "Three vertices of a triangle are represented by cube roots of unity".
Let $Q$ be the statement: "The triangle is an equilateral triangle".
The given statement is of the form $P \implies Q$.
The contrapositive of a conditional statement $P \implies Q$ is $\neg Q \implies \neg P$,which is logically equivalent to the original statement.
Here,$\neg Q$ is: "The triangle is not an equilateral triangle".
Here,$\neg P$ is: "The three vertices of the triangle are not represented by cube roots of unity".
Thus,the equivalent statement is $\neg Q \implies \neg P$,which is: "If a triangle is not an equilateral triangle,then the three vertices of the triangle cannot be represented by cube roots of unity".
Therefore,the correct option is $D$.

(C) Let $S$ be the statement $(p \wedge \sim q) \rightarrow (p \vee \sim q)$.
Recall that the negation of an implication $A \rightarrow B$ is $A \wedge \sim B$.
Here,$A = (p \wedge \sim q)$ and $B = (p \vee \sim q)$.
So,the negation is $(p \wedge \sim q) \wedge \sim (p \vee \sim q)$.
Using De Morgan's Law,$\sim (p \vee \sim q) \equiv \sim p \wedge \sim (\sim q) \equiv \sim p \wedge q$.
Thus,the negation is $(p \wedge \sim q) \wedge (\sim p \wedge q)$.
By the associative and commutative properties,this is $(p \wedge \sim p) \wedge (\sim q \wedge q)$.
Since $p \wedge \sim p$ is a contradiction $(F)$ and $\sim q \wedge q$ is a contradiction $(F)$,the expression becomes $F \wedge F$,which is $F$.
$A$ statement that is always false is called a contradiction.

(D) The implication $(p \wedge q) \rightarrow (r \vee \sim s)$ is false only when the antecedent $(p \wedge q)$ is true and the consequent $(r \vee \sim s)$ is false.
For $(p \wedge q)$ to be true,both $p$ and $q$ must be true $(T)$.
For $(r \vee \sim s)$ to be false,both $r$ and $\sim s$ must be false $(F)$.
Since $\sim s$ is false,$s$ must be true $(T)$.
Therefore,the truth values are $p = T, q = T, r = F, s = T$.

(A) Given expression: $[\sim(\sim p \vee q) \vee (p \wedge r) \wedge (\sim q \wedge r)]$.
Applying De Morgan's Law to the first part: $\sim(\sim p \vee q) \equiv (p \wedge \sim q)$.
Now the expression becomes: $[(p \wedge \sim q) \vee ((p \wedge r) \wedge (\sim q \wedge r))]$.
Using the associative and commutative properties: $(p \wedge r) \wedge (\sim q \wedge r) \equiv (p \wedge \sim q) \wedge (r \wedge r) \equiv (p \wedge \sim q) \wedge r$.
Substituting this back: $(p \wedge \sim q) \vee ((p \wedge \sim q) \wedge r)$.
Using the absorption law: $A \vee (A \wedge B) \equiv A$,where $A = (p \wedge \sim q)$ and $B = r$.
Therefore,the expression simplifies to $(p \wedge \sim q)$.

(B) Let the given logical statement be $L = [\{q \wedge (\sim q \vee r)\} \wedge \{\sim p \vee (p \wedge \sim r)\}] \vee (p \wedge r)$.
Using the distributive law,$\{q \wedge (\sim q \vee r)\} = (q \wedge \sim q) \vee (q \wedge r) = F \vee (q \wedge r) = (q \wedge r)$.
Using the distributive law,$\{\sim p \vee (p \wedge \sim r)\} = (\sim p \vee p) \wedge (\sim p \vee \sim r) = T \wedge (\sim p \vee \sim r) = (\sim p \vee \sim r)$.
Substituting these back into $L$,we get $L = [(q \wedge r) \wedge (\sim p \vee \sim r)] \vee (p \wedge r)$.
Using the distributive law,$(q \wedge r) \wedge (\sim p \vee \sim r) = (q \wedge r \wedge \sim p) \vee (q \wedge r \wedge \sim r) = (q \wedge r \wedge \sim p) \vee F = (q \wedge r \wedge \sim p)$.
Now,$L = (q \wedge r \wedge \sim p) \vee (p \wedge r) = r \wedge [(q \wedge \sim p) \vee p]$.
Using the distributive law,$(q \wedge \sim p) \vee p = (q \vee p) \wedge (\sim p \vee p) = (q \vee p) \wedge T = (q \vee p)$.
Thus,$L = r \wedge (p \vee q)$.
In terms of switches,$r$ is $S_3$,$p$ is $S_1$,and $q$ is $S_2$.
So,$L = S_3 \wedge (S_1 \vee S_2)$.
This represents switch $S_3$ in series with the parallel combination of switches $S_1$ and $S_2$.

(D) conditional statement $A \rightarrow B$ is $False$ if and only if $A$ is $True$ and $B$ is $False$.
Here,$A = \{(p \wedge \sim q) \wedge (p \wedge r)\}$ and $B = (\sim p \vee q)$.
For $B = (\sim p \vee q)$ to be $False$,both $\sim p$ and $q$ must be $False$.
This implies $p = True$ and $q = False$.
Now,substitute these values into $A$:
$A = \{(T \wedge \sim F) \wedge (T \wedge r)\} = \{(T \wedge T) \wedge (T \wedge r)\} = \{T \wedge (T \wedge r)\} = (T \wedge r)$.
For $A$ to be $True$,$r$ must be $True$.
Therefore,the truth values are $p = True, q = False, r = True$.

(D) Let us evaluate each statement:
$A$: The cube roots of unity are $1, \omega, \omega^2$. They form a Geometric Progression with common ratio $\omega$. Their sum is $1+\omega+\omega^2 = 0$. The statement says the sum is $1$,which is false. So,$A$ is $F$.
$B$: $4+7 > 10$ is $11 > 10$ $(T)$. $2+8 < 10$ is $10 < 10$ $(F)$. $T \iff F$ is $F$. So,$B$ is $F$.
$C$: $\exists x \in N$ such that $x^2-3x+2=0$. The roots are $x=1$ and $x=2$,both are in $N$. This part is $T$. $\exists n \in N$ such that $n$ is an odd number is $T$ (e.g.,$n=1$). $T \land T$ is $T$. So,$C$ is $T$.
$D$: $3+i$ is a complex number $(T)$. $\sqrt{2}+\sqrt{3}=\sqrt{5}$ is false $(F)$. $T \lor F$ is $T$. So,$D$ is $T$.
Therefore,both $C$ and $D$ have the truth value $T$.

(D) The given statement pattern is $[p \wedge \sim r] \rightarrow [\sim r \wedge q]$ and its truth value is $F$.
An implication $A \rightarrow B$ is $F$ only when $A$ is $T$ and $B$ is $F$.
So,$[p \wedge \sim r] = T$ and $[\sim r \wedge q] = F$.
From $[p \wedge \sim r] = T$,we get $p = T$ and $\sim r = T$,which implies $r = F$.
Since $r = F$,$\sim r = T$.
Substituting $\sim r = T$ into $[\sim r \wedge q] = F$,we get $[T \wedge q] = F$,which implies $q = F$.
Thus,the truth values are $p = T, q = F, r = F$.
Now,check the options:
$A$: $(p \vee r)$ $\rightarrow \sim r \implies (T \vee F)$ $\rightarrow T \implies T$ $\rightarrow T = T$.
$B$: $(r \vee q)$ $\rightarrow \sim p \implies (F \vee F)$ $\rightarrow F \implies F$ $\rightarrow F = T$.
$C$: $\sim(p \vee q)$ $\rightarrow \sim r \implies \sim(T \vee F)$ $\rightarrow T \implies \sim T$ $\rightarrow T \implies F$ $\rightarrow T = T$.
$D$: $\sim(r \vee q)$ $\rightarrow \sim p \implies \sim(F \vee F)$ $\rightarrow F \implies \sim F$ $\rightarrow F \implies T$ $\rightarrow F = F$.
Therefore,the statement with truth value $F$ is $\sim(r \vee q) \rightarrow \sim p$.

(D) The circuit consists of two parallel branches connected in series with the lamp $L$.
Branch $1$ consists of switch $S_1$ in series with a parallel combination of $S_2$ and $S_3$. The symbolic form for this branch is $p \wedge (q \vee r)$.
Branch $2$ consists of switches $S_3$,$S_2$,and $S_1$ all in series. The symbolic form for this branch is $r \wedge q \wedge p$.
Since the two branches are in parallel,the total symbolic form is $(p \wedge (q \vee r)) \vee (r \wedge q \wedge p)$.

(C) The contrapositive of a conditional statement $P \implies Q$ is $\neg Q \implies \neg P$.
For statement $p$: $P$ is '$7$ is an odd number' (True),$Q$ is '$7$ is divisible by $2$' (False). The contrapositive is 'If $7$ is not divisible by $2$,then $7$ is not an odd number'. Since $7$ is not divisible by $2$ (True) and $7$ is an odd number (True),the statement 'If True,then False' is False. Thus,$V_1 = F$.
For statement $q$: $P$ is '$7$ is a prime number' (True),$Q$ is '$7$ is an odd number' (True). The contrapositive is 'If $7$ is not an odd number,then $7$ is not a prime number'. Since $7$ is an odd number (True),the antecedent 'If $7$ is not an odd number' is False. $A$ conditional statement with a False antecedent is always True. Thus,$V_2 = T$.
Therefore,$(V_1, V_2) = (F, T)$.

(D) The given statement is $S = \sim p \vee (q \wedge \sim r)$.
We know that the implication $A \rightarrow B$ is logically equivalent to $\sim A \vee B$.
Thus,the statement $S$ can be written as $p \rightarrow (q \wedge \sim r)$.
The contrapositive of an implication $p \rightarrow Q$ is $\sim Q \rightarrow \sim p$.
Here,$Q = (q \wedge \sim r)$.
Therefore,$\sim Q = \sim (q \wedge \sim r) = \sim q \vee \sim (\sim r) = \sim q \vee r$.
So,the contrapositive is $(\sim q \vee r) \rightarrow \sim p$.

(C) The given statement is of the form $\forall M > 0, \exists x \in S$ such that $P(x, M)$,where $P(x, M)$ is $x \geqslant M$.
To find the negation of a statement involving quantifiers,we replace $\forall$ with $\exists$ and $\exists$ with $\forall$,and negate the predicate.
The negation of $\forall M > 0, \exists x \in S, (x \geqslant M)$ is $\exists M > 0, \forall x \in S, \neg(x \geqslant M)$.
Since the negation of $x \geqslant M$ is $x < M$,the negated statement is $\exists M > 0$ such that $x < M$ for all $x \in S$.

(B) The given circuit consists of two parallel branches connected in series with a battery and a lamp $L$.
Let the switches be represented by variables $S_1, S_2, S_3$. The first branch is $S_1 \land (S_2' \lor S_3')$.
The second branch is $S_2 \land S_3 \land S_1$.
The total circuit expression is $P = [S_1 \land (S_2' \lor S_3')] \lor (S_2 \land S_3 \land S_1)$.
Using the distributive law,we can factor out $S_1$:
$P = S_1 \land [(S_2' \lor S_3') \lor (S_2 \land S_3)]$.
By De Morgan's law,$(S_2' \lor S_3')$ is equivalent to $(S_2 \land S_3)'$.
So,$P = S_1 \land [(S_2 \land S_3)' \lor (S_2 \land S_3)]$.
Since $(X' \lor X)$ is always true (a tautology),the expression simplifies to $P = S_1 \land T = S_1$.
Thus,the equivalent circuit is just a single switch $S_1$ in series with the lamp.
The number of switches in the equivalent circuit is $1$.

(C) In logic,a conditional statement $P \implies Q$ is true if $P$ is false,regardless of the truth value of $Q$.
$(A)$ $4+3=8$ is false. Since the antecedent is false,the implication $(A)$ is true.
$(B)$ $6+4=10$ is true,but the consequent 'the moon is flat' is false. Since a true antecedent implies a false consequent,$(B)$ is false.
$(C)$ The antecedent is 'both $(A)$ and $(B)$ are true'. Since $(B)$ is false,the conjunction 'both $(A)$ and $(B)$ are true' is false. $A$ conditional with a false antecedent is true. Thus,$(C)$ is true.
Therefore,$(A)$ and $(C)$ are true,while $(B)$ is false.

(A) Let $p$ be the statement "The triangle is an equilateral triangle".
Let $q$ be the statement "The triangle is an isosceles triangle".
Let $r$ be the statement "The triangle is right angled".
The given statement is $(p \lor q) \land (\neg q \land r)$.
We need to find the negation: $\neg((p \lor q) \land (\neg q \land r))$.
Using De Morgan's Law,$\neg(A \land B) = \neg A \lor \neg B$,we get:
$\neg(p \lor q) \lor \neg(\neg q \land r)$.
Applying De Morgan's Law again:
$(\neg p \land \neg q) \lor (q \lor \neg r)$.
This translates to: "The triangle is not equilateral and not isosceles,or the triangle is isosceles or it is not right angled".

(A) Given that $p = T$,$q = F$,and $r = T$.
First,evaluate the statement pattern $S = [\sim q \wedge (p \vee \sim q) \wedge \sim r] \vee p$.
Substituting the values:
$S = [\sim F \wedge (T \vee \sim F) \wedge \sim T] \vee T$
$S = [T \wedge (T \vee T) \wedge F] \vee T$
$S = [T \wedge T \wedge F] \vee T$
$S = F \vee T = T$.
Next,find the dual statement $S^*$. To find the dual,replace $\wedge$ with $\vee$,$\vee$ with $\wedge$,$T$ with $F$,and $F$ with $T$.
The dual statement is $S^* = [\sim q \vee (p \wedge \sim q) \vee \sim r] \wedge p$.
Substituting the values:
$S^* = [\sim F \vee (T \wedge \sim F) \vee \sim T] \wedge T$
$S^* = [T \vee (T \wedge T) \vee F] \wedge T$
$S^* = [T \vee T \vee F] \wedge T$
$S^* = T \wedge T = T$.
Thus,the truth values are $T$ and $T$.

(A) The circuit consists of two main branches connected in parallel.
$1$. The upper branch contains switch $S_1$ in series with a parallel combination of $S_2$ and $S_3$. The symbolic form for this branch is $p \wedge (q \vee r)$.
$2$. The lower branch contains switches $S_3'$,$S_2'$,and $S_1$ in series. Since $S_3'$ is the complement of $S_3$ and $S_2'$ is the complement of $S_2$,the symbolic form for this branch is $(\neg r \wedge \neg q \wedge p)$.
$3$. Since the two branches are in parallel,the total symbolic form is the disjunction of the two branches: $p \wedge (q \vee r) \vee (\neg r \wedge \neg q \wedge p)$.

(A) In logic,a conditional statement $P \implies Q$ is false only when $P$ is true and $Q$ is false. Otherwise,it is true.
$(A)$ $P: 3+2=7$ (False),$Q: 4+3=8$ (False). Since $P$ is false,the implication $P \implies Q$ is true.
$(B)$ $P: 5+2=7$ (True),$Q: \text{earth is flat}$ (False). Since $P$ is true and $Q$ is false,the implication $P \implies Q$ is false.
$(C)$ $P: (A) \text{ is true AND } (B) \text{ is true}$,$Q: 5+6=11$ (True). Since $(A)$ is true and $(B)$ is false,the condition $P$ is false. An implication with a false antecedent is true. Thus,$(C)$ is true.
Therefore,$(A)$ and $(C)$ are true,while $(B)$ is false.

(D) The given circuit consists of two main parts connected in series with switch $S_3$. Let the first part be $C_1$ and the second part be $C_2$. The circuit is $C_1 \wedge r \wedge C_2$.
For $C_1$: It has three parallel branches: $(p' \wedge q')$,$p$,and $q$. So,$C_1 = (p' \wedge q') \vee p \vee q$.
Using the law of absorption and distributive laws:
$C_1 = (p' \wedge q') \vee (p \vee q) = (p' \vee (p \vee q)) \wedge (q' \vee (p \vee q)) = (T \vee q) \wedge (p \vee (q' \vee q)) = T \wedge (p \vee T) = T \wedge T = T$.
For $C_2$: It has two parallel branches: $(p \wedge q)$ and $(p' \wedge q)$. So,$C_2 = (p \wedge q) \vee (p' \wedge q)$.
Using the distributive law:
$C_2 = (p \vee p') \wedge q = T \wedge q = q$.
Thus,the total circuit expression is $T \wedge r \wedge q = q \wedge r$.
The simplified circuit is a series connection of switches $S_2$ and $S_3$,which corresponds to the logical statement $(q \wedge r)$.
Therefore,the correct option is $D$.