Mathematical logic Questions in English

(C) The given expression is $x \leftrightarrow \sim y$.
We know that $p \leftrightarrow q \equiv (p$ $\rightarrow q) \wedge (q$ $\rightarrow p)$.
So,$x \leftrightarrow \sim y \equiv (x$ $\rightarrow \sim y) \wedge (\sim y$ $\rightarrow x)$.
Using the identity $p \rightarrow q \equiv \sim p \vee q$,we get:
$x \leftrightarrow \sim y \equiv (\sim x \vee \sim y) \wedge (y \vee x)$.
Now,we find the negation:
$\sim(x \leftrightarrow \sim y) \equiv \sim((\sim x \vee \sim y) \wedge (x \vee y))$.
By De Morgan's Law,$\sim(A \wedge B) \equiv \sim A \vee \sim B$:
$\sim(x \leftrightarrow \sim y) \equiv \sim(\sim x \vee \sim y) \vee \sim(x \vee y)$.
Applying De Morgan's Law again:
$\sim(x \leftrightarrow \sim y) \equiv (x \wedge y) \vee (\sim x \wedge \sim y)$.

(B) The contrapositive of a conditional statement $(p \rightarrow q)$ is defined as $(\sim q \rightarrow \sim p)$.
Here,$p$ is the statement "$n^{3}-1$ is even" and $q$ is the statement "$n$ is odd".
The negation $\sim q$ is "$n$ is not odd",which means "$n$ is even".
The negation $\sim p$ is "$n^{3}-1$ is not even",which means "$n^{3}-1$ is odd".
Therefore,the contrapositive statement is: "For an integer $n$,if $n$ is even,then $n^{3}-1$ is odd."

(C) Given expression: $p \vee (\sim p \wedge q)$
Using the distributive law: $p \vee (\sim p \wedge q) = (p \vee \sim p) \wedge (p \vee q)$
Since $(p \vee \sim p) = T$ (Tautology):
$= T \wedge (p \vee q) = p \vee q$
Now,we need to find the negation of $(p \vee q)$:
$\sim (p \vee q) = \sim p \wedge \sim q$
Therefore,the correct option is $C$.

(C) Let the outer numbers in each quadrant be $(a, b)$ and the inner number be $k$. The pattern is $k = (a - b)^{n!}$,where $n$ is the number of digits or a specific index related to the quadrant.
For the first quadrant: $(2 - 1)^{1!} = 1^{1} = 1$.
For the second quadrant: $(12 - 8)^{4!} = 4^{24}$.
For the third quadrant: $(7 - 4)^{3!} = 3^{6}$.
For the fourth quadrant: $(5 - 3)^{2!} = 2^{2} = 4$.
Thus,the missing value is $4$.

(A) We test the options to see which expression results in a tautology $(t)$:
Option $A$: $(p \wedge q)$ $\rightarrow (p$ $\rightarrow q)$
$= \sim(p \wedge q) \vee (\sim p \vee q)$
$= (\sim p \vee \sim q) \vee (\sim p \vee q)$
$= \sim p \vee (\sim q \vee q)$
$= \sim p \vee t$
$= t$ (This is a tautology).
Option $B$: $(p \wedge q) \wedge (p \vee q) = (p \wedge q)$ (Not a tautology).
Option $C$: $(p \wedge q) \vee (p \rightarrow q) = (p \wedge q) \vee (\sim p \vee q) = \sim p \vee q$ (Not a tautology).
Option $D$: $(p \wedge q) \wedge (p \rightarrow q) = (p \wedge q) \wedge (\sim p \vee q) = p \wedge q$ (Not a tautology).
Thus,the correct option is $A$.

(B) Let us simplify the antecedent $((P \Rightarrow Q) \wedge \sim Q)$ for all options:
$((P \Rightarrow Q) \wedge \sim Q) \equiv ((\sim P \vee Q) \wedge \sim Q)$
Using the distributive law: $(\sim P \wedge \sim Q) \vee (Q \wedge \sim Q)$
Since $(Q \wedge \sim Q) \equiv F$,we have $(\sim P \wedge \sim Q) \vee F \equiv \sim P \wedge \sim Q$.
Now,check each option:
$(A) (\sim P \wedge \sim Q)$ $\Rightarrow Q \equiv \sim(\sim P \wedge \sim Q) \vee Q \equiv (P \vee Q) \vee Q \equiv P \vee Q$ (Not a tautology)
$(B) (\sim P \wedge \sim Q)$ $\Rightarrow \sim P \equiv \sim(\sim P \wedge \sim Q) \vee \sim P \equiv (P \vee Q) \vee \sim P \equiv (P \vee \sim P) \vee Q \equiv T \vee Q \equiv T$ (Tautology)
$(C) (\sim P \wedge \sim Q)$ $\Rightarrow P \equiv \sim(\sim P \wedge \sim Q) \vee P \equiv (P \vee Q) \vee P \equiv P \vee Q$ (Not a tautology)
$(D) (\sim P \wedge \sim Q) \Rightarrow (P \wedge Q) \equiv (P \vee Q) \vee (P \wedge Q) \equiv P \vee Q$ (Not a tautology)
Therefore,the correct option is $B$.

(D) To check if the statement $[A \wedge (A$ $\rightarrow B)]$ $\rightarrow B$ is a tautology,we simplify it using logical laws:
$[A \wedge (A$ $\rightarrow B)]$ $\rightarrow B$
$= [A \wedge (\sim A \vee B)] \rightarrow B$
$= [(A \wedge \sim A) \vee (A \wedge B)] \rightarrow B$
$= [F \vee (A \wedge B)] \rightarrow B$
$= (A \wedge B) \rightarrow B$
$= \sim (A \wedge B) \vee B$
$= (\sim A \vee \sim B) \vee B$
$= \sim A \vee (\sim B \vee B)$
$= \sim A \vee T$
$= T$
Since the final result is $T$ (a tautology),the statement $[A \wedge (A$ $\rightarrow B)]$ $\rightarrow B$ is a tautology.

(D) The given statement is $A$ $\rightarrow (B$ $\rightarrow A)$.
Using the implication law $p \rightarrow q \equiv \sim p \vee q$,we get:
$A$ $\rightarrow (B$ $\rightarrow A) \equiv \sim A \vee (\sim B \vee A)$
By the associative and commutative laws:
$\equiv (\sim A \vee A) \vee \sim B$
$\equiv T \vee \sim B \equiv T$ (Tautology).
Now,check the options for a tautology or equivalent form:
Option $D$ is $A \rightarrow (A \vee B) \equiv \sim A \vee (A \vee B) \equiv (\sim A \vee A) \vee B \equiv T \vee B \equiv T$.
Since both the original expression and option $D$ are tautologies,they are equivalent.

(B) To determine if a statement is a tautology,we construct a truth table for each.
For statement $(a)$: $(\sim q \wedge (p$ $\rightarrow q))$ $\rightarrow \sim p$

$p, q$	$\sim q$	$p \rightarrow q$	$\sim q \wedge (p \rightarrow q)$	$\sim p$	$(a)$
$T, T$	$F$	$T$	$F$	$F$	$T$
$T, F$	$T$	$F$	$F$	$F$	$T$
$F, T$	$F$	$T$	$F$	$T$	$T$
$F, F$	$T$	$T$	$T$	$T$	$T$

Since all values in the final column are $T$,$(a)$ is a tautology.
For statement $(b)$: $((p \vee q) \wedge \sim p) \rightarrow q$

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

Since all values in the final column are $T$,$(b)$ is also a tautology.
Therefore,both $(a)$ and $(b)$ are tautologies.

(B) We want to find the negation of the statement $\sim p \wedge (p \vee q)$.
Applying De Morgan's Law: $\sim (\sim p \wedge (p \vee q)) \equiv \sim (\sim p) \vee \sim (p \vee q)$.
Using the law of double negation and De Morgan's Law: $p \vee (\sim p \wedge \sim q)$.
Applying the Distributive Law: $(p \vee \sim p) \wedge (p \vee \sim q)$.
Since $(p \vee \sim p) \equiv T$ (Tautology),we have $T \wedge (p \vee \sim q)$.
Therefore,the result is $p \vee \sim q$.

(C) The contrapositive of a conditional statement $p \rightarrow q$ is defined as $\sim q \rightarrow \sim p$.
Here,the statement $p$ is "You will work" and $q$ is "You will earn money".
Therefore,the contrapositive $\sim q \rightarrow \sim p$ is "If you will not earn money,you will not work".

(C) For $F_{1}(A, B, C) = (A \wedge \sim B) \vee [\sim C \wedge (A \vee B)] \vee \sim A$:
Using the distributive law:
$F_{1} = [(A \wedge \sim B) \vee \sim A] \vee [\sim C \wedge (A \vee B)]$
$F_{1} = [(A \vee \sim A) \wedge (\sim B \vee \sim A)] \vee [\sim C \wedge (A \vee B)]$
Since $(A \vee \sim A) = t$ (tautology):
$F_{1} = [t \wedge (\sim A \vee \sim B)] \vee [\sim C \wedge (A \vee B)]$
$F_{1} = (\sim A \vee \sim B) \vee [\sim C \wedge (A \vee B)]$.
This expression depends on the values of $A, B, C$,so it is not a tautology.
For $F_{2}(A, B) = (A \vee B) \vee (B \rightarrow \sim A)$:
Using the implication rule $(P \rightarrow Q) = (\sim P \vee Q)$:
$F_{2} = (A \vee B) \vee (\sim B \vee \sim A)$
By commutative and associative laws:
$F_{2} = (A \vee \sim A) \vee (B \vee \sim B)$
$F_{2} = t \vee t = t$.
Thus,$F_{2}$ is a tautology.

(C) The implication $A \rightarrow B$ is false if and only if $A$ is true and $B$ is false.
Here,$A = (p \vee q) \wedge (q \rightarrow r) \wedge (\sim r)$ and $B = (p \wedge q)$.
For $B = (p \wedge q)$ to be false,at least one of $p$ or $q$ must be false.
For $A$ to be true,all components $(p \vee q)$,$(q \rightarrow r)$,and $(\sim r)$ must be true.
Since $(\sim r)$ is true,$r$ must be false.
Since $(q \rightarrow r)$ is true and $r$ is false,$q$ must be false.
Since $(p \vee q)$ is true and $q$ is false,$p$ must be true.
Thus,the truth values are $p=T, q=F, r=F$.
Checking option $C$: $p=T, q=F, r=F$ gives $A = (T \vee F) \wedge (F \rightarrow F) \wedge (\sim F) = T \wedge T \wedge T = T$ and $B = (T \wedge F) = F$.
Since $T \rightarrow F$ is false,option $C$ is correct.

(C) For $(S1): (p$ $\rightarrow q) \vee (\sim q$ $\rightarrow p)$
Using the implication law $(a \rightarrow b \equiv \sim a \vee b)$,we get:
$(\sim p \vee q) \vee (q \vee p)$
$= (q \vee \sim p) \vee (q \vee p) = q \vee (\sim p \vee p) = q \vee t = t$.
Thus,$(S1)$ is a tautology.
For $(S2): (p \wedge \sim q) \wedge (\sim p \vee q)$
Using De Morgan's Law,$(\sim p \vee q) \equiv \sim (p \wedge \sim q)$.
So,$(S2) = (p \wedge \sim q) \wedge \sim (p \wedge \sim q) = C$ (Contradiction/Fallacy).
Thus,both $(S1)$ and $(S2)$ are true.

(A) Let the statement be $S = (p \wedge (p$ $\rightarrow q) \wedge (q$ $\rightarrow r))$ $\rightarrow r$.
Using the implication law $a \rightarrow b \equiv \sim a \vee b$,we have:
$S \equiv \sim (p \wedge (p$ $\rightarrow q) \wedge (q$ $\rightarrow r)) \vee r$
Using De Morgan's law $\sim (A \wedge B \wedge C) \equiv \sim A \vee \sim B \vee \sim C$:
$S \equiv \sim p \vee \sim (p$ $\rightarrow q) \vee \sim (q$ $\rightarrow r) \vee r$
Since $\sim (a \rightarrow b) \equiv a \wedge \sim b$:
$S \equiv \sim p \vee (p \wedge \sim q) \vee (q \wedge \sim r) \vee r$
Using distributive law $\sim p \vee (p \wedge \sim q) \equiv (\sim p \vee p) \wedge (\sim p \vee \sim q) \equiv T \wedge (\sim p \vee \sim q) \equiv \sim p \vee \sim q$:
$S \equiv (\sim p \vee \sim q) \vee (q \wedge \sim r) \vee r$
Using associative and distributive laws:
$S \equiv \sim p \vee \sim q \vee (q \vee r) \wedge (\sim r \vee r)$
$S \equiv \sim p \vee \sim q \vee (q \vee r) \wedge T$
$S \equiv \sim p \vee (\sim q \vee q) \vee r$
$S \equiv \sim p \vee T \vee r \equiv T$
Since the result is always true,the statement is a tautology.

(A) Given expression: $(p \wedge q) \Rightarrow ((r \wedge q) \wedge p)$
Using the implication law $A \Rightarrow B \equiv \sim A \vee B$:
$\sim(p \wedge q) \vee ((r \wedge q) \wedge p)$
Using the associative and commutative properties:
$\sim(p \wedge q) \vee ((r \wedge p) \wedge (p \wedge q))$
Using the distributive law $X \vee (Y \wedge Z) \equiv (X \vee Y) \wedge (X \vee Z)$:
$(\sim(p \wedge q) \vee (r \wedge p)) \wedge (\sim(p \wedge q) \vee (p \wedge q))$
Since $\sim A \vee A \equiv t$ (tautology):
$(\sim(p \wedge q) \vee (r \wedge p)) \wedge t$
$\equiv \sim(p \wedge q) \vee (r \wedge p)$
Converting back to implication form:
$(p \wedge q) \Rightarrow (r \wedge p)$

(C) We test the expression $(p \wedge \sim q) \Rightarrow (p \vee q)$ using a truth table:

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

Since all values in the final column are $T$,the expression is a tautology when $* = \wedge$ and $\square = \vee$.

(A) The negation of an implication $A \Rightarrow B$ is given by $A \wedge \sim B$.
Therefore,$\sim((p \vee r) \Rightarrow (q \vee r)) = (p \vee r) \wedge \sim(q \vee r)$.
Using De Morgan's Law,$\sim(q \vee r) = (\sim q \wedge \sim r)$.
So,the expression becomes $(p \vee r) \wedge (\sim q \wedge \sim r)$.
By the distributive law,this is $((p \vee r) \wedge \sim r) \wedge \sim q$.
Since $(p \vee r) \wedge \sim r = (p \wedge \sim r) \vee (r \wedge \sim r) = (p \wedge \sim r) \vee F = p \wedge \sim r$.
Thus,the final expression is $(p \wedge \sim r) \wedge \sim q$,which is $p \wedge \sim q \wedge \sim r$.

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

(A) To find the equivalent expression,we simplify the given Boolean expression using logical laws:
Given expression: $(p \wedge \sim q) \Rightarrow (q \vee \sim p)$
Using the implication law $A \Rightarrow B \equiv \sim A \vee B$:
$\equiv \sim (p \wedge \sim q) \vee (q \vee \sim p)$
Apply De Morgan's Law $\sim (p \wedge \sim q) \equiv \sim p \vee q$:
$\equiv (\sim p \vee q) \vee (q \vee \sim p)$
By Associative and Commutative laws:
$\equiv (\sim p \vee \sim p) \vee (q \vee q)$
$\equiv \sim p \vee q$
Since $\sim p \vee q \equiv p \Rightarrow q$,the expression is equivalent to $p \Rightarrow q$.

(A) In logic,a conditional statement $P \rightarrow Q$ is false only when $P$ is true and $Q$ is false. Otherwise,it is true.
Statement $(A)$: $P: 3+3=7$ (False),$Q: 4+3=8$ (False). Since $P$ is false,$P \rightarrow Q$ is True.
Statement $(B)$: $P: 5+3=8$ (True),$Q: \text{earth is flat}$ (False). Since $P$ is true and $Q$ is false,$P \rightarrow Q$ is False.
Statement $(C)$: $P: (A) \text{ is true and } (B) \text{ is true}$ (False,because $(B)$ is false),$Q: 5+6=17$ (False). Since $P$ is false,$P \rightarrow Q$ is True.
Thus,$(A)$ is true,$(B)$ is false,and $(C)$ is true.

(A) We evaluate each expression using the identity $A \Rightarrow B \equiv \sim A \vee B$:
$A) (\sim p$ $\Rightarrow q) \vee (\sim q$ $\Rightarrow p) \equiv (p \vee q) \vee (q \vee p) \equiv p \vee q$. This is not a tautology as it depends on the truth values of $p$ and $q$.
$B) (q$ $\Rightarrow p) \vee (\sim q$ $\Rightarrow p) \equiv (\sim q \vee p) \vee (q \vee p) \equiv (\sim q \vee q) \vee p \equiv T \vee p \equiv T$. This is a tautology.
$C) (p$ $\Rightarrow q) \vee (\sim q$ $\Rightarrow p) \equiv (\sim p \vee q) \vee (q \vee p) \equiv (\sim p \vee p) \vee q \equiv T \vee q \equiv T$. This is a tautology.
$D) (p$ $\Rightarrow \sim q) \vee (\sim q$ $\Rightarrow p) \equiv (\sim p \vee \sim q) \vee (q \vee p) \equiv (\sim p \vee p) \vee (\sim q \vee q) \equiv T \vee T \equiv T$. This is a tautology.
Thus,the expression in option $A$ is not a tautology.

(D) We are given the expression $(p$ $\Rightarrow q) \wedge (q$ $\Rightarrow \sim p)$.
Using the logical equivalence $(A \Rightarrow B) \equiv (\sim A \vee B)$,we rewrite the expression:
$(\sim p \vee q) \wedge (\sim q \vee \sim p)$
Using the commutative property,we rearrange the terms:
$(\sim p \vee q) \wedge (\sim p \vee \sim q)$
Applying the distributive property,we factor out $(\sim p \vee)$:
$\sim p \vee (q \wedge \sim q)$
Since $(q \wedge \sim q)$ is a contradiction (always false),the expression becomes:
$\sim p \vee F \equiv \sim p$
Therefore,the expression is equivalent to $\sim p$.

(D) Let $p$ be the statement "The weather is good".
Let $q$ be the statement "The ground is not wet".
Let $r$ be the statement "The match will be played".
The given statement is $r \implies (p \wedge q)$.
The negation of $r \implies (p \wedge q)$ is $\sim(r \implies (p \wedge q)) \equiv r \wedge \sim(p \wedge q)$.
Using De Morgan's Law,$\sim(p \wedge q) \equiv \sim p \vee \sim q$.
Thus,the negation is "The match will be played and (the weather is not good or the ground is wet)".

(B) To determine the equivalence,we construct the truth table for the expression $(P \vee Q) \wedge (\sim P) \Rightarrow Q$.

$P$	$Q$	$P \vee Q$	$\sim P$	$(P \vee Q) \wedge (\sim P)$	$(P \vee Q) \wedge (\sim P) \Rightarrow Q$
$T$	$T$	$T$	$F$	$F$	$T$
$T$	$F$	$T$	$F$	$F$	$T$
$F$	$T$	$T$	$T$	$T$	$T$
$F$	$F$	$F$	$T$	$F$	$T$

The final column shows that the statement is a tautology (always $T$).
Checking the options,option $B$ is $\sim(P \Rightarrow Q) \Leftrightarrow P \wedge \sim Q$. Since $\sim(P \Rightarrow Q) \equiv P \wedge \sim Q$,the biconditional $\sim(P \Rightarrow Q) \Leftrightarrow P \wedge \sim Q$ is also a tautology.
Thus,the statement is equivalent to option $B$.

(C) Let the statement $P$ be: "For all $M > 0$,there exists $x \in S$ such that $x \geq M$."
The negation of a statement involving quantifiers follows these rules:
$1$. The negation of "for all" $(\forall)$ is "there exists" $(\exists)$.
$2$. The negation of "there exists" $(\exists)$ is "for all" $(\forall)$.
$3$. The negation of $x \geq M$ is $x < M$.
Applying these rules to $P$:
$\sim P$: "There exists $M > 0$ such that for all $x \in S$,$x < M$."
Thus,the correct option is $C$.

(A) Let the given statement be $P \rightarrow B$,where $P = (A \wedge C)$.
The statement is $(A \wedge C) \rightarrow B$.
The negation of an implication $P \rightarrow Q$ is $P \wedge (\sim Q)$.
Here,$P = (A \wedge C)$ and $Q = B$.
Therefore,the negation is $(A \wedge C) \wedge (\sim B)$.

(B) Let $S$ be the statement $(p \Delta q) \Rightarrow ((p \Delta \sim q) \vee ((\sim p) \Delta q))$.
We test each operator for $\Delta$:
$1$. If $\Delta = \wedge$: $(p \wedge q) \Rightarrow ((p \wedge \sim q) \vee (\sim p \wedge q))$. This is not a tautology (e.g.,if $p=T, q=T$,then $T \Rightarrow (F \vee F) = F$).
$2$. If $\Delta = \vee$: $(p \vee q) \Rightarrow ((p \vee \sim q) \vee (\sim p \vee q))$. This simplifies to $(p \vee q) \Rightarrow (T) = T$,which is a tautology.
$3$. If $\Delta = \Rightarrow$: $(p$ $\Rightarrow q)$ $\Rightarrow ((p$ $\Rightarrow \sim q) \vee (\sim p$ $\Rightarrow q))$. If $p=T, q=T$,then $(T$ $\Rightarrow T)$ $\Rightarrow ((T$ $\Rightarrow F) \vee (F$ $\Rightarrow T)) = T$ $\Rightarrow (F \vee T) = T$ $\Rightarrow T = T$. If $p=T, q=F$,then $(T$ $\Rightarrow F)$ $\Rightarrow ((T$ $\Rightarrow T) \vee (F$ $\Rightarrow F)) = F$ $\Rightarrow (T \vee T) = T$. If $p=F, q=T$,then $(F$ $\Rightarrow T)$ $\Rightarrow ((F$ $\Rightarrow F) \vee (T$ $\Rightarrow T)) = T$ $\Rightarrow (T \vee T) = T$. If $p=F, q=F$,then $(F$ $\Rightarrow F)$ $\Rightarrow ((F$ $\Rightarrow T) \vee (T$ $\Rightarrow F)) = T$ $\Rightarrow (T \vee F) = T$. Thus,it is a tautology.
$4$. If $\Delta = \Leftrightarrow$: $(p \Leftrightarrow q) \Rightarrow ((p \Leftrightarrow \sim q) \vee (\sim p \Leftrightarrow q))$. If $p=T, q=T$,then $(T \Leftrightarrow T)$ $\Rightarrow ((T \Leftrightarrow F) \vee (F \Leftrightarrow T)) = T$ $\Rightarrow (F \vee F) = F$. Not a tautology.
Therefore,there are $2$ choices: $\vee$ and $\Rightarrow$.

(C) Let the given expression be $S = ((\sim q) \wedge p) \Rightarrow ((\sim p) \vee q)$.
Using the implication rule $A \Rightarrow B \equiv \sim A \vee B$,we have:
$S \equiv \sim((\sim q) \wedge p) \vee ((\sim p) \vee q)$.
Applying De Morgan's law: $\sim((\sim q) \wedge p) \equiv q \vee (\sim p)$.
So,$S \equiv (q \vee \sim p) \vee (\sim p \vee q) \equiv \sim p \vee q$.
We know that $\sim p \vee q \equiv p \Rightarrow q$.
Therefore,the negation of the expression is $\sim(p \Rightarrow q)$.

(C) Given the proposition $p \rightarrow ((\sim p) \vee q)$ is $FALSE$.
An implication $A \rightarrow B$ is $FALSE$ only when $A$ is $TRUE$ and $B$ is $FALSE$.
Therefore,$p$ must be $TRUE$ and $((\sim p) \vee q)$ must be $FALSE$.
Since $p$ is $TRUE$,$\sim p$ is $FALSE$.
For $(\sim p \vee q)$ to be $FALSE$,both $\sim p$ and $q$ must be $FALSE$. Thus,$q$ is $FALSE$.
Now,evaluate $P_1$ and $P_2$ for $p = TRUE, q = FALSE$:
$P_1 = \sim(p$ $\rightarrow \sim q) = \sim(T$ $\rightarrow \sim F) = \sim(T$ $\rightarrow T) = \sim(T) = FALSE$.
$P_2 = (p \wedge \sim q) \wedge ((\sim p) \vee q) = (T \wedge \sim F) \wedge ((\sim T) \vee F) = (T \wedge T) \wedge (F \vee F) = T \wedge F = FALSE$.
Thus,both $P_1$ and $P_2$ are $FALSE$.

(C) To determine which $r$ makes the statement $r \vee (\sim p) \Rightarrow (p \wedge q) \vee r$ a tautology,we test each option using a truth table.
For $r = \sim p$:
The statement becomes $(\sim p \vee \sim p) \Rightarrow (p \wedge q) \vee \sim p$.
This simplifies to $\sim p \Rightarrow (p \wedge q) \vee \sim p$.
Using the law of absorption,$(p \wedge q) \vee \sim p$ is equivalent to $(\sim p \vee p) \wedge (\sim p \vee q)$,which is $T \wedge (\sim p \vee q) = \sim p \vee q$.
So the statement is $\sim p \Rightarrow (\sim p \vee q)$.
Since $\sim p \Rightarrow (\sim p \vee q)$ is equivalent to $\neg(\sim p) \vee (\sim p \vee q) = p \vee \sim p \vee q = T \vee q = T$,the statement is a tautology.
Thus,$r = \sim p$ is the correct option.

(A) Given the expression $(p \nabla q) \Rightarrow ((p \nabla q) \nabla r)$ is a tautology.
$Case-I$: If $\nabla \equiv \wedge$,then $(p \wedge q) \Rightarrow ((p \wedge q) \wedge r)$. This is not a tautology because if $p=T, q=T, r=F$,the expression becomes $T \Rightarrow F$,which is $F$.
$Case-II$: If $\nabla \equiv \vee$,then $(p \vee q) \Rightarrow ((p \vee q) \vee r)$. This is a tautology because if the antecedent $(p \vee q)$ is $T$,then the consequent $((p \vee q) \vee r)$ is also $T$.
Since $\nabla \equiv \vee$,we need to evaluate $(p \nabla q) \Delta r$,which is $(p \vee q) \Delta r$.
If $\Delta \equiv \vee$,then $(p \vee q) \vee r \equiv (p \vee r) \vee q$,which matches option $(A)$ as $(p \Delta r) \vee q$.
If $\Delta \equiv \wedge$,then $(p \vee q) \wedge r$,which does not match the options.
Thus,the expression is logically equivalent to $(p \Delta r) \vee q$.

(C) Given expression: $(\sim(p \wedge q)) \vee q$
Using De Morgan's Law: $(\sim p \vee \sim q) \vee q$
Using Associative Law: $\sim p \vee (\sim q \vee q)$
Since $(\sim q \vee q) = t$ (tautology),the expression becomes $\sim p \vee t = t$.
Thus,the expression is a tautology.
Now,check the options:
$A. q$ $\rightarrow (p \wedge q) \equiv \sim q \vee (p \wedge q) \equiv (\sim q \vee p) \wedge (\sim q \vee q) \equiv \sim q \vee p$ (Not a tautology)
$B. p \rightarrow q \equiv \sim p \vee q$ (Not a tautology)
$C. p \rightarrow (p \vee q) \equiv \sim p \vee (p \vee q) \equiv (\sim p \vee p) \vee q \equiv t \vee q \equiv t$ (Tautology)
$D. p$ $\rightarrow (p$ $\rightarrow q) \equiv \sim p \vee (\sim p \vee q) \equiv \sim p \vee q$ (Not a tautology)
Therefore,the expression is equivalent to $p \rightarrow (p \vee q)$.

(A) Let the given propositions be $C_1, C_2, \dots, C_9$. We test the truth values for $p, q, r, s$ to maximize the number of true propositions.
If we assign $p=F, q=F, r=T, s=F$:
$C_1: F \vee T \vee F = T$
$C_2: F \vee F \vee T = T$
$C_3: F \vee T \vee F = T$
$C_4: T \vee F \vee F = T$
$C_5: T \vee F \vee T = T$
$C_6: T \vee F \vee T = T$
$C_7: F \vee T \vee T = T$
$C_8: F \vee F \vee T = T$
$C_9: T \vee T \vee T = T$
All $9$ propositions are true for this assignment.

(C) Given $S_{1}: ((\sim p) \vee q) \vee ((\sim p) \vee r)$.
Using the associative and idempotent laws,we have $S_{1} \equiv (\sim p \vee \sim p) \vee (q \vee r) \equiv \sim p \vee (q \vee r)$.
Given $S_{2}: p \rightarrow (q \vee r)$.
Using the conditional law $p \rightarrow x \equiv \sim p \vee x$,we have $S_{2} \equiv \sim p \vee (q \vee r)$.
Since $S_{1} \equiv S_{2}$,they always have the same truth value.
Therefore,if $S_{2}$ is False,$S_{1}$ must also be False.
Thus,the statement 'If $S_{2}$ is False,then $S_{1}$ is True' is $NOT$ true.

(C) Let the statement be $S = (p \vee q) \Rightarrow ((\sim r) \vee p)$.
Using the implication rule $A \Rightarrow B \equiv \sim A \vee B$,we get:
$S \equiv \sim (p \vee q) \vee ((\sim r) \vee p)$
$S \equiv (\sim p \wedge \sim q) \vee (\sim r \vee p)$
Using the distributive law $(A \wedge B) \vee C \equiv (A \vee C) \wedge (B \vee C)$:
$S \equiv (\sim p \vee \sim r \vee p) \wedge (\sim q \vee \sim r \vee p)$
Since $(\sim p \vee p) \equiv T$ (Tautology),we have:
$S \equiv (T \vee \sim r) \wedge (\sim q \vee \sim r \vee p)$
$S \equiv T \wedge (\sim q \vee \sim r \vee p) \equiv \sim q \vee \sim r \vee p$
Now,the negation of $S$ is $\sim (\sim q \vee \sim r \vee p)$.
Applying De Morgan's Law,$\sim (A \vee B \vee C) \equiv \sim A \wedge \sim B \wedge \sim C$:
$\sim S \equiv q \wedge r \wedge \sim p$
Rearranging,we get $(\sim p) \wedge q \wedge r$.

(C) First,simplify the expression $(p \vee q) \Rightarrow q$:
$(p \vee q) \Rightarrow q \equiv \sim(p \vee q) \vee q$
$\equiv (\sim p \wedge \sim q) \vee q$
$\equiv (\sim p \vee q) \wedge (\sim q \vee q)$
$\equiv (\sim p \vee q) \wedge t \equiv \sim p \vee q$.
Now,we test the options for $(p \wedge q) \Delta (\sim p \vee q)$.
For option $C$ $(\Rightarrow)$:
$(p \wedge q) \Rightarrow (\sim p \vee q) \equiv \sim(p \wedge q) \vee (\sim p \vee q)$
$\equiv (\sim p \vee \sim q) \vee (\sim p \vee q)$
$\equiv \sim p \vee (\sim q \vee q)$
$\equiv \sim p \vee t \equiv t$.
Since the result is a tautology,$\Delta$ is $\Rightarrow$.

(D) Given statements:
$P$: Ramu is intelligent
$Q$: Ramu is rich
$R$: Ramu is not honest
The statement "Ramu is intelligent and honest if and only if Ramu is not rich" is represented as $(P \wedge \sim R) \Leftrightarrow \sim Q$.
The negation of the statement is $\sim[(P \wedge \sim R) \Leftrightarrow \sim Q]$.
Using the identity $\sim(A \Leftrightarrow B) \equiv (A \wedge \sim B) \vee (B \wedge \sim A)$,where $A = (P \wedge \sim R)$ and $B = \sim Q$:
$= ((P \wedge \sim R) \wedge \sim(\sim Q)) \vee (\sim Q \wedge \sim(P \wedge \sim R))$
$= ((P \wedge \sim R) \wedge Q) \vee (\sim Q \wedge (\sim P \vee \sim(\sim R)))$
$= ((P \wedge \sim R) \wedge Q) \vee (\sim Q \wedge (\sim P \vee R))$
Thus,the correct option is $D$.

(D) statement is a tautology if its truth value is $T$ for all possible truth values of its components.
Let us evaluate option $B$: $p \Rightarrow ((\sim p) \vee q)$.
This is equivalent to $(\sim p) \vee ((\sim p) \vee q)$ by the implication law $A \Rightarrow B \equiv (\sim A) \vee B$.
Using the associative law,this becomes $(\sim p \vee \sim p) \vee q$,which simplifies to $(\sim p) \vee q$.
Since this is not always $T$ (e.g.,if $p=T, q=F$,then $\sim p \vee q = F$),let us re-examine the options.
Actually,option $D$ is $q \Rightarrow ((\sim p) \vee q)$.
This is equivalent to $(\sim q) \vee ((\sim p) \vee q) = (\sim q \vee q) \vee (\sim p) = T \vee (\sim p) = T$.
Since the result is always $T$,option $D$ is a tautology.

(D) We want to find the negation of the expression $p \Leftrightarrow (q \Rightarrow p)$.
Let $S = p \Leftrightarrow (q \Rightarrow p)$.
Using the property $\sim(A \Leftrightarrow B) = (A \wedge \sim B) \vee (B \wedge \sim A)$,we have:
$\sim S = (p \wedge \sim(q$ $\Rightarrow p)) \vee ((q$ $\Rightarrow p) \wedge \sim p)$.
Now,simplify the first part: $p \wedge \sim(q \Rightarrow p) = p \wedge (q \wedge \sim p) = (p \wedge \sim p) \wedge q = F \wedge q = F$ (where $F$ is a contradiction).
Simplify the second part: $(q$ $\Rightarrow p) \wedge \sim p = (\sim q \vee p) \wedge \sim p = (\sim q \wedge \sim p) \vee (p \wedge \sim p) = (\sim p \wedge \sim q) \vee F = \sim p \wedge \sim q$.
Combining these,$\sim S = F \vee (\sim p \wedge \sim q) = \sim p \wedge \sim q$.

(C) We are given the expression $(\sim( p \Leftrightarrow \sim q )) \wedge q$.
First,recall that $\sim( p \Leftrightarrow \sim q )$ is equivalent to $p \Leftrightarrow q$ because $\sim( p \Leftrightarrow \sim q ) \equiv p \Leftrightarrow \sim(\sim q) \equiv p \Leftrightarrow q$.
Now,the expression becomes $( p \Leftrightarrow q ) \wedge q$.
We know that $p \Leftrightarrow q$ is $( p$ $\Rightarrow q ) \wedge ( q$ $\Rightarrow p )$.
So,$( p \Leftrightarrow q ) \wedge q \equiv ( p$ $\Rightarrow q ) \wedge ( q$ $\Rightarrow p ) \wedge q$.
Since $( q \Rightarrow p ) \wedge q \equiv q \wedge p$,the expression simplifies to $( p \Rightarrow q ) \wedge ( p \wedge q ) \equiv p \wedge q$.
Alternatively,checking the truth table for $( p \Leftrightarrow q ) \wedge q$:
If $p=T, q=T$,then $(T \Leftrightarrow T) \wedge T = T \wedge T = T$.
If $p=F, q=T$,then $(F \Leftrightarrow T) \wedge T = F \wedge T = F$.
If $p=T, q=F$,then $(T \Leftrightarrow F) \wedge F = F \wedge F = F$.
If $p=F, q=F$,then $(F \Leftrightarrow F) \wedge F = T \wedge F = F$.
Comparing this with option $(C)$: $( p \Rightarrow q ) \wedge q$.
If $p=T, q=T$,then $(T \Rightarrow T) \wedge T = T \wedge T = T$.
If $p=F, q=T$,then $(F \Rightarrow T) \wedge T = T \wedge T = T$.
This does not match.
Wait,let's re-evaluate the original expression: $(\sim( p \Leftrightarrow \sim q )) \wedge q$.
Since $\sim( p \Leftrightarrow \sim q ) \equiv p \Leftrightarrow q$,the expression is $( p \Leftrightarrow q ) \wedge q$.
This is equivalent to $p \wedge q$.
Looking at the options,none of the provided options are $p \wedge q$. Let us re-check the logic.
Actually,$( p \Rightarrow q ) \wedge q$ is equivalent to $q \wedge p$ if $p$ is true,but generally $( p \Rightarrow q ) \wedge q \equiv q \wedge p$ is not true. Wait,$( p \Rightarrow q ) \wedge q \equiv (\sim p \vee q) \wedge q \equiv q$.
Let's re-examine the expression: $(\sim( p \Leftrightarrow \sim q )) \wedge q \equiv (p \Leftrightarrow q) \wedge q$. This is $p \wedge q$.
Given the options,there might be a typo in the question or options. However,based on standard logic,$(p \Leftrightarrow q) \wedge q \equiv p \wedge q$.

(D) The implication $X \rightarrow Y$ is $F$ if and only if $X$ is $T$ and $Y$ is $F$.
Given $(P \wedge (\sim R)) \rightarrow ((\sim R) \wedge Q)$ is $F$,we have:
$P \wedge (\sim R) = T \implies P = T$ and $\sim R = T \implies R = F$.
$(\sim R) \wedge Q = F$. Since $\sim R = T$,we must have $Q = F$.
So,the truth values are $P = T, Q = F, R = F$.
Now,check the options:
$(A) P \vee Q$ $\rightarrow \sim R = (T \vee F)$ $\rightarrow T = T$ $\rightarrow T = T$.
$(B) R \vee Q$ $\rightarrow \sim P = (F \vee F)$ $\rightarrow F = F$ $\rightarrow F = T$.
$(C) \sim(P \vee Q)$ $\rightarrow \sim R = \sim(T \vee F)$ $\rightarrow T = \sim T$ $\rightarrow T = F$ $\rightarrow T = T$.
$(D) \sim(R \vee Q)$ $\rightarrow \sim P = \sim(F \vee F)$ $\rightarrow F = \sim F$ $\rightarrow F = T$ $\rightarrow F = F$.
Thus,the statement in option $(D)$ is $F$.

(C) We are given the logical equivalence: $(p \wedge r) \Leftrightarrow (p \wedge (\sim q)) \equiv (\sim p)$.
Let us test the options by substituting $r = \sim q$:
$(p \wedge (\sim q)) \Leftrightarrow (p \wedge (\sim q)) \equiv T$ (Tautology),which is not $(\sim p)$.
Let us test $r = q$:
$(p \wedge q) \Leftrightarrow (p \wedge (\sim q))$.
If $p = T$,then $(T \wedge q) \Leftrightarrow (T \wedge (\sim q)) \equiv q \Leftrightarrow (\sim q)$,which is $F$.
If $p = F$,then $(F \wedge q) \Leftrightarrow (F \wedge (\sim q)) \equiv F \Leftrightarrow F$,which is $T$.
This matches the truth table for $(\sim p)$.
Thus,$r = q$ is the correct choice.

(D) Given statements are:
$p$: Ramesh listens to music.
$q$: Ramesh is out of his village.
$r$: It is Sunday.
$s$: It is Saturday.
We need to translate the statement: "Ramesh listens to music only if he is in his village and it is Sunday or Saturday".
$1$. "Ramesh is in his village" is the negation of "Ramesh is out of his village",which is $\sim q$.
$2$. "It is Sunday or Saturday" is $r \vee s$.
$3$. The phrase "$p$ only if $A$" is logically equivalent to $p \Rightarrow A$.
Therefore,the statement becomes $p \Rightarrow ((\sim q) \wedge (r \vee s))$.

(B) We check each option by evaluating the expression $(p * q) \odot (p \odot \sim q)$.
For option $A$: $* = \vee, \odot = \wedge$.
The expression becomes $(p \vee q) \wedge (p \wedge \sim q)$.
Using the distributive law: $p \vee (q \wedge \sim q) \wedge (p \wedge \sim q)$ is not correct. Let's re-evaluate: $(p \vee q) \wedge (p \wedge \sim q) \equiv (p \wedge p \wedge \sim q) \vee (q \wedge p \wedge \sim q) \equiv (p \wedge \sim q) \vee (F) \equiv p \wedge \sim q$,which is not a tautology.
For option $B$: $* = \vee, \odot = \vee$.
The expression becomes $(p \vee q) \vee (p \vee \sim q)$.
By associativity and commutativity: $p \vee p \vee (q \vee \sim q) \equiv p \vee T \equiv T$.
Since the result is $T$ (a tautology),option $B$ is correct.

(B) Given statement: $(p$ $\Rightarrow q) \vee (p$ $\Rightarrow r)$
Using the identity $(p \Rightarrow q) \equiv (\sim p \vee q)$:
$= (\sim p \vee q) \vee (\sim p \vee r)$
$= \sim p \vee (q \vee r)$
$= p \Rightarrow (q \vee r)$
This matches option $(C)$.
Now check other options:
Option $(A): (p \wedge \sim r)$ $\Rightarrow q
\equiv \sim(p \wedge \sim r) \vee q
\equiv (\sim p \vee r) \vee q
\equiv \sim p \vee (q \vee r)
\equiv p$ $\Rightarrow (q \vee r)$. (Equivalent)
Option $(D): (p \wedge \sim q)$ $\Rightarrow r
\equiv \sim(p \wedge \sim q) \vee r
\equiv (\sim p \vee q) \vee r
\equiv \sim p \vee (q \vee r)
\equiv p$ $\Rightarrow (q \vee r)$. (Equivalent)
Option $(B): (\sim q)$ $\Rightarrow ((\sim r) \vee p)
\equiv \sim(\sim q) \vee (\sim r \vee p)
\equiv q \vee \sim r \vee p
\equiv p \vee q \vee \sim r$. (Not equivalent to $p \Rightarrow (q \vee r)$).

(D) Given statement: $(p \wedge q) \Rightarrow (p \wedge r)$
Using the logical equivalence $A \Rightarrow B \equiv \sim A \vee B$:
$\sim(p \wedge q) \vee (p \wedge r)$
Apply De Morgan's Law:
$(\sim p \vee \sim q) \vee (p \wedge r)$
Using the distributive law $(\sim p \vee \sim q) \vee (p \wedge r) \equiv ((\sim p \vee \sim q) \vee p) \wedge ((\sim p \vee \sim q) \vee r)$:
$((\sim p \vee p) \vee \sim q) \wedge (\sim p \vee \sim q \vee r)$
Since $(\sim p \vee p) \equiv T$ (Tautology):
$(T \vee \sim q) \wedge (\sim p \vee \sim q \vee r)$
$T \wedge (\sim p \vee \sim q \vee r) \equiv \sim p \vee \sim q \vee r$
Rearranging the terms:
$(\sim p \vee \sim q) \vee r$
Using De Morgan's Law again:
$\sim(p \wedge q) \vee r$
Converting back to implication:
$(p \wedge q) \Rightarrow r$

(B) Let the given statement be $r \Rightarrow s$,where $r = (\sim(P \wedge Q)) \vee ((\sim P) \wedge Q)$ and $s = ((\sim P) \wedge (\sim Q))$.
We construct the truth table for the statement:
| $P$ | $Q$ | $\sim(P \wedge Q)$ | $(\sim P) \wedge Q$ | $r$ | $s$ | $r \Rightarrow s$ |
|---|---|---|---|---|---|---|
| $T$ | $T$ | $F$ | $F$ | $F$ | $F$ | $T$ |
| $T$ | $F$ | $T$ | $F$ | $T$ | $F$ | $F$ |
| $F$ | $T$ | $T$ | $T$ | $T$ | $F$ | $F$ |
| $F$ | $F$ | $T$ | $F$ | $T$ | $T$ | $T$ |
Comparing the truth values of $r \Rightarrow s$ with the options:
For option $(B)$,$(\sim Q) \vee P$:
- If $P=T, Q=T$,$(\sim T) \vee T = F \vee T = T$.
- If $P=T, Q=F$,$(\sim F) \vee T = T \vee T = T$.
- If $P=F, Q=T$,$(\sim T) \vee F = F \vee F = F$.
- If $P=F, Q=F$,$(\sim F) \vee F = T \vee F = T$.
This does not match. Let's re-evaluate $r \Rightarrow s$:
$r = (\sim P \vee \sim Q) \vee (\sim P \wedge Q) = \sim P \vee (\sim Q \vee (\sim P \wedge Q)) = \sim P \vee (\sim Q \vee \sim P) \wedge (\sim Q \vee Q) = \sim P \vee \sim Q = \sim(P \wedge Q)$.
$s = \sim P \wedge \sim Q = \sim(P \vee Q)$.
So,$r \Rightarrow s$ is $\sim(P \wedge Q) \Rightarrow \sim(P \vee Q)$.
This is equivalent to $\sim(\sim(P \vee Q)) \Rightarrow \sim(\sim(P \wedge Q))$,which is $(P \vee Q) \Rightarrow (P \wedge Q)$.
This is only true when $P$ and $Q$ have the same truth value,i.e.,$P \Leftrightarrow Q$.

(C) We are given the expression $\sim(p \wedge (p \rightarrow \sim q))$.
Using De Morgan's Law,$\sim(A \wedge B) \equiv \sim A \vee \sim B$:
$\sim(p \wedge (p$ $\rightarrow \sim q)) \equiv \sim p \vee \sim(p$ $\rightarrow \sim q)$.
Since $p \rightarrow q \equiv \sim p \vee q$,we have $p \rightarrow \sim q \equiv \sim p \vee \sim q$.
Substituting this into the expression:
$\equiv \sim p \vee \sim(\sim p \vee \sim q)$.
Applying De Morgan's Law again:
$\equiv \sim p \vee (p \wedge q)$.
Using the Distributive Law,$A \vee (B \wedge C) \equiv (A \vee B) \wedge (A \vee C)$:
$\equiv (\sim p \vee p) \wedge (\sim p \vee q)$.
Since $\sim p \vee p \equiv t$ (tautology):
$\equiv t \wedge (\sim p \vee q)$.
$\equiv \sim p \vee q$.

(B) Let the given statement be $S = (p \wedge \sim q)$ $\Rightarrow (p$ $\Rightarrow \sim q)$.
Using the implication law $A \Rightarrow B \equiv \sim A \vee B$,we get:
$S \equiv \sim (p \wedge \sim q) \vee (p \Rightarrow \sim q)$
$S \equiv (\sim p \vee \sim (\sim q)) \vee (\sim p \vee \sim q)$
$S \equiv (\sim p \vee q) \vee (\sim p \vee \sim q)$
Using the associative and commutative laws:
$S \equiv \sim p \vee (q \vee \sim q)$
Since $(q \vee \sim q)$ is a tautology $(t)$:
$S \equiv \sim p \vee t$
$S \equiv t$
Therefore,the statement is a tautology.