Mathematical logic Questions in English

(B) 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.
From $(\sim r) = T$,we get $r = F$.
Substituting $r = F$ into $(q \rightarrow r) = T$,we get $(q \rightarrow F) = T$,which implies $q = F$.
Now,substitute $q = F$ into $(p \vee q) = T$,we get $(p \vee F) = T$,which implies $p = T$.
Checking $B = (p \wedge q) = (T \wedge F) = F$,which satisfies the condition.
Thus,the truth values are $p = T, q = F, r = F$.

(D) $1$. Evaluate the truth value of each statement:
- $p$: For matrices $A$ and $B$,$(A-B)(A+B) = A^2 + AB - BA - B^2$. Since $AB \neq BA$,$A^2 - B^2 \neq (A-B)(A+B)$. Thus,$p$ is False $(F)$.
- $q$: $5 \leqslant 5$ is true. Thus,$q$ is True $(T)$.
- $r$: We know $\sum_{k=0}^{n} { }^n C_k = 2^n$. Here,${ }^8 C_0 + { }^8 C_1 + \ldots + { }^8 C_8 = 2^8 = 256$. Since ${ }^8 C_0 = 1$,the sum ${ }^8 C_1 + \ldots + { }^8 C_8 = 256 - 1 = 255$. Thus,$r$ is False $(F)$.
- $s$: The maximum value of ${ }^n C_r$ is ${ }^n C_{n/2}$ for even $n$. For $n=8$,it is ${ }^8 C_4 = \frac{8 \times 7 \times 6 \times 5}{4 \times 3 \times 2 \times 1} = 70$. Thus,$s$ is True $(T)$.
$2$. Truth values: $p=F, q=T, r=F, s=T$.
$3$. Check options:
- $A: (F \wedge \sim F) \vee (\sim T \wedge \sim T) = (F \wedge T) \vee (F \wedge F) = F \vee F = F$.
- $B: (F \vee \sim T) \leftrightarrow (\sim F$ $\rightarrow T) = (F \vee F) \leftrightarrow (T$ $\rightarrow T) = F \leftrightarrow T = F$.
- $C: (F \leftrightarrow T) \wedge (\sim F \vee \sim T) = F \wedge (T \vee F) = F \wedge T = F$.
- $D: (T \vee \sim F) \leftrightarrow (\sim F \wedge \sim F) = (T \vee T) \leftrightarrow (T \wedge T) = T \leftrightarrow T = T$.
Therefore,option $D$ is true.

(D) tautology is a statement that is always true for all possible truth values of its components.
We evaluate option $D$: $(\sim q \wedge p) \vee (p \vee \sim p)$.
We know that $(p \vee \sim p)$ is a tautology (always true,denoted as $T$).
Thus,the expression becomes $(\sim q \wedge p) \vee T$.
Since any statement $X \vee T$ is always $T$,the entire expression is a tautology.
Therefore,the correct option is $D$.

(A) Step $1$: Evaluate statement $p$.
$2$ is indeed an even prime number. So,$p$ is True $(T)$.
Step $2$: Evaluate statement $q$.
Given $z_1 = 2 - i$ and $z_2 = -2 + i$,then $\bar{z}_2 = -2 - i$.
$z_1 \bar{z}_2 = (2 - i)(-2 - i) = -4 - 2i + 2i + i^2 = -4 - 1 = -5$.
Then $\frac{1}{z_1 \bar{z}_2} = \frac{1}{-5} = -\frac{1}{5} + 0i$.
The imaginary part $\operatorname{Im}\left[\frac{1}{z_1 \bar{z}_2}\right] = 0$.
Since $0 \neq -\frac{1}{5}$,statement $q$ is False $(F)$.
Step $3$: Evaluate statement $r$.
$\tan(-945^{\circ}) = -\tan(945^{\circ}) = -\tan(2 \times 360^{\circ} + 225^{\circ}) = -\tan(225^{\circ}) = -\tan(180^{\circ} + 45^{\circ}) = -\tan(45^{\circ}) = -1$.
So,$r$ is True $(T)$.
Step $4$: Check the options.
$p = T, q = F, r = T$.
$(A)$ $(T$ $\rightarrow F) \leftrightarrow (F \wedge T)$ $\Rightarrow F \leftrightarrow F = T$.
$(B)$ $F \leftrightarrow T = F$.
$(C)$ $T \rightarrow F = F$.
$(D)$ $(T$ $\rightarrow T) \leftrightarrow F$ $\Rightarrow T \leftrightarrow F = F$.
Thus,the correct option is $(A)$.

(B) The circuit consists of switch $s_1$ in series with a parallel combination of $s_1'$ and $s_2'$,which is then in series with $s_2$.
Symbolically,this is represented as $p \wedge (\sim p \vee \sim q) \wedge q$.
Using the distributive law: $p \wedge ((\sim p \vee \sim q) \wedge q) = p \wedge ((\sim p \wedge q) \vee (\sim q \wedge q))$.
Since $(\sim q \wedge q)$ is a contradiction $(F)$,we have $p \wedge ((\sim p \wedge q) \vee F) = p \wedge (\sim p \wedge q)$.
By associative law: $(p \wedge \sim p) \wedge q = F \wedge q = F$.
Since the symbolic form simplifies to a contradiction $(F)$,the lamp is always off.

(C) Let the given statement be $S = (p \wedge \sim q) \rightarrow (p \vee \sim q)$.
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,$\sim S = (p \wedge \sim q) \wedge \sim (p \vee \sim q)$.
Using De Morgan's Law,$\sim (p \vee \sim q) = (\sim p \wedge \sim (\sim q)) = (\sim p \wedge q)$.
Thus,$\sim S = (p \wedge \sim q) \wedge (\sim p \wedge q)$.
By the associative and commutative properties,$\sim S = (p \wedge \sim p) \wedge (\sim q \wedge q)$.
Since $(p \wedge \sim p) = F$ (a contradiction) and $(\sim q \wedge q) = F$,we have $\sim S = F \wedge F = F$.
$A$ statement that is always false is called a contradiction.

(C) Let $p$: The work is finished on time.
Let $q$: The contractor is in trouble.
$Statement-I$ is $\sim p \rightarrow q$.
Using the logical equivalence $\sim p \rightarrow q \equiv p \vee q$,we get:
$Statement-I \equiv p \vee q$.
$Statement-II$ is given as: Either the work is finished on time or the contractor is in trouble,which is $p \vee q$.
Since $Statement-I \equiv p \vee q$ and $Statement-II \equiv p \vee q$,both statements are equivalent.

(C) The statement $p \vee \sim(q \wedge r)$ is false if and only if both components of the disjunction are false.
Therefore,$p = F$ and $\sim(q \wedge r) = F$.
Since $\sim(q \wedge r) = F$,it implies $(q \wedge r) = T$.
For the conjunction $(q \wedge r)$ to be true,both $q$ and $r$ must be true.
Thus,$q = T$ and $r = T$.
So,the truth values are $p = F, q = T, r = T$.

(D) The symbolic form of the given circuit is $(p \land q) \lor (p \land r)$.
By the distributive law of logic,$(p \land q) \lor (p \land r) \equiv p \land (q \lor r)$.
Comparing this with the given options:
Circuit $(i)$ is $(p \land q) \lor (p \land r)$.
Circuit $(iii)$ is $p \land (q \lor r)$.
Since $(p \land q) \lor (p \land r) \equiv p \land (q \lor r)$,the circuits $(i)$ and $(iii)$ are equivalent.

(B) The contrapositive of the statement $p \rightarrow (\sim q \wedge r)$ is $\sim (\sim q \wedge r) \rightarrow \sim p$.
By De Morgan's law,this is equivalent to $(q \vee \sim r) \rightarrow \sim p$.
The negation of the contrapositive is $\sim [(q \vee \sim r) \rightarrow \sim p]$.
Using the logical equivalence $\sim (A \rightarrow B) \equiv A \wedge \sim B$,we get $(q \vee \sim r) \wedge \sim (\sim p)$.
This simplifies to $(q \vee \sim r) \wedge p$.

(C) Let $p$ be the switch $S_1$ is closed,$q$ be the switch $S_2$ is closed,and $r$ be the switch $S_3$ is closed.
Then $\sim p$ represents the switch $S_1'$ is closed,$\sim q$ represents the switch $S_2'$ is closed,and $\sim r$ represents the switch $S_3'$ is closed.
The given statement is $(p \wedge q) \vee (\sim p \wedge (\sim q \vee p \vee r))$.
This represents two main branches connected in parallel:
$1$. The first branch is $(p \wedge q)$,which corresponds to switches $S_1$ and $S_2$ in series.
$2$. The second branch is $(\sim p \wedge (\sim q \vee p \vee r))$,which corresponds to switch $S_1'$ in series with a parallel combination of switches $S_2'$,$S_1$,and $S_3$.
Comparing this with the given options,the circuit diagram in option $C$ matches this description.

(C) Step $1$: Identify the definitions: $A$ contradiction is always false,a tautology is always true,and a contingency can be either true or false depending on the truth values of its components.
Step $2$: Evaluate option $(A)$: $p \wedge \sim p$ is always false (contradiction).
Step $3$: Evaluate option $(B)$: $p \vee \sim p$ is always true (tautology).
Step $4$: Evaluate option $(C)$: $p \vee q$ can be true or false depending on the truth values of $p$ and $q$. For example,if $p = F$ and $q = F$,then $p \vee q = F$. If $p = T$ and $q = F$,then $p \vee q = T$. Thus,it is a contingency.
Step $5$: Evaluate option $(D)$: $(p \wedge (p$ $\rightarrow q))$ $\rightarrow q$ is a tautology (Modus Ponens).
Final Answer: $C$)

(C) We use the logical equivalence $A \rightarrow B \equiv \sim A \vee B$ and De Morgan's Laws.
Given expression: $\sim[(p \vee \sim q) \rightarrow (p \wedge \sim q)]$
Applying the conditional law: $\sim[\sim(p \vee \sim q) \vee (p \wedge \sim q)]$
Applying De Morgan's law: $(p \vee \sim q) \wedge \sim(p \wedge \sim q)$
Applying De Morgan's law again: $(p \vee \sim q) \wedge (\sim p \vee q)$
Thus,the correct option is $C$.

(C) The implication $A \rightarrow B$ is false only when $A$ is true and $B$ is false.
Given $(p \wedge \sim q) \wedge (p \wedge r) \rightarrow \sim p \vee q \equiv F$.
This implies $(p \wedge \sim q) \wedge (p \wedge r) \equiv T$ and $\sim p \vee q \equiv F$.
From $\sim p \vee q \equiv F$,we get $\sim p \equiv F$ and $q \equiv F$,which means $p \equiv T$ and $q \equiv F$.
Now substitute these into the first part: $(T \wedge \sim F) \wedge (T \wedge r) \equiv T$.
$(T \wedge T) \wedge (T \wedge r) \equiv T$.
$T \wedge (T \wedge r) \equiv T$.
This requires $T \wedge r \equiv T$,which implies $r \equiv T$.
Thus,the truth values are $p = T, q = F, r = T$.

(D) Given expression: $[p \wedge (q \vee r)] \vee [\sim r \wedge \sim q \wedge p]$
$\equiv [p \wedge (q \vee r)] \vee [p \wedge (\sim r \wedge \sim q)]$ (Commutative Law)
$\equiv [p \wedge (q \vee r)] \vee [p \wedge \sim (r \vee q)]$ (De Morgan's Law)
$\equiv p \wedge [(q \vee r) \vee \sim (q \vee r)]$ (Distributive Law)
$\equiv p \wedge T$ (Complement Law)
$\equiv p$ (Identity Law)

(B) Given truth values: $p = F, q = T, r = F$.
For the first statement pattern $(p \wedge \sim q) \rightarrow r$:
$(F \wedge \sim T) \rightarrow F$
$= (F \wedge F) \rightarrow F$
$= F \rightarrow F$
$= T$.
For the second statement pattern $(p \vee q) \rightarrow r$:
$(F \vee T) \rightarrow F$
$= T \rightarrow F$
$= F$.
Thus,the truth values are $T$ and $F$ respectively.

(D) The converse of a conditional statement $A \rightarrow B$ is defined as $B \rightarrow A$.
Given the statement $[p \wedge (\sim q)] \rightarrow r$,the converse is $r \rightarrow [p \wedge (\sim q)]$.
However,looking at the provided options,there seems to be a misunderstanding in the original problem's solution logic.
If we consider the contrapositive of the converse,or if the question intended to ask for the inverse,the options provided do not match the standard definition of a converse.
Given the structure of option $D$,$r \rightarrow (p \wedge \sim q)$ is the correct converse.

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

(B) In logic,an implication $p \rightarrow q$ is false only when $p$ is true and $q$ is false. Otherwise,it is true.
$(A)$ Let $p: 3+3=7$ (False) and $q: 4+3=8$ (False). Since $p$ is false,$p \rightarrow q$ is true.
$(B)$ Let $p: 5+3=8$ (True) and $q: \text{the earth is flat}$ (False). Since $p$ is true and $q$ is false,$p \rightarrow q$ is false.
$(C)$ Let $p: \text{Both } (A) \text{ and } (B) \text{ are true}$ (False,because $B$ is false) and $q: 5+6=17$ (False). Since $p$ is false,$p \rightarrow q$ is true.
Therefore,$(A)$ and $(C)$ are true,while $(B)$ is false.

(A) Let $p$: Horses have wings.
Let $q$: Crows have tails.
The given statement is $p \leftrightarrow q$.
The negation of a biconditional statement is $\sim(p \leftrightarrow q) \equiv (p \wedge \sim q) \vee (q \wedge \sim p)$.
This translates to: 'Horses have wings and crows do not have tails,or crows have tails and horses do not have wings.'
Comparing this with the given options,option $A$ represents this logical equivalence.

(D) The inverse of a conditional statement $A \rightarrow B$ is defined as $\sim A \rightarrow \sim B$.
For the statement $p$ $\rightarrow (q$ $\rightarrow r)$,the inverse is $\sim p$ $\rightarrow \sim (q$ $\rightarrow r)$.
Using the property that $\sim (q \rightarrow r) \equiv q \wedge \sim r$,this does not immediately match the options.
However,let us re-examine the logical equivalence.
Given $p$ $\rightarrow (q$ $\rightarrow r)$,the inverse is $\sim p$ $\rightarrow \sim (q$ $\rightarrow r)$.
Using the contrapositive rule,$\sim A$ $\rightarrow \sim B \equiv B$ $\rightarrow A$.
Let $A = p$ and $B = (q \rightarrow r)$.
Then $\sim p$ $\rightarrow \sim (q$ $\rightarrow r) \equiv (q$ $\rightarrow r)$ $\rightarrow p$.
Thus,the inverse is logically equivalent to $(q$ $\rightarrow r)$ $\rightarrow p$.

(D) Let $p$ : The payment will be made.
Let $q$ : The work is finished in time.
The given statement is a biconditional statement: $p \leftrightarrow q$,which is equivalent to $(p \rightarrow q) \wedge (q \rightarrow p)$.
The negation of $(p \leftrightarrow q)$ is $\sim(p \leftrightarrow q)$,which is equivalent to $(p \wedge \sim q) \vee (q \wedge \sim p)$.
This translates to: "The payment is made and the work is not finished in time,or the work is finished in time and the payment is not made."
Therefore,Option $(D)$ is correct.

(C) The implication $A \rightarrow B$ is $False$ only when $A$ is $True$ and $B$ is $False$.
Given $(p \wedge \sim r) \rightarrow (\sim p \vee q) \equiv F$.
This implies $(p \wedge \sim r) \equiv T$ and $(\sim p \vee q) \equiv F$.
From $(\sim p \vee q) \equiv F$,we get $\sim p \equiv F$ and $q \equiv F$,which means $p \equiv T$ and $q \equiv F$.
Substituting $p \equiv T$ into $(p \wedge \sim r) \equiv T$,we get $(T \wedge \sim r) \equiv T$,which implies $\sim r \equiv T$,so $r \equiv F$.
Thus,the truth values are $p=T, q=F, r=F$.
Therefore,Option $(C)$ is correct.

(A) Let $p: 3$ is a prime number.
Let $q: 3$ is odd.
The given statement is $p \rightarrow q$.
The converse of a conditional statement $p \rightarrow q$ is $q \rightarrow p$.
Therefore,the converse is: "If $3$ is odd,then $3$ is a prime number."

(D) First,simplify the expression $((p \wedge q) \vee (p \vee \sim q))$.
Using the Associative and Commutative laws,we have $(p \wedge q) \vee (p \vee \sim q) \equiv p \vee (q \vee \sim q) \equiv p \vee T \equiv T$.
Now,substitute this back into the original expression:
$T \wedge (\sim p \wedge \sim q) \equiv \sim p \wedge \sim q$.
Thus,the expression is equivalent to $\sim p \wedge \sim q$.

(B) The inverse of a conditional statement $p \rightarrow q$ is defined as $\sim p \rightarrow \sim q$.
Given $p$: $A$ man is a judge,then $\sim p$: $A$ man is not a judge.
Given $q$: He is honest,then $\sim q$: He is not honest.
Therefore,the inverse is: If a man is not a judge,then he is not honest.

(D) Given $r: q \vee \sim p$.
Using the logical equivalence $p \rightarrow q \equiv \sim p \vee q$,we can rewrite the expression:
$q \vee \sim p \equiv \sim p \vee q$.
By the contrapositive rule,$\sim p \vee q \equiv \sim q \rightarrow \sim p$.
Translating this back into words:
$\sim q$: $X$ is not an isosceles triangle.
$\sim p$: $X$ is not an equilateral triangle.
Thus,$\sim q \rightarrow \sim p$ means: "If $X$ is not an isosceles triangle,then $X$ is not an equilateral triangle."
Therefore,Option $(D)$ is correct.

(D) We evaluate the logical expression using the laws of logic:
$(p$ $\rightarrow q)$ $\rightarrow ((\sim p$ $\rightarrow q)$ $\rightarrow q)$
$\equiv (p$ $\rightarrow q)$ $\rightarrow ((p \vee q)$ $\rightarrow q)$
$\equiv (p$ $\rightarrow q)$ $\rightarrow (\sim(p \vee q) \vee q)$
$\equiv (p$ $\rightarrow q)$ $\rightarrow ((\sim p \wedge \sim q) \vee q)$
$\equiv (p$ $\rightarrow q)$ $\rightarrow ((\sim p \vee q) \wedge (\sim q \vee q))$
$\equiv (p$ $\rightarrow q)$ $\rightarrow ((\sim p \vee q) \wedge T)$
$\equiv (p$ $\rightarrow q)$ $\rightarrow (\sim p \vee q)$
$\equiv (p$ $\rightarrow q)$ $\rightarrow (p$ $\rightarrow q)$
$\equiv T$
Since the result is always true,the statement is a tautology.

(C) Let $p$ be the statement: 'Two numbers are not equal'.
Let $q$ be the statement: 'Their squares are not equal'.
The given statement is $p \rightarrow q$.
The contrapositive of $p \rightarrow q$ is defined as $\sim q \rightarrow \sim p$.
Here,$\sim q$ is: 'The squares of two numbers are equal'.
And $\sim p$ is: 'The numbers are equal'.
Therefore,the contrapositive is: 'If the squares of two numbers are equal,then the numbers are equal'.

(C) The contrapositive of a conditional statement $p \rightarrow q$ is defined as $\sim q \rightarrow \sim p$.
Given statement $p$: 'Two numbers are equal' and $q$: 'Their squares are equal'.
Therefore,the contrapositive $\sim q \rightarrow \sim p$ is: 'If the squares of two numbers are not equal,then the numbers are not equal'.

(C) First,determine the truth values of the statements:
$p$: The prime numbers between $2$ and $100$ are $25$,not $26$. So,$p$ is $F$.
$q$: Zero $(0)$ can be written as $0 + 0i$,which is a complex number. So,$q$ is $T$.
$r$: The $L$.$C$.$M$. of $6$ and $7$ is $42$,not $6$. So,$r$ is $F$.
Now evaluate the options:
$(A)$ $(p \wedge q)$ $\rightarrow r \equiv (F \wedge T)$ $\rightarrow F \equiv F$ $\rightarrow F \equiv T$.
$(B)$ $(p$ $\rightarrow q)$ $\rightarrow r \equiv (F$ $\rightarrow T)$ $\rightarrow F \equiv T$ $\rightarrow F \equiv F$.
$(C)$ $(p \vee q) \leftrightarrow r \equiv (F \vee T) \leftrightarrow F \equiv T \leftrightarrow F \equiv F$.
$(D)$ $(p$ $\rightarrow q)$ $\rightarrow (q$ $\rightarrow p) \equiv (F$ $\rightarrow T)$ $\rightarrow (T$ $\rightarrow F) \equiv T$ $\rightarrow F \equiv F$.
Thus,option $(C)$ is correct.

(B) The inverse of a conditional statement $A \rightarrow B$ is $\sim A \rightarrow \sim B$.
For the statement $p$ $\rightarrow (p$ $\rightarrow q)$,the inverse is $\sim p$ $\rightarrow \sim (p$ $\rightarrow q)$.
The contrapositive of a conditional statement $A \rightarrow B$ is $\sim B \rightarrow \sim A$.
Applying this to the inverse $\sim p$ $\rightarrow \sim (p$ $\rightarrow q)$,we get:
$\sim [\sim (p$ $\rightarrow q)]$ $\rightarrow \sim (\sim p)$
$= (p$ $\rightarrow q)$ $\rightarrow p$
Since $p \rightarrow q \equiv \sim p \vee q$,the expression becomes $(\sim p \vee q) \rightarrow p$.

(B) Using the distributive law: $(\sim p) \vee (p \wedge \sim q) \equiv (\sim p \vee p) \wedge (\sim p \vee \sim q)$.
Since $(\sim p \vee p) \equiv T$ (Tautology),the expression becomes $T \wedge (\sim p \vee \sim q)$.
By the identity law,$T \wedge (\sim p \vee \sim q) \equiv \sim p \vee \sim q$.
Using the logical equivalence $p \rightarrow q \equiv \sim p \vee q$,we can rewrite $\sim p \vee \sim q$ as $p \rightarrow (\sim q)$.

(A) To determine the nature of the statement $\sim(p \leftrightarrow \sim q)$,we construct a truth table:
| $p$ | $q$ | $\sim q$ | $p \leftrightarrow \sim q$ | $\sim(p \leftrightarrow \sim q)$ | $p \leftrightarrow q$ |
| :--- | :--- | :--- | :--- | :--- | :--- |
| $T$ | $T$ | $F$ | $F$ | $T$ | $T$ |
| $T$ | $F$ | $T$ | $T$ | $F$ | $F$ |
| $F$ | $T$ | $F$ | $T$ | $F$ | $F$ |
| $F$ | $F$ | $T$ | $F$ | $T$ | $T$ |
Comparing column $5$ and column $6$,we observe that the truth values are identical for all combinations of $p$ and $q$.
Therefore,$\sim(p \leftrightarrow \sim q) \equiv p \leftrightarrow q$.

(A) Given that the truth value of $p \rightarrow r$ is $F$,we must have $p \equiv T$ and $r \equiv F$.
Given that the truth value of $p \leftrightarrow q$ is $F$,and we know $p \equiv T$,it follows that $q \equiv F$.
Now,evaluate the first expression: $(\sim p \vee q) \rightarrow (p \vee \sim q)$
$= (\sim T \vee F) \rightarrow (T \vee \sim F)$
$= (F \vee F) \rightarrow (T \vee T)$
$= F \rightarrow T \equiv T$.
Now,evaluate the second expression: $(p \wedge \sim q) \rightarrow (\sim p \wedge q)$
$= (T \wedge \sim F) \rightarrow (\sim T \wedge F)$
$= (T \wedge T) \rightarrow (F \wedge F)$
$= T \rightarrow F \equiv F$.
Thus,the truth values are $T, F$.

(C) To find the negation of the statement $(p \wedge q) \rightarrow (\sim p \vee r)$,we use the logical equivalence $\sim(A \rightarrow B) \equiv A \wedge \sim B$.
Let $A = (p \wedge q)$ and $B = (\sim p \vee r)$.
Then,$\sim(A \rightarrow B) \equiv (p \wedge q) \wedge \sim(\sim p \vee r)$.
Applying De Morgan's Law,$\sim(\sim p \vee r) \equiv \sim(\sim p) \wedge \sim r \equiv p \wedge \sim r$.
So,the expression becomes $(p \wedge q) \wedge (p \wedge \sim r)$.
By the Associative and Idempotent Laws,$(p \wedge q) \wedge p \wedge \sim r \equiv p \wedge q \wedge \sim r$.

(B) Given: $p = T, q = T, r = F, s = F$.
$a: \sim(p \wedge \sim r) \vee(\sim q \vee s)$
$\equiv \sim(T \wedge \sim F) \vee(\sim T \vee F)$
$\equiv \sim(T \wedge T) \vee(F \vee F)$
$\equiv \sim(T) \vee(F)$
$\equiv F \vee F \equiv F$
$b: (\sim q \wedge \sim r) \leftrightarrow(p \vee s)$
$\equiv (\sim T \wedge \sim F) \leftrightarrow(T \vee F)$
$\equiv (F \wedge T) \leftrightarrow(T)$
$\equiv F \leftrightarrow T \equiv F$
$c: (\sim p \vee q) \rightarrow(r \wedge \sim s)$
$\equiv (\sim T \vee T) \rightarrow(F \wedge \sim F)$
$\equiv (F \vee T) \rightarrow(F \wedge T)$
$\equiv T \rightarrow F \equiv F$
Thus,the truth values are $F, F, F$.

(C) We analyze the logical expression $[p$ $\rightarrow (q$ $\rightarrow r)] \leftrightarrow [(p \wedge q)$ $\rightarrow r]$.
First,consider the left side: $p$ $\rightarrow (q$ $\rightarrow r)$.
Using the implication law $A \rightarrow B \equiv \neg A \vee B$,we get:
$p$ $\rightarrow (q$ $\rightarrow r) \equiv \neg p \vee (\neg q \vee r)$.
By the associative law,this is equivalent to $(\neg p \vee \neg q) \vee r$.
By De Morgan's law,$\neg p \vee \neg q \equiv \neg (p \wedge q)$.
Thus,$p$ $\rightarrow (q$ $\rightarrow r) \equiv \neg (p \wedge q) \vee r$.
Again,using the implication law,$\neg (p \wedge q) \vee r \equiv (p \wedge q) \rightarrow r$.
Since both sides of the biconditional are logically equivalent,the expression $[p$ $\rightarrow (q$ $\rightarrow r)] \leftrightarrow [(p \wedge q)$ $\rightarrow r]$ is always true.
Therefore,it is a tautology.

(A) We need to find the negation of the statement $\sim s \vee (\sim r \wedge s)$.
Applying the negation operator:
$\sim (\sim s \vee (\sim r \wedge s))$
Using De Morgan's law,$\sim (p \vee q) \equiv \sim p \wedge \sim q$:
$\equiv s \wedge \sim (\sim r \wedge s)$
Using De Morgan's law again,$\sim (p \wedge q) \equiv \sim p \vee \sim q$:
$\equiv s \wedge (r \vee \sim s)$
Using the Distributive law,$p \wedge (q \vee r) \equiv (p \wedge q) \vee (p \wedge r)$:
$\equiv (s \wedge r) \vee (s \wedge \sim s)$
Using the Complement law,$s \wedge \sim s \equiv F$:
$\equiv (s \wedge r) \vee F$
Using the Identity law,$p \vee F \equiv p$:
$\equiv s \wedge r$

(D) Given that $p = T$,$q = T$,$r = F$,and $s = F$.
For the first statement pattern $(p \wedge q) \vee r$:
$(T \wedge T) \vee F \equiv T \vee F \equiv T$.
For the second statement pattern $(p \vee s) \leftrightarrow (q \wedge r)$:
$(T \vee F) \leftrightarrow (T \wedge F) \equiv T \leftrightarrow F \equiv F$.
Thus,the truth values are $T$ and $F$ respectively.

(B) We use the logical equivalence $p \rightarrow q \equiv \sim p \vee q$.
$p \rightarrow \sim(p \wedge \sim q)$
$\equiv \sim p \vee \sim(p \wedge \sim q)$
$\equiv \sim p \vee (\sim p \vee \sim(\sim q))$
$\equiv \sim p \vee (\sim p \vee q)$
$\equiv (\sim p \vee \sim p) \vee q$
$\equiv \sim p \vee q$

(C) The implication $(p \leftrightarrow \sim q) \rightarrow (\sim p \wedge q)$ is false only when the antecedent is true and the consequent is false.
So,$(p \leftrightarrow \sim q) \equiv T$ and $(\sim p \wedge q) \equiv F$.
From $(\sim p \wedge q) \equiv F$,we know that the conjunction is false.
Now,consider the antecedent $(p \leftrightarrow \sim q) \equiv T$. This means $p$ and $\sim q$ must have the same truth value.
If we test option $C$ $(p=T, q=F)$:
$\sim q = \sim F = T$.
Then $(p \leftrightarrow \sim q) = (T \leftrightarrow T) = T$.
And $(\sim p \wedge q) = (\sim T \wedge F) = (F \wedge F) = F$.
Since the antecedent is $T$ and the consequent is $F$,the implication is $T \rightarrow F = F$.
Thus,the truth values are $p=T$ and $q=F$.

(D) Let $p$ be the statement: "$x$ and $y$ are integers such that $x y$ is odd".
Let $q$ be the statement: "both $x$ and $y$ are odd".
The given statement is $p \rightarrow q$.
The contrapositive of $p \rightarrow q$ is $\sim q \rightarrow \sim p$.
Here,$\sim q$ is "it is not the case that both $x$ and $y$ are odd",which means "$x$ and $y$ are not both odd".
$\sim p$ is "$x y$ is not odd".
Therefore,the contrapositive is "If $x$ and $y$ are not both odd,then $x y$ is not odd".
Thus,option $D$ is correct.