Mathematical logic Questions in English

(D) statement is a declarative sentence that is either true or false,but not both.
$(a)$,$(b)$,and $(c)$ are imperative sentences (commands),which are not statements.
$(d)$ $2 + 2 = 4$ is a declarative sentence that is true,hence it is a statement.

(C) statement in logic is a declarative sentence that is either true or false,but not both.
$(a)$ $May \text{ you live long!}$ is an optative sentence.
$(b)$ $May \text{ God bless you!}$ is an optative sentence.
$(c)$ $The \text{ sun is a star.}$ is a declarative sentence that is true,hence it is a statement.
$(d)$ $Hurrah! \text{ We have won the match.}$ is an exclamatory sentence.
Therefore,the correct option is $C$.

(D) statement in logic is a declarative sentence that is either true or false,but not both.
$(a)$ $Roses \text{ are red}$ is a declarative sentence (true).
$(b)$ $New \text{ Delhi is in India}$ is a declarative sentence (true).
$(c)$ $Every \text{ square is a rectangle}$ is a declarative sentence (true).
$(d)$ $Alas! I have failed$ is an exclamatory sentence,which expresses strong emotion and cannot be classified as true or false. Therefore,it is not a statement.

(C) statement is a declarative sentence that is either true or false,but not both.
$(a)$ "Every set is a finite set" is a false statement.
$(b)$ "$8$ is less than $6$" is a false statement.
$(c)$ "Where are you going?" is an interrogative sentence,not a declarative one,so it is not a statement.
$(d)$ "The sum of interior angles of a triangle is $180^{\circ}$" is a true statement.
Therefore,the correct option is $C$.

(A) statement in logic is a declarative sentence that is either true or false,but not both.
$(A)$ "Please do me a favour" is an imperative sentence (a request),which cannot be classified as true or false. Therefore,it is not a statement.
$(B)$ "$2$ is an even integer" is a true statement.
$(C)$ "$2 + 1 = 3$" is a true statement.
$(D)$ "The number $17$ is prime" is a true statement.
Thus,the correct option is $A$.

(A) statement in logic is a declarative sentence that is either true or false,but not both.
$(A)$ "Give me a glass of water" is an imperative sentence (a request),not a declarative sentence,so it is not a statement.
$(B)$ "Asia is a continent" is a declarative sentence that is true.
$(C)$ "The earth revolves round the sun" is a declarative sentence that is true.
$(D)$ "The number $6$ has two prime factors $2, 3$" is a declarative sentence that is true.
Therefore,the correct option is $A$.

(A) An open statement is a declarative sentence containing one or more variables such that it becomes a true or false statement depending on the values assigned to the variables.
$(a)$ "$x$ is a natural number" is an open statement because its truth value depends on the value of $x$.
$(b)$,$(c)$,and $(d)$ are imperative or exclamatory sentences,which are not considered statements in mathematical logic.

(B) Let $p$ : Paris is in France.
Let $q$ : London is in England.
The given statement is $p \wedge q$.
The negation of a conjunction is given by De Morgan's Law: $\sim (p \wedge q) = \sim p \vee \sim q$.
Here,$\sim p$ is "Paris is not in France" and $\sim q$ is "London is not in England".
Therefore,the negation is "Paris is not in France or London is not in England".

(C) Let $p$ be the statement "$2 + 3 = 5$" and $q$ be the statement "$8 < 10$".
The given compound statement is $p \wedge q$.
The negation of a conjunction is given by De Morgan's Law: $\sim(p \wedge q) = \sim p \vee \sim q$.
Here,$\sim p$ is "$2 + 3 \neq 5$" and $\sim q$ is "$8 \nless 10$".
Therefore,the negation is "$2 + 3 \neq 5$ or $8 \nless 10$".

(D) Let $p$: Ram is in Class $X$.
Let $q$: Rashmi is in Class $XII$.
The given statement is $p \vee q$.
The negation of a disjunction is given by De Morgan's Law: $\sim (p \vee q) = \sim p \wedge \sim q$.
Thus,the negation is: "Ram is not in Class $X$ $AND$ Rashmi is not in Class $XII$".
Comparing this with the given options,none of the options $A$,$B$,or $C$ match this result.
Therefore,the correct option is $D$.

(A) To determine if the statement $(p \wedge q) \implies p$ is a tautology,we construct a truth table:

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

Since the final column contains only $T$ (True) values for all possible combinations of truth values of $p$ and $q$,the statement is a tautology.

(A) statement is a contradiction if its truth value is $F$ (False) for all possible truth values of its components.
Let us construct the truth table for $(p \wedge q) \wedge \sim (p \vee q)$:

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

Since the final column contains only $F$,the expression $(p \wedge q) \wedge \sim (p \vee q)$ is a contradiction.

(A) According to De Morgan's Law in mathematical logic,the negation of a conjunction is the disjunction of the negations.
Therefore,$\sim (p \wedge q) \equiv \sim p \vee \sim q$.

(B) By the Law of Double Negation,we know that $\sim (\sim p) \equiv p$.
Therefore,the given expression $(\sim (\sim p)) \wedge q$ simplifies to $p \wedge q$.

(B) Using De Morgan's Law,$\sim (p \vee r) \equiv \sim p \wedge \sim r$.
Applying this to the given expression:
$\sim (p \vee (\sim q)) \equiv \sim p \wedge \sim (\sim q)$.
Since $\sim (\sim q) \equiv q$,the expression simplifies to:
$\sim p \wedge q$.

(A) Using De Morgan's Law,$\sim (A \wedge B) \equiv (\sim A) \vee (\sim B)$.
Applying this to the given expression:
$\sim ((\sim p) \wedge q) \equiv \sim (\sim p) \vee (\sim q)$.
Since $\sim (\sim p) \equiv p$,the expression simplifies to:
$p \vee (\sim q)$.

(C) The biconditional statement $p \Leftrightarrow q$ is defined as $(p$ $\Rightarrow q) \wedge (q$ $\Rightarrow p)$.
Since $p \Rightarrow q \equiv \sim p \vee q$,we have $p \Leftrightarrow q \equiv (\sim p \vee q) \wedge (\sim q \vee p)$.
Taking the negation: $\sim (p \Leftrightarrow q) \equiv \sim ((\sim p \vee q) \wedge (\sim q \vee p))$.
Using De Morgan's law: $\sim (p \Leftrightarrow q) \equiv \sim (\sim p \vee q) \vee \sim (\sim q \vee p)$.
This simplifies to $(p \wedge \sim q) \vee (q \wedge \sim p)$.

(B) The logical implication $p \Rightarrow q$ is equivalent to the disjunction $\sim p \vee q$.
This is a standard logical equivalence where the conditional statement is false only when $p$ is true and $q$ is false,which matches the truth table for $\sim p \vee q$.

(A) Given the truth values: $p = T, q = F, r = T$.
First,evaluate $\sim p$: $\sim T = F$.
Next,evaluate $\sim r$: $\sim T = F$.
Now,evaluate $(\sim p \vee q)$: $F \vee F = F$.
Then,evaluate $(\sim p \vee q) \wedge \sim r$: $F \wedge F = F$.
Finally,evaluate the implication $[(\sim p \vee q) \wedge \sim r] \Rightarrow p$: $F \Rightarrow T$.
Since the implication $F \Rightarrow T$ is $True$,the final truth value is $True$.

(A) The implication $(p \wedge \sim r) \Rightarrow (q \vee r)$ is false only when the antecedent $(p \wedge \sim r)$ is true and the consequent $(q \vee r)$ is false.
Given that $q$ and $r$ are both false,the consequent $(q \vee r)$ becomes $(F \vee F) = F$,which is consistent with the implication being false.
For the antecedent $(p \wedge \sim r)$ to be true,both $p$ and $\sim r$ must be true.
Since $r$ is false,$\sim r$ is true.
Therefore,$p \wedge T = T$ implies that $p$ must be true.

(B) The given logical expression is $(p \wedge q) \wedge (q \wedge r) = T$.
For the conjunction of two statements to be true,both components must be true.
Therefore,$(p \wedge q) = T$ and $(q \wedge r) = T$.
For $(p \wedge q) = T$,both $p$ and $q$ must be true.
For $(q \wedge r) = T$,both $q$ and $r$ must be true.
Since $q$ is true in both cases,and $p$ and $r$ must also be true,it follows that $p, q, r$ are all true.

(C) The given statement is $\sim (p \Rightarrow q) \Leftrightarrow \sim p \vee \sim q$.
By the laws of logic,$\sim (p \Rightarrow q) \equiv p \wedge \sim q$.
Thus,the expression becomes $(p \wedge \sim q) \Leftrightarrow (\sim p \vee \sim q)$.
Constructing the truth table for the expression:

$p$	$q$	$p \Rightarrow q$	$\sim (p \Rightarrow q)$	$\sim p$	$\sim q$	$\sim p \vee \sim q$	$\sim (p \Rightarrow q) \Leftrightarrow (\sim p \vee \sim q)$
$T$	$T$	$T$	$F$	$F$	$F$	$F$	$T$
$T$	$F$	$F$	$T$	$F$	$T$	$T$	$T$
$F$	$T$	$T$	$F$	$T$	$F$	$T$	$F$
$F$	$F$	$T$	$F$	$T$	$T$	$T$	$F$

Since the final column contains both $T$ and $F$,the statement is neither a tautology nor a contradiction. Therefore,the correct option is $C$.

(A) The given expression is $(p \wedge \sim q) \wedge (\sim p \vee q)$.
We construct the truth table to evaluate the expression:

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

Since the final column contains only $F$ (False) for all possible truth values of $p$ and $q$,the statement is a contradiction.

(B) Let $P$ be the statement "$A$ number is real" and $Q$ be the statement "$A$ number is rational or irrational".
The given proposition is $P \implies Q$.
The contrapositive of $P \implies Q$ is $\neg Q \implies \neg P$.
Here,$\neg Q$ is "$A$ number is neither rational nor irrational" and $\neg P$ is "$A$ number is not real".
Thus,the contrapositive is: "If a number is neither rational nor irrational,then it is not real".
Option $D$ is the original statement $P \implies Q$.
Option $C$ is the contrapositive $\neg Q \implies \neg P$.
Option $A$ is also equivalent to the contrapositive.
Option $B$ states: "If a number is not rational or not irrational,then it is not real". This is logically incorrect because a number can be irrational (not rational) and still be real,or rational (not irrational) and still be real. Thus,$B$ is not logically equivalent.

(A) Given propositions are:
$p$: It rains today
$q$: $I$ go to school
$r$: $I$ shall meet my friend
$s$: $I$ shall go for a movie
The statement is: "If it does not rain or if $I$ do not go to school,then $I$ shall meet my friend and go for a movie."
"It does not rain" is $\sim p$.
"$I$ do not go to school" is $\sim q$.
"It does not rain or if $I$ do not go to school" is $(\sim p \vee \sim q)$.
By De Morgan's Law,$(\sim p \vee \sim q) \equiv \sim (p \wedge q)$.
"$I$ shall meet my friend and go for a movie" is $(r \wedge s)$.
Thus,the proposition is $\sim (p \wedge q) \Rightarrow (r \wedge s)$.
Therefore,the correct option is $A$.

(D) The given proposition is $S = p \vee (\sim p \vee q)$.
We need to find the negation $\sim S = \sim [p \vee (\sim p \vee q)]$.
Using De Morgan's Law,$\sim (A \vee B) \equiv \sim A \wedge \sim B$,we get:
$\sim S \equiv \sim p \wedge \sim (\sim p \vee q)$.
Applying De Morgan's Law again to the second part,$\sim (\sim p \vee q) \equiv \sim (\sim p) \wedge \sim q \equiv p \wedge \sim q$.
Therefore,$\sim S \equiv \sim p \wedge (p \wedge \sim q)$.
Since the conjunction is associative,this is equivalent to $(\sim p \wedge p) \wedge \sim q$.
Since $\sim p \wedge p$ is a contradiction $(F)$,the expression simplifies to $F \wedge \sim q$,which is $F$ (False).
Looking at the options,none of the expressions provided are logically equivalent to $F$. Thus,the correct option is $D$.

(C) We know that the negation of an implication is given by $\sim (p \Rightarrow q) \equiv p \wedge \sim q$.
Applying this to the expression in option $C$:
$\sim (\sim p \Rightarrow \sim q) \equiv (\sim p) \wedge \sim (\sim q)$
$\equiv \sim p \wedge q$.
Therefore,the statement in option $C$ is true.

(A) Using De Morgan's Law,$\sim (p \vee q) \equiv (\sim p \wedge \sim q)$.
Therefore,the expression becomes $(\sim p \wedge \sim q) \vee (\sim p \wedge q)$.
By the Distributive Law,we can factor out $\sim p$:
$\sim p \wedge (\sim q \vee q)$.
Since $(\sim q \vee q) \equiv T$ (a tautology),
$\sim p \wedge T \equiv \sim p$.

(A) In the given circuit,switch $p$ is in series with a parallel combination of two switches labeled $q$,and this whole combination is in series with switch $r$.
For the current to flow,the circuit must be complete.
This requires switch $p$ to be closed,at least one of the switches $q$ to be closed,and switch $r$ to be closed.
Therefore,the condition for current to flow is $p \land (q \lor q) \land r$,which simplifies to $p \land q \land r$.
Thus,$p, q,$ and $r$ must all be closed.

(B) To find the negation of the statement $q \vee \sim (p \wedge r)$,we apply De Morgan's Law: $\sim (A \vee B) = \sim A \wedge \sim B$.
Applying this,we get: $\sim (q \vee \sim (p \wedge r)) = \sim q \wedge \sim (\sim (p \wedge r))$.
Using the law of double negation $\sim (\sim P) = P$,we simplify the expression: $\sim q \wedge (p \wedge r)$.
Thus,the correct option is $B$.

(C) The logical equivalence $(p \Rightarrow q) \equiv (\sim p \vee q)$ is a standard identity.
Applying the negation operator to both sides:
$\sim (p \Rightarrow q) \equiv \sim (\sim p \vee q)$
Using De Morgan's Law,$\sim (A \vee B) \equiv (\sim A \wedge \sim B)$:
$\sim (p \Rightarrow q) \equiv (\sim (\sim p) \wedge \sim q)$
Since $\sim (\sim p) \equiv p$,we get:
$\sim (p \Rightarrow q) \equiv (p \wedge \sim q)$
Both options $A$,$C$,and $D$ are logically valid identities. However,in standard logic textbooks,these are all tautologies. Given the structure,$C$ is a fundamental definition of the negation of a conditional statement.

(C) The contrapositive of an implication $P \Rightarrow Q$ is defined as $\sim Q \Rightarrow \sim P$.
Given the implication $(p \vee q) \Rightarrow r$,we identify $P = (p \vee q)$ and $Q = r$.
Applying the rule,the contrapositive is $\sim r \Rightarrow \sim (p \vee q)$.
Using De Morgan's Law,$\sim (p \vee q)$ is equivalent to $(\sim p \wedge \sim q)$.
Therefore,the contrapositive is $\sim r \Rightarrow (\sim p \wedge \sim q)$.

(B) The statement $(p$ $\Rightarrow q) \Leftrightarrow (\sim q$ $\Rightarrow \;\sim p)$ is a tautology because $p \Rightarrow q$ is logically equivalent to its contrapositive $\sim q \Rightarrow \;\sim p$.
Since it is a tautology,it cannot be a contradiction.
Therefore,the statement in option $B$ is false.

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

(C) statement is a declarative sentence that is either true or false,but not both.
Option $(A)$ is a false statement.
Option $(B)$ is a true statement.
Option $(D)$ is a false statement.
Option $(C)$ 'Mathematics is interesting' is not a statement because it is subjective; it may be interesting for some people and not for others. Therefore,it cannot be classified as strictly true or false.

(B) Given expression: $(p \wedge \sim q) \wedge (\sim p \wedge q)$
By using the associative and commutative laws of logic,we can rearrange the terms:
$= (p \wedge \sim p) \wedge (\sim q \wedge q)$
We know that for any proposition $p$,$(p \wedge \sim p) = F$ (a contradiction).
$= F \wedge F = F$
Since the final truth value is always $F$ (False),the statement is a contradiction.

(D) The logical implication $q \to p$ is equivalent to $\sim q \vee p$.
Taking the negation of both sides,we get $\sim (q \to p) \equiv \sim (\sim q \vee p)$.
By De Morgan's Law,$\sim (\sim q \vee p) \equiv (\sim \sim q) \wedge (\sim p)$.
This simplifies to $q \wedge \sim p$,which is equivalent to $\sim p \wedge q$.
Therefore,$\sim p \wedge q \equiv \sim (q \to p)$.

(B) Let $p$ be the statement: "$A$ number is a prime."
Let $q$ be the statement: "It is odd."
The given proposition is $p \Rightarrow q$.
The inverse of a conditional statement $p \Rightarrow q$ is defined as $\sim p \Rightarrow \sim q$.
Here,$\sim p$ is: "$A$ number is not a prime."
$\sim q$ is: "It is not odd."
Therefore,the inverse is: "If a number is not a prime then it is not odd."
Thus,the correct option is $B$.

(D) Let $p$,$q$,and $r$ be the following statements:
$p$: $A$ number is divisible by $15$.
$q$: $A$ number is divisible by $5$.
$r$: $A$ number is divisible by $3$.
The given statement is of the form $p \rightarrow (q \wedge r)$.
The negation of an implication $p \rightarrow S$ is $p \wedge (\sim S)$.
Therefore,the negation is $p \wedge \sim(q \wedge r)$.
Using De Morgan's Law,$\sim(q \wedge r) \equiv (\sim q \vee \sim r)$.
Thus,the negation is $p \wedge (\sim q \vee \sim r)$,which means: "$A$ number is divisible by $15$ and it is not divisible by $5$ or it is not divisible by $3$."
However,looking at the logical structure provided in the options,the standard negation of "If $p$ then $q$ and $r$" is "$p$ and not ($q$ and $r$)",which is "$p$ and (not $q$ or not $r$)".
Given the options provided,option $D$ is the intended answer representing the logical negation $p \wedge \sim(q \wedge r)$.

(B) The statement $p$ $\rightarrow (q$ $\rightarrow p)$ is a tautology because it is only false if $p$ is true and $(q \rightarrow p)$ is false,which is impossible since $p$ is true.
Now,let us evaluate $p \rightarrow (p \vee q)$.
This statement is false only if $p$ is true and $(p \vee q)$ is false.
If $p$ is true,then $(p \vee q)$ must be true (by the definition of disjunction).
Thus,$p \rightarrow (p \vee q)$ is also a tautology.
Since both statements are tautologies,they are logically equivalent.
Therefore,$p$ $\rightarrow (q$ $\rightarrow p) \equiv p$ $\rightarrow (p \vee q)$.

(B) statement in mathematical logic is a declarative sentence that is either true or false,but not both.
$(A)$ 'Ahmedabad is situated on the banks of the Sabarmati river' is a true statement.
$(B)$ 'Tomorrow is a holiday' is not a statement because its truth value depends on the context of 'tomorrow',which is ambiguous.
$(C)$ 'For non-zero numbers $x$ and $y$,$x^2 + y^2 \neq 0$' is a true statement.
$(D)$ 'Pythagoras was a mathematician' is a true statement.
Therefore,option $B$ is not a statement.

(B) The statement $(p$ $\rightarrow q) \equiv (\sim q$ $\rightarrow \sim p)$ is the contrapositive of the implication,which is logically equivalent.
Using the definition of implication $p \rightarrow q \equiv \sim p \vee q$:
For the contrapositive: $\sim q \rightarrow \sim p \equiv \sim (\sim q) \vee \sim p$
$\equiv q \vee \sim p$
$\equiv \sim p \vee q$
$\equiv p \rightarrow q$
Thus,$(p$ $\rightarrow q) \equiv (\sim q$ $\rightarrow \sim p)$ is true.

(B) For Statement-$I$: $(p \wedge \sim q) \wedge (\sim p \wedge q) = p \wedge (\sim q \wedge \sim p) \wedge q = (p \wedge \sim p) \wedge (\sim q \wedge q) = F \wedge F = F$. Since the result is always false,it is a contradiction. Thus,Statement-$I$ is true.
For Statement-$II$: The expression $(p$ $\rightarrow q) \Leftrightarrow (\sim q$ $\rightarrow \sim p)$ represents the equivalence between a conditional statement and its contrapositive. Since a conditional statement is always logically equivalent to its contrapositive,this is a tautology. Thus,Statement-$II$ is true.
Since Statement-$II$ does not explain why Statement-$I$ is a contradiction,option $B$ is correct.

(A) Given $p$: $x$ is irrational,so $\sim p$: $x$ is rational.
Given $q$: $y$ is transcendental.
Statement $r$ is '$x$ is rational or $y$ is transcendental',which is $\sim p \lor q$.
Let's evaluate the truth tables for the given statements:

$p$	$q$	$\sim p$	$\sim q$	$r = (\sim p \lor q)$	$q \lor p$	$(p \Leftrightarrow \sim q)$
$T$	$T$	$F$	$F$	$T$	$T$	$F$
$T$	$F$	$F$	$T$	$F$	$T$	$T$
$F$	$T$	$T$	$F$	$T$	$T$	$T$
$F$	$F$	$T$	$T$	$T$	$F$	$F$

Comparing the columns:
$r$ is $(\sim p \lor q)$.
Statement-$1$: $r \equiv q \lor p$. From the table,$r$ and $q \lor p$ are not equivalent. So,Statement-$1$ is false.
Statement-$2$: $r \equiv (p \Leftrightarrow \sim q)$. From the table,$r$ and $(p \Leftrightarrow \sim q)$ are not equivalent. So,Statement-$2$ is false.
Wait,re-evaluating the logic: $r$ is $\sim p \lor q$.
$q \lor p$ is not $\sim p \lor q$.
$(p \Leftrightarrow \sim q)$ is true when $p$ and $\sim q$ have the same truth value,i.e.,$p$ and $q$ have opposite truth values.
Thus,both statements are false.

(D) By definition,the biconditional statement $p \Leftrightarrow q$ is defined as the conjunction of the conditional statements $p \Rightarrow q$ and $q \Rightarrow p$.
Thus,$p \Leftrightarrow q \equiv (p$ $\Rightarrow q) \wedge (q$ $\Rightarrow p)$.

(C) Let $p : ab = 0$,$q : a = 0$,and $r : b = 0$.
The given statement is $p \Rightarrow (q \lor r)$.
The contrapositive of $p \Rightarrow (q \lor r)$ is $\sim (q \lor r) \Rightarrow \sim p$.
Using De Morgan's Law,$\sim (q \lor r) \equiv (\sim q \land \sim r)$.
Thus,the contrapositive is $(\sim q \land \sim r) \Rightarrow \sim p$.
Substituting the values,we get: If $(a \neq 0 \text{ and } b \neq 0)$,then $ab \neq 0$.

(A) Let $p, q, r$ be the statements such that $p: x = 5$,$q: y = -2$,and $r: x - 2y = 9$.
The given statement is $(p \wedge q) \rightarrow r$.
The contrapositive of a conditional statement $A \rightarrow B$ is $\sim B \rightarrow \sim A$.
Here,the contrapositive is $\sim r \rightarrow \sim (p \wedge q)$.
By De Morgan's Law,$\sim (p \wedge q) \equiv \sim p \vee \sim q$.
Thus,the contrapositive is $\sim r \rightarrow (\sim p \vee \sim q)$,which translates to: If $x - 2y \neq 9$,then $x \neq 5$ or $y \neq -2$.

(D) $1$. $p \rightarrow q \equiv \sim p \vee q$ is a standard logical equivalence.
$2$. For $(p \vee q) \wedge (q \vee r)$ to be true,both $(p \vee q)$ and $(q \vee r)$ must be true. If $p=T, q=F, r=T$,then $(T \vee F) \wedge (F \vee T) = T \wedge T = T$. This is correct.
$3$. By De Morgan's Law,$\sim (p \wedge (q \vee r)) \equiv \sim p \vee \sim (q \vee r) \equiv \sim p \vee (\sim q \wedge \sim r) \equiv (\sim p \vee \sim q) \wedge (\sim p \vee \sim r)$. This is correct.
$4$. For $p \wedge \sim (p \vee q)$,we have $p \wedge (\sim p \wedge \sim q) \equiv (p \wedge \sim p) \wedge \sim q \equiv F \wedge \sim q \equiv F$. Thus,the statement that it is always $T$ is false.

(D) We evaluate the statement $((p \wedge p)$ $\rightarrow q)$ $\rightarrow p$.
Since $p \wedge p \equiv p$,the expression simplifies to $(p$ $\rightarrow q)$ $\rightarrow p$.
If $p$ is $F$ (False) and $q$ is $T$ (True) or $F$ (False),then $(p \rightarrow q)$ is $T$.
Then $(p$ $\rightarrow q)$ $\rightarrow p$ becomes $T \rightarrow F$,which is $F$.
Since the statement can be false,it is not a tautology.
Therefore,option $D$ is false.

(D) Given statements:
$p :$ It is raining today.
$q :$ $I$ go to school.
$r :$ $I$ will meet my friends.
$s :$ $I$ will go to watch a movie.
Negations:
$\sim p :$ It does not rain today.
$\sim q :$ $I$ do not go to school.
The statement is: 'If it does not rain today or $I$ do not go to school,then $I$ will meet my friends and go to watch a movie'.
This can be written as: $(\sim p \vee \sim q) \Rightarrow (r \wedge s)$.
Since $(\sim p \vee \sim q)$ is equivalent to $\sim (p \wedge q)$ by De Morgan's Law,the statement can also be represented as $\sim (p \wedge q) \Rightarrow (r \wedge s)$.