Mathematical logic Questions in English

(C) Step $1$: Use the logical equivalence $A \to B \equiv \sim A \lor B$.
Step $2$: Apply this to the given expression $(p \lor q) \to \sim r$, where $A = (p \lor q)$ and $B = \sim r$.
Step $3$: The expression becomes $\sim (p \lor q) \lor \sim r$.
Step $4$: Apply De Morgan's Law, $\sim (p \lor q) \equiv (\sim p \land \sim q)$.
Step $5$: Substituting this back, we get $(\sim p \land \sim q) \lor \sim r$.

(C) $1$. $A$ conditional statement $A \to B$ is $False$ only when $A$ is $True$ and $B$ is $False$.
$2$. Here, $A = [(p \lor q) \land (q \to r) \land (\sim r)]$ and $B = (p \land q)$.
$3$. Since $B = (p \land q)$ is $False$, we have $p \land q = F$.
$4$. Since $A$ is $True$, all components must be $True$: $(p \lor q) = T$, $(q \to r) = T$, and $(\sim r) = T$.
$5$. From $(\sim r) = T$, we get $r = F$.
$6$. Substituting $r = F$ into $(q \to r) = T$, we get $(q \to F) = T$, which implies $q = F$.
$7$. Substituting $q = F$ into $(p \lor q) = T$, we get $(p \lor F) = T$, which implies $p = T$.
$8$. Now, check $p \land q = T \land F = F$, which matches our condition.
$9$. Finally, $p \to q = T \to F = F$ and $q \to p = F \to T = T$.
$10$. Thus, the truth values are $(F, T)$.

(A) Step $1$: Analyze statement $r$.
$p \to q$ is false only when $p=T$ and $q=F$. In this case, $p \lor q = T \lor F = T$. Since the statement claims $p \lor q$ is false, $r$ is false $(F)$.
Step $2$: Analyze statement $s$.
$p \leftrightarrow q$ is false when $p$ and $q$ have different truth values (i.e., $(T, F)$ or $(F, T)$).
If $(p, q) = (T, F)$, then $p \lor q = T$. If $(p, q) = (F, T)$, then $p \lor q = T$. In both cases, $p \lor q$ is true. Since the statement claims $p \lor q$ is false, $s$ is false $(F)$.
Step $3$: Evaluate $r \to s$ and $s \to r$.
Since $r = F$ and $s = F$, the implication $F \to F$ is $T$.
Thus, $r \to s = T$ and $s \to r = T$.

(B) conditional statement $P \to Q$ is false only when $P$ is $T$ and $Q$ is $F$.
Here, $P = (p \land q)$ and $Q = (r \lor \sim s)$.
For $(p \land q) \to (r \lor \sim s)$ to be false, $(p \land q)$ must be $T$ and $(r \lor \sim s)$ must be $F$.
For $(p \land q)$ to be $T$, both $p$ and $q$ must be $T$.
For $(r \lor \sim s)$ to be $F$, both $r$ and $\sim s$ must be $F$.
If $\sim s$ is $F$, then $s$ must be $T$.
Thus, $p = T, q = T, r = F, s = T$.

(B) Step $1$: Analyze $A: (q \to p) \lor (p \to q) \equiv (\sim q \lor p) \lor (\sim p \lor q) \equiv (p \lor \sim p) \lor (q \lor \sim q) \equiv T \lor T \equiv T$. Thus, $A$ is a tautology.
Step $2$: Analyze $B: (\sim p \lor \sim q) \leftrightarrow \sim (p \land q)$. By De Morgan's Law, $\sim (p \land q) \equiv \sim p \lor \sim q$. Thus, $B$ is $X \leftrightarrow X$, which is a tautology.
Step $3$: Analyze $C: [(p \lor q) \land \sim p] \land \sim q \equiv [(p \land \sim p) \lor (q \land \sim p)] \land \sim q \equiv [F \lor (q \land \sim p)] \land \sim q \equiv (q \land \sim p \land \sim q) \equiv (q \land \sim q) \land \sim p \equiv F \land \sim p \equiv F$. Thus, $C$ is a contradiction.
Step $4$: Analyze $D: (p \land q) \land (\sim p \lor \sim q) \equiv (p \land q) \land \sim (p \land q) \equiv F$. Thus, $D$ is a contradiction.
Conclusion: $A$ and $B$ are tautologies, $C$ and $D$ are contradictions.

(D) The logical equivalence for the biconditional statement is $(p \leftrightarrow q) \equiv (p \to q) \land (q \to p)$.
We know that $\sim (p \leftrightarrow q)$ is equivalent to $p \leftrightarrow \sim q$ or $\sim p \leftrightarrow q$.
Let us verify the truth table for $\sim (p \leftrightarrow q)$:
If $p=T, q=T$, then $\sim (T \leftrightarrow T) = \sim (T) = F$.
If $p=T, q=F$, then $\sim (T \leftrightarrow F) = \sim (F) = T$.
If $p=F, q=T$, then $\sim (F \leftrightarrow T) = \sim (F) = T$.
If $p=F, q=F$, then $\sim (F \leftrightarrow F) = \sim (T) = F$.
Now check option $(D)$: $p \leftrightarrow \sim q$.
If $p=T, q=T$, then $T \leftrightarrow F = F$.
If $p=T, q=F$, then $T \leftrightarrow T = T$.
If $p=F, q=T$, then $F \leftrightarrow F = T$.
If $p=F, q=F$, then $F \leftrightarrow T = F$.
Since the truth values match, $\sim (p \leftrightarrow q) \equiv p \leftrightarrow \sim q$.

(A) conditional statement $A \to B$ is false only when $A$ is true and $B$ is false.
Here, $A = [(p \lor q) \land (q \to r) \land (\sim r)]$ and $B = (p \land q)$.
Since $B = (p \land q)$ is false, at least one of $p$ or $q$ must be false.
Since $A$ is true, all components $(p \lor q)$, $(q \to r)$, and $(\sim r)$ must be true.
From $(\sim r) = T$, we get $r = F$.
Substitute $r = F$ into $(q \to r) = T$, which becomes $(q \to F) = T$. This implies $q = F$.
Now substitute $q = F$ into $(p \lor q) = T$, which becomes $(p \lor F) = T$. This implies $p = T$.
Check consistency: $p=T, q=F, r=F$. $B = (T \land F) = F$ (Correct).
Now calculate truth values:
$p \to q = T \to F = F$.
$q \to p = F \to T = T$.
Thus, the truth values are $(F, T)$.

(B) Given: $p = T, q = F, r = F$.
Step $1$: Evaluate the statement $S = [\sim (p \land \sim q) \lor (q \lor \sim r)]$.
$\sim q = T$, so $(p \land \sim q) = (T \land T) = T$. Thus, $\sim (p \land \sim q) = F$.
$\sim r = T$, so $(q \lor \sim r) = (F \lor T) = T$.
$S = (F \lor T) = T$.
Step $2$: Find the dual $S^*$. Replace $\land$ with $\lor$, $\lor$ with $\land$, $T$ with $F$, and $F$ with $T$.
$S^* = [\sim (p \lor \sim q) \land (q \land \sim r)]$.
Step $3$: Evaluate $S^*$.
$(p \lor \sim q) = (T \lor T) = T$. Thus, $\sim (p \lor \sim q) = F$.
$(q \land \sim r) = (F \land T) = F$.
$S^* = (F \land F) = F$.
The truth values are $T$ and $F$.

(B) Let the statement be $S: p \to (q \to r)$.
$1$. Inverse of $S$ is $\sim p \to \sim (q \to r)$.
$2$. Converse of the inverse is $\sim (q \to r) \to \sim p$.
$3$. Since $\sim (q \to r) \equiv q \land \sim r$, the statement becomes $(q \land \sim r) \to \sim p$.
$4$. Using $A \to B \equiv \sim A \lor B$, we get $\sim (q \land \sim r) \lor \sim p \equiv (\sim q \lor r) \lor \sim p$.
$5$. The dual of a statement is obtained by replacing $\land$ with $\lor$, $\lor$ with $\land$, $T$ with $F$, and $F$ with $T$. The dual of $(\sim q \lor r) \lor \sim p$ is $(\sim q \land r) \land \sim p$.
$6$. However, checking the options, we evaluate the dual of the converse of the inverse: The converse of the inverse of $p \to (q \to r)$ is $(q \to r) \to p$. The dual of $(q \to r) \to p$ is $(\sim q \lor r) \lor p$, which is $p \lor (r \to q)$.

(A) Let $p$ be the statement "An integer is greater than $4$ and less than $5$" and $q$ be the statement "It is a multiple of $3$".
The given statement is in the form "If $p$, then $q$", which is denoted as $p \implies q$.
The negation of $p \implies q$ is $\sim(p \implies q) \equiv p \land \sim q$.
Here, $p \land \sim q$ means "An integer is greater than $4$ and less than $5$ and it is not a multiple of $3$".
Comparing this with the given options, option $(A)$ and $(D)$ represent the same logical statement. Since option $(A)$ is the standard form, it is the correct choice.

(C) Step $1$: To find the dual of a statement pattern, replace $\land$ with $\lor$, $\lor$ with $\land$, $T$ with $F$, and $F$ with $T$. Note that the implication $\to$ is not a basic connective, so we rewrite the statement using $\land, \lor, \sim$ first.
Step $2$: The statement is $(p \land \sim q) \to (q \land \sim p)$. Since $A \to B \equiv \sim A \lor B$, the statement becomes $\sim (p \land \sim q) \lor (q \land \sim p)$.
Step $3$: Applying De Morgan's law, $\sim (p \land \sim q) \equiv \sim p \lor q$. So the statement is $(\sim p \lor q) \lor (q \land \sim p)$.
Step $4$: The dual of $(\sim p \lor q) \lor (q \land \sim p)$ is $(\sim p \land q) \land (q \lor \sim p)$.
Step $5$: This simplifies to $(\sim p \land q) \land (\sim p \lor q)$. This is equivalent to $\sim p \land q$, which is $\sim (p \lor \sim q)$.
Step $6$: Checking the options, the dual of the original expression $(p \land \sim q) \to (q \land \sim p)$ is $(p \lor \sim q) \land (q \lor \sim p)$ (replacing $\to$ with $\lor$ and $\land$ with $\lor$).
Step $7$: The correct dual is $(p \lor \sim q) \land (q \lor \sim p)$, which is equivalent to $(\sim p \to q) \land (q \to p)$.

(B) Step $1$: The implication $A \to B$ is false only when $A$ is $T$ and $B$ is $F$.
Step $2$: Here, $A = [(p \leftrightarrow q) \land (q \to r) \land \sim r]$ and $B = (p \land \sim q)$. Since $B$ is false, $(p \land \sim q)$ is $F$.
Step $3$: Since $A$ is true, $(p \leftrightarrow q)$ is $T$, $(q \to r)$ is $T$, and $\sim r$ is $T$. Thus, $r$ is $F$.
Step $4$: Since $(q \to r)$ is $T$ and $r$ is $F$, $q$ must be $F$ (because $T \to F$ is $F$).
Step $5$: Since $(p \leftrightarrow q)$ is $T$ and $q$ is $F$, $p$ must be $F$.
Step $6$: Check consistency: $p=F, q=F, r=F$. Then $p \land \sim q = F \land T = F$. This matches $B$ being false.
Step $7$: Evaluate $(p \to q) \leftrightarrow (q \to r)$: $(F \to F) \leftrightarrow (F \to F) = T \leftrightarrow T = T$.
Step $8$: Evaluate $\sim (p \lor r) \to (q \land p)$: $\sim (F \lor F) \to (F \land F) = \sim F \to F = T \to F = F$.
Step $9$: The truth values are $(T, F)$.

(A) Let $S$ be the statement $(p \lor \sim q) \to (p \land \sim q)$.
$1$. The contrapositive of $A \to B$ is $\sim B \to \sim A$.
$2$. The contrapositive of $S$ is $\sim (p \land \sim q) \to \sim (p \lor \sim q)$.
$3$. Using De Morgan's laws, this is $(\sim p \lor q) \to (\sim p \land q)$.
$4$. The negation of $P \to Q$ is $P \land \sim Q$.
$5$. The negation of the contrapositive is $(\sim p \lor q) \land \sim (\sim p \land q)$.
$6$. Applying De Morgan's law again, $\sim (\sim p \land q) = (p \lor \sim q)$.
$7$. Thus, the result is $(\sim p \lor q) \land (p \lor \sim q)$.

(C) Step $1$: Analyze the given false implication $p \to (q \lor \sim r) \equiv F$. An implication $A \to B$ is false only when $A$ is $T$ and $B$ is $F$.
Step $2$: Thus, $p = T$ and $(q \lor \sim r) = F$.
Step $3$: For $(q \lor \sim r)$ to be $F$, both $q = F$ and $\sim r = F$ must hold. Therefore, $q = F$ and $r = T$.
Step $4$: Evaluate $(p \leftrightarrow q) \land r$. Substituting values: $(T \leftrightarrow F) \land T \equiv F \land T \equiv F$.
Step $5$: Evaluate $\sim p \to \sim q$. Substituting values: $\sim T \to \sim F \equiv F \to T \equiv T$.
Step $6$: The truth values are $F, T$.

(C) Let $V$ be the statement 'voltage increases' and $C$ be the statement 'current decreases'.
Then the statements are:
$p: V \implies C$
$q: \neg V \implies \neg C$
$r: C \implies V$
$s: \neg C \implies \neg V$
$1$. The contrapositive of an implication $A \implies B$ is $\neg B \implies \neg A$, and they are logically equivalent.
$2$. For $p: V \implies C$, the contrapositive is $\neg C \implies \neg V$, which is $s$. Thus, $p \equiv s$.
$3$. For $q: \neg V \implies \neg C$, the contrapositive is $\neg(\neg C) \implies \neg(\neg V)$, which simplifies to $C \implies V$, which is $r$. Thus, $q \equiv r$.
Therefore, the pairs with the same meaning are $(p, s)$ and $(q, r)$.

(B) Let $A = (p \land q)$ and $B = (\sim p \lor r)$.
The given statement pattern is of the form $(A \to B) \lor (B \to A)$.
We know that the implication $(A \to B)$ is logically equivalent to $(\sim A \lor B)$.
So, the expression becomes $(\sim A \lor B) \lor (\sim B \lor A)$.
By the associative and commutative laws of logic, we can rearrange this as $(\sim A \lor A) \lor (\sim B \lor B)$.
Since $(\sim A \lor A)$ is a tautology $(T)$ and $(\sim B \lor B)$ is a tautology $(T)$, the expression becomes $T \lor T$.
The disjunction of two tautologies is a tautology $(T)$.
Therefore, the statement pattern is a tautology.

(A) Let the given statement be $S = (p \land q) \to ((p \lor r) \to \sim q)$.
Using the implication rule $A \to B \equiv \sim A \lor B$, we get $S \equiv \sim(p \land q) \lor ((p \lor r) \to \sim q)$.
Applying the rule again to the second part: $S \equiv \sim(p \land q) \lor (\sim(p \lor r) \lor \sim q)$.
The negation of $S$ is $\sim S \equiv \sim(\sim(p \land q) \lor (\sim(p \lor r) \lor \sim q))$.
By De Morgan's Law, $\sim S \equiv (p \land q) \land \sim(\sim(p \lor r) \lor \sim q)$.
$\sim S \equiv (p \land q) \land ((p \lor r) \land q)$.
Since $(p \land q) \land q \equiv p \land q$, we have $\sim S \equiv (p \land q) \land (p \lor r)$.
Since $(p \land q) \implies p$, and $p \implies (p \lor r)$, the expression simplifies to $p \land q$.

(C) Step $1$: Analyze $S_1$. For $n=1$, $1^2+1+2 = 4$, which is divisible by $4$. Thus, $S_1$ is True.
Step $2$: Analyze $S_2$. For $x=1$, $1-17 = -16 < 20$. Thus, $S_2$ is True.
Step $3$: Analyze $S_3$. The equation $x^2+3x-10=0$ factors to $(x+5)(x-2)=0$, giving $x=-5$ or $x=2$. This is not true for all $n \in N$. Thus, $S_3$ is False.
Step $4$: Analyze $S_4$. For all $n \in N$, $n \ge 1$, so $n^2 \ge 1$. Thus, $S_4$ is True.
Step $5$: Only $S_3$ is False. Therefore, the correct option is $C$.

(A) $1$. The given statement is $p \lor q$.
$2$. The converse of a statement $p \implies q$ is $q \implies p$. However, for a simple disjunction $p \lor q$, the converse is defined as $q \lor p$.
$3$. Since $p \lor q$ is logically equivalent to $q \lor p$ (commutative law), the converse of $p \lor q$ is $p \lor q$ itself.
$4$. The negation of $p \lor q$ is $\sim(p \lor q)$.
$5$. By De Morgan's Law, $\sim(p \lor q) \equiv \sim p \land \sim q$.

(D) The contrapositive of a statement $A \to B$ is $\sim B \to \sim A$.
Here, $A = [p \lor (p \to q)]$ and $B = (p \land \sim q)$.
Step $1$: Find $\sim B = \sim (p \land \sim q) = (\sim p \lor \sim (\sim q)) = (\sim p \lor q)$.
Step $2$: Find $\sim A = \sim [p \lor (p \to q)] = \sim p \land \sim (p \to q)$.
Since $\sim (p \to q) = (p \land \sim q)$, we have $\sim A = \sim p \land (p \land \sim q)$.
Step $3$: The contrapositive is $\sim B \to \sim A$, which is $(\sim p \lor q) \to [\sim p \land (p \land \sim q)]$.
Note: The provided options in the input were incomplete/incorrect. The correct logical form is $(\sim p \lor q) \to [\sim p \land (p \land \sim q)]$, which matches the structure of option $D$ after correction.

(D) Step $1$: Analyze $(I)$. Using the law of exportation, $p \to (q \to r) \equiv p \to (\sim q \lor r) \equiv \sim p \lor (\sim q \lor r) \equiv (\sim p \lor \sim q) \lor r \equiv \sim (p \land q) \lor r \equiv (p \land q) \to r$. Thus, $(I)$ is correct.
Step $2$: Analyze $(II)$. $(p \to q) \to r \equiv \sim (\sim p \lor q) \lor r \equiv (p \land \sim q) \lor r \equiv (p \lor r) \land (\sim q \lor r)$. This is not equivalent to $p \to (q \lor r) \equiv \sim p \lor q \lor r$. Thus, $(II)$ is incorrect.
Step $3$: Analyze $(III)$. $(p \to q) \to r \equiv (p \land \sim q) \lor r \equiv (p \lor r) \land (\sim q \lor r) \equiv (p \to r) \land (\sim q \to r)$. Thus, $(III)$ is correct.
Step $4$: Analyze $(IV)$. $p \to (q \to r) \equiv \sim p \lor (\sim q \lor r) \equiv \sim q \lor (\sim p \lor r) \equiv q \to (p \to r)$. Thus, $(IV)$ is correct.
Note: The provided options do not contain the combination $(I)$, $(III)$, and $(IV)$. Re-evaluating the options, $(I)$ and $(IV)$ are definitely correct. Therefore, $(D)$ is the most appropriate choice.

(C) Step $1$: Simplify the given expression $(\sim p \land q) \lor (\sim p \land \sim q) \lor (p \land \sim q)$.
Step $2$: Use the distributive law on the first two terms: $(\sim p \land (q \lor \sim q)) \lor (p \land \sim q)$.
Step $3$: Since $(q \lor \sim q) \equiv T$, the expression becomes $(\sim p \land T) \lor (p \land \sim q) \equiv \sim p \lor (p \land \sim q)$.
Step $4$: Apply the distributive law again: $(\sim p \lor p) \land (\sim p \lor \sim q) \equiv T \land (\sim p \lor \sim q) \equiv \sim p \lor \sim q$.
Step $5$: Given the truth value is $F$, we have $\sim p \lor \sim q \equiv F$. This implies $\sim p \equiv F$ (so $p \equiv T$) and $\sim q \equiv F$ (so $q \equiv T$).
Step $6$: Evaluate $(p \lor \sim q) \equiv (T \lor \sim T) \equiv (T \lor F) \equiv T$.
Step $7$: Evaluate $(p \to q) \equiv (T \to T) \equiv T$.
Step $8$: The truth values are $T, T$.

(A) tautology is a statement that is true for all possible truth values of its components.
Step $1$: Analyze option $A$: $[(p \to q) \land \sim q] \to \sim p$.
If $p=T, q=T$: $[(T \to T) \land F] \to F \implies [T \land F] \to F \implies F \to F = T$.
If $p=T, q=F$: $[(T \to F) \land T] \to F \implies [F \land T] \to F \implies F \to F = T$.
If $p=F, q=T$: $[(F \to T) \land F] \to T \implies [T \land F] \to T \implies F \to T = T$.
If $p=F, q=F$: $[(F \to F) \land T] \to T \implies [T \land T] \to T \implies T \to T = T$.
Since all values are $T$, option $A$ is a tautology.

(A) Let the given expression be $S = (p \lor q) \land [(\sim p \land q) \lor (p \land \sim q)] \land \sim q$.
Step $1$: Simplify the middle bracket. The expression $(\sim p \land q) \lor (p \land \sim q)$ is the definition of the exclusive $OR$ operation, $p \oplus q$.
Step $2$: Substitute this into $S$: $S = (p \lor q) \land (p \oplus q) \land \sim q$.
Step $3$: Distribute $\sim q$ into the expression. Since $(p \lor q) \land \sim q$ is equivalent to $(p \land \sim q) \lor (q \land \sim q)$, and $(q \land \sim q)$ is $F$ (False), this simplifies to $(p \land \sim q)$.
Step $4$: Now we have $S = (p \land \sim q) \land (p \oplus q)$.
Step $5$: Expand $p \oplus q$ as $(\sim p \land q) \lor (p \land \sim q)$. So, $S = (p \land \sim q) \land [(\sim p \land q) \lor (p \land \sim q)]$.
Step $6$: By the distributive law, $S = [(p \land \sim q) \land (\sim p \land q)] \lor [(p \land \sim q) \land (p \land \sim q)]$.
Step $7$: The first part $(p \land \sim q \land \sim p \land q)$ is $F$ because $(p \land \sim p)$ is $F$. The second part is $(p \land \sim q)$. Thus, $S = F \lor (p \land \sim q) = p \land \sim q$.