The statement sim(p leftrightarrow sim q) is | vedclass

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(C) To determine the nature of the statement $\sim(p \leftrightarrow \sim q)$,we construct a truth table:$1$. The biconditional $p \leftrightarrow \si

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equivalent to $(p \leftrightarrow q)$

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(C) To determine the nature of the statement $\sim(p \leftrightarrow \sim q)$,we construct a truth table:$1$. The biconditional $p \leftrightarrow \sim q$ is true when $p$ and $\sim q$ have the same truth value.$2$. The negation $\sim(p \leftrightarrow \sim q)$ is true when $p \leftrightarrow \sim q$ is false.Truth Table:$p$ | $q$ | $\sim q$ | $p \leftrightarrow \sim q$ | $\sim(p \leftrightarrow \sim q)$$T$ | $T$ | $F$ | $F$ | $T$$T$ | $F$ | $T$ | $T$ | $F$$F$ | $T$ | $F$ | $T$ | $F$$F$ | $F$ | $T$ | $F$ | $T$Comparing this to $p \leftrightarrow q$:$p$ | $q$ | $p \leftrightarrow q$$T$ | $T$ | $T$$T$ | $F$ | $F$$F$ | $T$ | $F$$F$ | $F$ | $T$Since the truth values of $\sim(p \leftrightarrow \sim q)$ and $(p \leftrightarrow q)$ are identical for all combinations of $p$ and $q$,the statement is equivalent to $(p \leftrightarrow q)$.

The statement $\sim(p \leftrightarrow \sim q)$ is :

Similar Questions

$(\sim (\sim p)) \wedge q$ is equal to .........

If truth values of statements $p, q$ are true,and $r, s$ are false,then the truth values of the following statement patterns are respectively:
$a: \sim(p \wedge \sim r) \vee(\sim q \vee s)$
$b: (\sim q \wedge \sim r) \leftrightarrow(p \vee s)$
$c: (\sim p \vee q) \rightarrow(r \wedge \sim s)$

If $p, q, r$ are simple propositions with truth values $T, F, T$ respectively, then which of the following is not a true statement?

The negation of $(p$ $\Rightarrow q)$ $\Rightarrow (q$ $\Rightarrow p)$ is

Consider the following statements:
$r: \text{If } p \to q \text{ is false, then } p \lor q \text{ is false.}$
$s: \text{If } p \leftrightarrow q \text{ is false, then } p \lor q \text{ is false.}$
The truth values of $r \to s$ and $s \to r$ are respectively:

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