# Find what $p \to \sim q$ can also be written as:

A. $p \to q$

B. $\left( { \sim p} \right) \vee \left( { \sim q} \right)$

C. $q \to q$

D. $ \sim q \to \sim p$

Last updated date: 22nd Mar 2023

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Answer

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Hint- Here, we will be using the theorem of propositional logic because of the given statement.

According to the theorem of propositional logic, $p$ implies $q$ is equivalent to not $p$ or q

i.e., $p \to q \equiv \left( { \sim p} \right) \vee q$

Now let us replace $q$ by not $q$ ($ \sim q$) then the above compound statement becomes

$p \to \sim q \equiv \left( { \sim p} \right) \vee \left( { \sim q} \right)$

From the above compound statement, we can say that $p$ implies not $q$ ($ \sim q$) is equivalent to not $p$ ($ \sim p$) or not $q$ ($ \sim q$).

Therefore, option B is correct.

Note- In these types of problems, the given statement is observed and an appropriate theorem is used which will give a compound statement where the given statement is equivalent to another statement.

According to the theorem of propositional logic, $p$ implies $q$ is equivalent to not $p$ or q

i.e., $p \to q \equiv \left( { \sim p} \right) \vee q$

Now let us replace $q$ by not $q$ ($ \sim q$) then the above compound statement becomes

$p \to \sim q \equiv \left( { \sim p} \right) \vee \left( { \sim q} \right)$

From the above compound statement, we can say that $p$ implies not $q$ ($ \sim q$) is equivalent to not $p$ ($ \sim p$) or not $q$ ($ \sim q$).

Therefore, option B is correct.

Note- In these types of problems, the given statement is observed and an appropriate theorem is used which will give a compound statement where the given statement is equivalent to another statement.

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