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# 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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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 $q$ ($\sim q$).