I know that the negation of $p\to q$ is ~p V q but I can’t seem to figure out the effect $\sim q$ will have on the negation. Also is their a way to check if something is the negation of a statement?
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1Welcome to Maths SX! I'm sorry, but the negation of $p\to q$ is $p\wedge ({\sim}q)$ since $p\to q$ is $({\sim}p)\vee q$. – Bernard Jan 22 '19 at 21:57
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2The negation for $p\to q$ is $p\land\lnot q$. So by substitution ... – Graham Kemp Jan 22 '19 at 21:59
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Caution. The negation of $p\to q$ is ${\sim (p}\to q)$ but it is not $({\sim p})\to q$. Which did you mean when you wrote ${\sim p}\to q$? So in the same way, the negation of $p \to \sim q$ is $\sim (p \to \sim q)$. – GEdgar Jan 22 '19 at 22:44
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Whoops meant to write the negation as ~p V q – Mikayla Jan 22 '19 at 23:09
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The definition of implication in propositional logic can be expressed with
\begin{equation}\tag{1} p\rightarrow q~:=~(\sim p)\lor q \end{equation}
By this definition, the negation of $p\rightarrow q$ is $p\land(\sim q)$. This means that the statement $p$ implies $q$, is false only when the antecedent or premise of the statement is true, and its consequent false.
Using the same definition, negation of the statement $p\rightarrow (\sim q)$ is
\begin{align} \sim(p\rightarrow (\sim q))~~~&Start\\ \sim((\sim p)\lor (\sim q))~~~&Substitute~(1)\\ p\land q~~~&De~Morgan's~law \end{align}
To check if any two statements in propositional logic are the negation of one another, it is sufficient to substitute all instances of implication with the first definition, then negate one, and apply De Morgan's law successively to it until one reduces to the other. If this is the case, then they are negations of one-another.
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