True or False: The negation of $\sim p \Rightarrow q$ is $p \Rightarrow q$

I understand this is a simple true or false question, but I'd like to understand the underlying reason why it is true or false.

So this is my first week in a Real Analysis class and we are reviewing logic, and I know the basics. A conditional is false only when the antecedent is true and the consequence is false.

But I don't understand how to apply that logic to the above question.

So breaking down the problem if the antecedent is false and the consequence is true, that means the result is true, but since I'm negating it it is actually false; but how do I apply that to the second part of the question to determine if it is true or false?

I assume the answer is true although the antecedent and consequence are different.

  • $\begingroup$ @JamesWalker I copied it exactly the way it is shown in my text, which means I would be negating $(\sim p \Rightarrow q)$ $\endgroup$ – hax0r_n_code Jun 4 '16 at 0:23
  • $\begingroup$ In the standard order of operations, $\sim$ binds before $\vee$ and $\wedge$, and these before logical connectives. $\endgroup$ – ncmathsadist Jun 4 '16 at 0:25

You could draw a truth table: $$ \begin{array}{|c|c|c|c|} \hline p& q & \neg p & p \rightarrow q & \neg p \rightarrow q & \neg( \neg p\rightarrow q)\\ \hline T & T & F & T&T&F\\ \hline T& F & F & F&T&F\\ \hline F & T & T & T&T&F\\ \hline F & F & T & T&F&T \\ \hline \end{array} $$

We see that $ \neg( \neg p\rightarrow q)$ is not the same as $p\rightarrow q$, as for some given truth values of $p$ and $q$, the two statements do not yield the same result.


$$ (\neg p)\Rightarrow q $$ is equivalent to $ p\vee q, $ the negation of which is $(\neg p)\wedge(\neg q)$.

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

  • $\begingroup$ So the answer to the question is true $\endgroup$ – hax0r_n_code Jun 4 '16 at 0:54
  • $\begingroup$ @free_mind: where did you get this question? Providing references is always a good idea. $\endgroup$ – Jack Jun 4 '16 at 0:59

False. We could have $q$ true whether $p$ is true or not, then both $p\to q$ and $\lnot(p)\to q $ are true.

Think of Morton's Fork.

  • $\begingroup$ Tried to make a "not" symbol, failed. How to get it? Thanks. $\endgroup$ – Oscar Lanzi Jun 4 '16 at 0:31
  • 1
    $\begingroup$ \sim for $\sim$ and \neg for $\neg$ $\endgroup$ – user336735 Jun 4 '16 at 0:38
  • 2
    $\begingroup$ \lnot for $\lnot$ also works and goes with \land , \lor for $\land, \lor$. also \to gives $\to$ . $\endgroup$ – Graham Kemp Jun 4 '16 at 0:40

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