What is the negation of the implication statement

It's because $A\to B$ is equivalent to $(\lnot A)\lor B$ and the negation of that is equivalent to $A\land \lnot B$.


All of this is loose, in layman's terms...

$P \Rightarrow Q$ means if P occurs, so does Q. $P$ either happens or it doesn't. So, either $\neg P$ happens or $P$ does, but the latter means $Q$ does, too. So, you can write out the truth table that $P \Rightarrow Q$ is the same as $\neg P \vee Q$. Now use DeMorgan's law on $\neg (P \Rightarrow Q)$, which is $\neg (\neg P \vee Q)$ as just explained. This is your right-hand side.


One way of putting this that might help is the following: what if I told you, "If the real number $x$ is irrational, then $x^2$ is irrational"? (Logically, writing $\mathbb I$ for the set of irrational numbers, this might look like $x \in \mathbb I \to x^2 \in \mathbb I$.) Of course, what I'm saying is false, so you object: "Actually, that's not true. For example $\sqrt{2}$ is irrational, yet $\sqrt{2}^2 = 2$ is rational." (Logically, this looks like $\sqrt{2} \in \mathbb I \land \neg (\sqrt{2}^2 \in \mathbb I)$.)

Do you see how your negation of my statement gave you the conjunction? Do you agree that giving an example where $x \in \mathbb I$ is false, such as $2 \notin \mathbb I \land 4 \notin \mathbb I$ would not have been a refutation of my statement?

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Logic

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