Negation of "if A then B" (how to prove that "if A then B" is false) I know that this topic has been discussed before, but I still couldn't find an answer to my particular question.
I know that the negation of "If A then B" is "A and NOT B".
But I wanted some clarification and what determines true/false for the statement A and NOT B.
For example, let's assume the statement "if A then B" is true. Then to my understanding, it would follow that "A and NOT B" must always be false.
However, let's assume the statement "if A then B" is false. Then would the statement "A and NOT B" always be true? Or is it that there is at least one case where "A and NOT B" is true?
Just to make my question even clearer, if I wanted to prove that "if A then B" was indeed false, would I need to show that "A and NOT B" always holds true, or is it sufficient to show just one case where it is true?
Thanks!
 A: Let's look at the truth table of $A \rightarrow B$, we have
$$
\begin{array}{|c|c|c|}
\hline 
A & B & A\rightarrow B \\
\hline 
 T &  T &       T         \\ 
 T &  F &       F           \\ 
 F &  T &         T         \\
 F &  F &         T         \\
\hline  
\end{array}
$$
The only case to get $False$ value is when $A$ is $True$ and $B$ is $False$. So to get this result you only need to show that $B$ is $False$. Hope that helps
A: 
For example, let's assume the statement "if A then B" is true. Then to my understanding, it would follow that "A and NOT B" must always be false.

To be true is different from to be a tautology, so it doesn't follows that "A and NOT B" must always be false. Instead suppose "if A then B" is a tautology, this implies its negation must always be false i.e. a contradiction.
Eidt: It's correct if you mean "A and NOT B" always be false in those cases that "if A then B" is true.

However, let's assume the statement "if A then B" is false. Then would the statement "A and NOT B" always be true? Or is it that there is at least one case where "A and NOT B" is true?

If we know that "if A then B" is false in some fixed cases, then "A and NOT B" must be true in these cases, and if these cases covers all the possible cases, then yes that
$$\text{($'$A and NOT B$'$ always be true) hold, i.e. this would be a tautology}$$
However, when we say "if A then B" is false, normally it means this is false in some specific case, say case C. That there is at least one case where "A and NOT B" is true hold. Be specifically, because it's true in case C.

Just to make my question even clearer, if I wanted to prove that "if A then B" was indeed false, would I need to show that "A and NOT B" always holds true, or is it sufficient to show just one case where it is true?

If we want to prove that "if A then B" is indeed false in some case C, then it's sufficient to show that in case C "A and NOT B" is true.
For the same reason, if we want to prove that "if A then B" is always false, then we need to show that "A and NOT B" is always true.
A: Here is the truth table for $(\neg(A\to B)\to (A \land \neg B))$:

As you can see, it is always true.
Logical implication is often defined as:

$A\to B~~\equiv ~~ \neg (A \land \neg B)$

This equivalence can also be formally proven from first principles using a form of natural deduction:


