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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!

punypaw
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    "or is it sufficient to show just one case where it is true" You seem to be missing / referring to quantifiers which you don't have in your statements. The negation of "For all $x$ if $A(x)$ then $B(x)$" is going to be "There exists $x$ such that $A(x)$ and not $B(x)$." In such a case, indeed, showing just one case is sufficient. If these quantifiers aren't present then "just one case" doesn't even make sense to talk about. – JMoravitz Aug 15 '20 at 21:08
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    There is a subtlety if e.g., over the reals, you start with "If $x < 5$ then $x < 1$" which has a free variable $x$. This conditional formula is, formally speaking, not false even though we would all say informally that it is false because we'd implicitly take it as a universally quantified sentence. Formally, it is not true of course, but it isn't false either (e.g. it holds when $x$ is 17) and its negation "$x < 5$ and $x >= 1$" is also neither true nor false. I'm not sure if this has anything to do with your question ... – Ned Aug 15 '20 at 21:20
  • This question does not entail quantifiers, @JMoravitz. – amWhy Aug 15 '20 at 21:31
  • I misread the question, if you saw my comment and/or answer - ignore it – Riemann'sPointyNose Aug 15 '20 at 21:33

3 Answers3

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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

Muhammad
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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.

Ethan
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  • I see, thank you for the answer! I understood most of what you're saying, but I still don't quite follow your first paragraph. What do you mean that to be true is different from a tautology? – punypaw Aug 16 '20 at 20:47
  • @punypaw For example, $A\to B$ is true when Case $1$: $A,B$ are both true, Case $2$: $A$ is false but $B$ is true, Case $3$: $A,B$ are both false. And $A\to B$ is false when Case $4$: $A$ is true but $B$ is false. To be a tautology $A\to B$ have to be true in all possible cases; on the other hand, we can talk about the true or false of $A\to B$ in each different cases, that's the difference. – Ethan Aug 17 '20 at 03:30
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Here is the truth table for $(\neg(A\to B)\to (A \land \neg B))$:

enter image description here

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:

enter image description here enter image description here

Dan Christensen
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