0
$\begingroup$

I don't get the idea behind it. There are some things in the truth table which disturbs me. As if $A$ is false, the $A\Rightarrow B$ is always true. I read that implication was the negation of ($A \land (\neg B) $). Thus indicating that if $A$ is true and $B$ false, then the implication is false and that is all we care about, the other cases are just a consequence of the definition. Is this right?

$\endgroup$

marked as duplicate by Mauro ALLEGRANZA logic Feb 11 at 8:00

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • 2
    $\begingroup$ Yes, the negation of (a ^ not b) is not a or b. For the implication to hold, either a is false (and not a true) or a is true, requiring b to be true. $\endgroup$ – RJM Feb 10 at 22:53
  • 1
    $\begingroup$ Think about it this way. When is $A\to B$ false? To put it in natural language and exemplifying, when will I have lied if I say that "If it rains, then I take an umbrella"? Do you really want to say that I lied when it doesn't rain? $\endgroup$ – Git Gud Feb 10 at 23:13
  • $\begingroup$ see also the questions and answers linked here math.stackexchange.com/questions/3082747/… $\endgroup$ – spaceisdarkgreen Feb 10 at 23:22
0
$\begingroup$

It is a matter of convenience. The idea is that $a\Rightarrow b$ is false only when $a$ is true and $b$ is false. A false premise corresponds (in terms of sets) to the empty set. It is convenient to consider the empty set as a subset of any set, and it is convenient to identify properties (like $a$) with the subset of elements having the property. For example, it is true that "if $T$ is a right equilateral triangle, then $T$ is red" just because there is no triangle satisfying your hypothesis.

You could define the implication otherwise, but then this correspondence is lost and many simple propositions would have exceptions, etc.

$\endgroup$
0
$\begingroup$

The teach us think "if conditions hold" then "Are these statements true" but I find it better and more into intuitive to think if these statements are true, what is possible.

If $A \implies B$ what is possible?

Is $A$ possible? yes but only if $B$ is; If $A$ then $B$.

Is $A$ being false possible? Sure, $A\implies B$ doesn't say anything about what happens if $A$ is false.

Is $B$ possible? Sure, if $A$ then $B$ must be true. And if not $A$ then anything is possible.

Is $B$ false possible? Well, not if $A$ is true it's not but if $A$ is false that's just fine.

What about $A$ and $B$ etc. Well, $A$ and $B$ and $A$ and not $B$ and not $A$ and not $B$ are all possible but only $A$ and not $B$ are impossible. That can't ever happen (if we accept $A \implies B$) so one way or another we are either going to not have $A$ or we will have $B$.

Now do the same for $\lnot A \lor B$. You get the same results. That two statements can only be true under the same conditions.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.