Why $A \rightarrow B \Leftrightarrow \textrm{( A is False or B is True )} $ I am reading a book about Artificial intelligence and Knowledge representation and there is a logic formula that I cannot explain
Why $A \rightarrow B \Leftrightarrow \textrm{( A is False or B is True )} $ ?
 A: $A \rightarrow B$ is False only when $A$ is True AND $B$ is False. So when is it True?
A: You can do this in two ways

*

*Check the truth table for the proposition

$$A \to B \leftrightarrow \neg A \vee B$$
And conclude it is a tautology, therefore $A \to B \iff \neg A \vee B$


*Or think if they are truly equivalent (if they behave in the same way)


Note that $A \to B$ is false only in the case that $A$ is true and $B$ is false. If $A$ is true, then $\neg A$ are false. So if $\neg A$ and $B$ are false, then $\neg A \vee B$ is false. And it is also the only way to get $\neg A \vee B$ false.
For other combination of truth values for $A$ and $B$ you see that they are true in the same cases.
Therefore they are equivalent because they behave in the same way.

A: This is called material implication. It is a boolean operator like AND or OR. The symbol isn't to be confused with 'metalogically implies' ($\implies$), which isn't an operator.
You can see the logic behind the definition if you read it as:

If A doesn't have it (material possession), then it doesn't matter whether B does or not. If A does have it, then B must too.

A: *

*Suppose $A\to B$ holds.


*$A$ is either false or it is true.  If it is true, then $A\to B$ entails that $B$ is true too.


*So the supposition of $A\to B$ entails that $A$ is false or $B$ is true.


*

*Suppose $A$ is false or $B$ is true

*

*In the case of $A$ being false, then $B$ can be derived under an assumption of $A$ by exploding the contradiction.

*In the case of $B$ being true, then $B$ is still true under an assumption of $A$.



*Therefore, in either of its cases, the supposition entails $A\to B$

Therefore $A\to B$ is equivalent to $A\vee\neg B$ in classical logic.
