Why $A \rightarrow B \Leftrightarrow \textrm{( A is False or B is True )} $

145 Views Asked by At

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 )} $ ?

4

There are 4 best solutions below

2
On BEST ANSWER

$A \rightarrow B$ is False only when $A$ is True AND $B$ is False. So when is it True?

2
On

You can do this in two ways

  1. Check the truth table for the proposition

$$A \to B \leftrightarrow \neg A \vee B$$

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

  1. 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.

2
On

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.

0
On
  • 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.