Question about a Deduction in Modal Logic

Is the following deduction valid, and why?

1) A → B
2) ◇A
∴ ◇B

Attempt:

1) in every world with A, you also have B.

2) there is at least one world with A.

The conclusion, there is at least one world with B.

• Perhaps you should give an attempt? – user400188 Jul 9 '17 at 13:29
• Okay, 1) in every world with A, you also have B. 2) there is at least one world with A. The conclusion, there is at least one world with B. – Philip White Jul 9 '17 at 13:31

The deduction is not valid. The problem with your attempt is your first step. You write: $A \rightarrow B$ and your explanation of this formula is: "in every world with A, you also have B." However, this explanation is the correct explanation for a different formula, namely $\Box (A \rightarrow B)$. The formula $A \rightarrow B$ does not say anything about other worlds because it is not a modal formula.
Another way of seeing that the deduction is not valid is that you can take the negation of your conclusion and add it to your premises without obtaining an inconsistent premise set. If you took the negation of $\Diamond B$ you would get $\neg \Diamond B$ which is the same as $\Box \neg B$. From $\Box \neg B$ you can infer that $\neg B$ must be true. If we know that $\neg B$ is true we can apply modus tollens to $A \rightarrow B$ which gives us $\neg A$. $\neg A$ allows you to derive $\Diamond \neg A$. And so now you can see that your premise set would contain both $\Diamond A$ and $\Diamond \neg A$ which is not inconsistent because there can be an accessible world at which A is true and an accessible world at which $\neg A$ is true without there being an accessible world at which both A and $\neg A$ are true.