I would like to receive some help about the next problem:
The problem:
In this book, at the page 12, i have a problem understanding how the first picture represents the implication:
http://rvukovic.net/teka/mat_1.pdf
It's about the flow of electric current regulated with switches.
1) If $A$ is on ($A$ is true) and $B$ is on ($B$ is true), the electricity will continue to flow ($\top \Rightarrow \top = \top$).
2) If $A$ is on and $B$ is off, the electricity shouldn't continue to flow ($\top \Rightarrow \bot = \bot$), but beacuse it flows through switch $A$ it will go around the switch $B$ and it will continue to flow.
3) If $A$ is off and $B$ is on, the electiricty will flow ($\bot \Rightarrow \top = \top$).
4) If $A$ is off and $B$ is off, i guess that electricity should flow ($\bot \Rightarrow \bot = \top$), but $A$ and $B$ are off.
The question:
Please, could you tell me if this picture of implication switch is correct?
Thank you for your time and help!

Here is the relevant picture:
Based on the diagram for the $\Leftrightarrow$, I have to believe that there is a difference between a line going horizontally through a variable box as opposed to vertically. Specifically, since we know that
$$A \Leftrightarrow B \equiv (A \land B) \lor (\neg A \land \neg B)$$
and since
$$A \Leftrightarrow B \equiv \neg A \lor B$$
it must be that the horizontal direction takes on the value of the variable as is, whereas the vertical line negates the value.
To be even more clear: if there are multiple paths going through the circuit, then that means a $\lor$: to go through the circuit, you can either take this path, or some other path.
So, for example, for the $\Leftrightarrow$, you can either horizontally go through $A$ after which you then also have to horizontally go through $B$ (so that corresponds to $A \land B$, or you can go vertically through $A$, but then you also have to vertically go through $B$. So, that path corresponds to $\neg a \land \neg B$. And so since we have a choice of paths, we get $(A \land B) \lor (\neg A \land \neg B)$, which is exactly right.
As such, the diagram for the $\Rightarrow$ is also correct: to go through the circuit, we have two options: either go vertically through $A$, so that's $\neg A$, or go horizontally through $B$, so that's just $B$, and with those two options you thus get $\neg A \lor B$, which again is what you want, because if you yourself indicated: the only way for the $\Rightarrow$ to be false is for neither $A$ to be false nor $B$ to be true.