Understanding the implication switch 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!
 A: 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.
A: I don't think that the implication switch is correct. It is the same as the OR switch on the previous page.
The following would be the implication switch:

Here $\overline A$ is the switch which is closed (leads the current) if $A$ is false. 
A: I didn't read well what is written in the books. Now that i read it properly, i can say that i'm even more confused:
This is what it says in the book:
"1.5 Switches
One of the unusual uses of algebraic logic is in the modelling of the electric circuits (sets, frames) with switches. In the electric power circuit, to the predicate TRUE, we assign the switch that allows flow of electricity. To the predicatee FALSE, we assing the off state of the switch. 
To the constant TRUE corresponds the horizontal and to the constant FALSE corresponds vertical position of the power guide in the switch. In this case, constant TURE in horizontally drawn conductor means that the power can run and in the vertically drawn conductor that the power can't run."
I guess that the power giude is the conductor (Some kind of wire in the real world, i don't know much about electricity. I hope that it isn't necessary in this case.).
How i understood it:
To the switch on the horizontal line we assign on state if $p = TRUE$ and off state if $p =FALSE$.
To the switch on the vertical line we assign on state if $\lnot p = TRUE$ and off state if $ \lnot p = FALSE$.
Please, could you tell me if i understood it correctly and if not where i amde a mistake?
