"The following are equivalent" What does it mean for several statements to be equivalent? And why does it suffice to prove a "cyclic" chain
$$A_1\implies A_2\implies \cdots\implies A_n\implies A_1$$
in order to show that the conditions $A_1, \dots, A_n$ are equivalent?
 A: To say that several statements are equivalent means that if one is true, then all the statements are true, and if one of them is false, they're all false.  Now look at the cyclic chain of implications that you wrote.  Suppose some statement, say $A_k$, is true.  Then $A_{k+1}$ is true, so $A_{k+2}$ is true, and so on.  That is, if one of them is true, they're all true.  Similarly, if $A_k$ is false then $A_{k-1}$ must be false, so $A_{k-2}$ must be false, and so on.  So we can check all of the statements by just checking one of them.
A: If you have shown $$A_1\Longrightarrow A_2 \Longrightarrow\ldots\Longrightarrow A_n $$
To show equivalence of all of these statements it suffices to finally show $A_n \Longrightarrow A_1$, because the former implications all hold and so you get equivalence between any two statements. 
For instance, $A_2\iff A_5$ because $A_2\Longrightarrow A_3\Longrightarrow\ldots\Longrightarrow A_5$ means you have $A_2\Longrightarrow A_5$
but by completing the cycle you also have $A_5\Longrightarrow A_6\Longrightarrow\ldots\Longrightarrow A_n\Longrightarrow A_1\Longrightarrow A_2$ which is $A_5\Longrightarrow A_2$.
A: Suppose we have shown that $A_1 \implies A_2 \implies \cdots \implies A_n \implies A_1$.
Let $t(\varphi) \in \{0,1\}$ denote the truth value of $\varphi$.
Then the infinite sequence given by
$$
t(A_1), t(A_2), t(A_3), \ldots, t(A_n), t(A_1), t(A_2), \ldots t(A_n), t(A_1), \ldots
$$
is periodic and non-decreasing, so it is constant.
That is, either all $A_i$ are true, or all $A_i$ are false.
A: Two statements are equivalent if
$$A_1\Rightarrow A_2\text{ and } A_2\Rightarrow A_1.$$
More than two statements are equivalents if any two of them are equivalent.
So if you have for instance 
$$A_1\Rightarrow A_2\Rightarrow A_3\Rightarrow A_1$$
then
$$A_2\Rightarrow A_1$$
so $A_1\iff A_2$. In the same way you could show that $A_2\iff A_3$ and $A_1\iff A_3$.
It is the basic idea, and it is working for any chain of length $n$ (you can prove it by induction).
