Two "expressions" one of them isn't a statement 
*

*$((A_1)) \rightarrow (A_2)$

*$((A_1) \rightarrow (A_2)) \rightarrow (A_3)$
My book says that 1 isn't a statement because:
The first symbol is $($. By counting the number of parenthesis, the "main place" is the sixth symbol (Which is the first place where our sum is $0$ if $($ adds one and $)$ subtracts one.  which is: $\rightarrow$). This symbol is separating two expressions in the form $( \alpha )$ and thus we can split it into:

The left expression $(A_1)$ is starting with $($, but the "weight" is $0$ only at the last parenthesis and thus 1 is not a statement. (Weight = we start counting '(' adds one to the sum and ')' subtracts one)
However, when we prove that 2 is a statement we say that we need to split it by the tenth place (second $\rightarrow$) into:

But I don't get it, why is $(A_3)$ can be just $A_3$ but when we had the $(A_1)$ above (first picture) it didn't become just $A_1$ ...
Why is that? I really don't get it... Thanks for helping!
 A: Your rules say that if $p$ and $q$ are statements, then so is $(p)\to(q)$, but nowhere do they say or imply that $(p)$ is a statement. They also say that the atoms $A_i$ are statements. Thus, using the rules you can form a statement $(A_1)\to(A_2)$; call that statement $p$. You can then go on and form a statement $(p)\to(A_3)$, which in full is $((A_1)\to(A_2))\to(A_3)$. But you cannot form $((A_1))\to(A_2)$. That would be justifiable only if $(A_1)$ and $A_2$ were both statements, and while $A_2$ is a statement, $(A_1)$ is not.
In short, you can have $(A_i)$, but only as a component of a compound statement involving $\to$.
A: When splitting your statements you're undoing the possible construction that led to that statement, i.e., splitting $(\alpha)\to(\beta)$ gives you $(\alpha)$ and $(\beta)$, from which you take $\alpha$ and $\beta$ as statements. Note that we removed $\textit{one occurrence}$ of the parentheses (the only one we can see there really).
In the first case you see that splitting the statement gives you $((A_1))$, which results in $(A_1)$ being a statement, since we just have to remove $\textit{one occurrence}$ of the parentheses. This is false, since $(A_1)$ comes from adding parenthesis to an atom, and that's not valid. Think that inside the first occurrence of a parenthesis there has to be a statement, and that's not the case.
In your second case splitting the statement gives you $(A_3)$, which results in $A_3$ being a statement, and there's nothing wrong with that.
