<s>[INST] What is the best notation to write ``equations with reasons''? I'm learning my analytic number theory course. When I make notes, I find that it is effective to use ``equations with reasons'' as follow 
$$\begin{array}{rll}
\vartheta(x)&\displaystyle=\sum_{p\leq x} \log p& \textrm{partial sum $\begin{cases}
1 & \textrm{$n$ is prime}\\
0 & \textrm{otherwise}
\end{cases}$ and $\log x$}\\
&\displaystyle =\pi(x)\log x-\int_1^x \frac{\pi(t)}{t}\mathrm{d} t & \textrm{Assume that $\displaystyle\pi(x)=\frac{x}{\log x}+O\left(\frac{x}{\log x}\right)$}\\
&\displaystyle=x-\int_1^x \frac{1}{\log t}\mathrm{d}  t+O\left(x-\int_2^x \frac{1}{\log t}\mathrm{d}  t\right)
& \displaystyle\because\int_2^x \frac{1}{\log t}\mathrm{d}  t\leq \frac{x}{\log x}\leq x \\
&\displaystyle=x+O(x)\\
\end{array}$$
My question is are there any standard on this notations? For example, the reason should put into which line, before the result of the result? And, is it acceptable  to use the symbol $\because$? 
 [/INST] Here are two examples of the notation I use, although I don't think there is just one way to do it.
$$\begin{align*}
c^2 & = a^2 + b^2 & \text{(by Pythagorean Theorem)} \\
& = 2a^2 & \text{(since } a = b \text{)} \\
& = 2 & \text{(since } a = 1 \text{)}
\end{align*}$$
and it's code:
\begin{align*}
    c^2 & = a^2 + b^2 & \text{(by Pythagorean Theorem)} \\
    & = 2a^2 & \text{(since } a = b \text{)} \\
    & = 2 & \text{(since } a = 1 \text{)}
\end{align*}

And the other example:
$$\begin{align*}
c^2 = a^2 + b^2 & \Rightarrow c^4 = (a^2 + b^2)^2 & \text{(square both sides)} \\
& \Rightarrow c^4 = a^4 + 2a^2b^2 + b^4 & \\
& \Rightarrow c^4 = 2a^4 + 2a^2a^2 & \text{(since } a = b \text{)} \\
& \Rightarrow c^4 = 4a^4 &  \\
& \Rightarrow a^4 = \frac{c^4}{4} &  \\
\end{align*}$$
and it's code:
\begin{align*}
    c^2 = a^2 + b^2 & \Rightarrow c^4 = (a^2 + b^2)^2 & \text{(square both sides)} \\
    & \Rightarrow c^4 = a^4 + 2a^2b^2 + b^4 & \\
    & \Rightarrow c^4 = 2a^4 + 2a^2a^2 & \text{(since } a = b \text{)} \\
    & \Rightarrow c^4 = 4a^4 &  \\
    & \Rightarrow a^4 = \frac{c^4}{4} &  \\
\end{align*}

</s>