I've posted questions about stationarity, but I cannot get answers satisfying me because of my vague question. Thus, I read more times about definitions about stationarity, summed them up, and brought out distinctly what I was concerning. I extremely and hopefully desire someone to get rid of my misconception and curiosities. I hope you to understand what my questions exactly are because I spend a large amount of time translating into English (, even if there are many grammatical errors).
$1^{st}$ question: Are there any errors in my summary?
A random process is said to be first-order stationary if it satisfies the following equation:$$\forall \tau\in\mathbb{R}, \quad f_X(x_{t_1}) = f_X(x_{t_1+\tau})$$ , which means that expected value of process is independent of any time shift. Therefore, $E\left[X(t)\right]$ is constant.
Next, a random process is said to be second-order stationary or strict-sense stationary (SSS) if it satisfies the following equation:$$\forall \tau\in\mathbb{R}, \quad f_X(x_{t_1}, x_{t_2}) = f_X(x_{t_1+\tau}, x_{t_2+\tau})$$ , which means that the process depends only on the time difference. Therefore, $R_X(t_1, t_2)=R_X(t_1-t_2)=R_X(\tau)$, where $\tau$ is the time difference between $t_1$ and $t_2$.
In addition, a random process is said to be wide-sense stationary (WSS) if it satisfies the following two equations: \begin{align} &1. \quad E\left[X(t)\right] = k, \quad \text{where } k \text{ is constant}\\ &2. \quad R_X(t_1, t_2)=R_X(t_1-t_2)=R_X(\tau), \quad \text{where } \tau \text{ is } t_1-t_2 \end{align}
Note that a second-order stationary (or SSS) process will always be WSS; however, the reverse will not always hold true.
$2^{nd}$ question: In SSS section, how come $R_X(t_1, t_2)$ becomes $R_X(\tau)$?
I tried to prove that $R_X(t_1, t_2)$ becomes $R_X(\tau)$. I think it's almost done but I do not agree with the last equality. How come two domains change into one domain? By verbal expression, which is less logical than mathematical expression, I agree with that it could be because it just depends only on $\tau$. However, I want to get to agree this from thorough mathematical expression.
\begin{align} R_X(t_1, t_2) = E\left[X(t_1)X(t_2)\right] &= \int_{-\infty}^{\infty}\int_{-\infty}^{\infty}{xy f_{X_{t_1},X_{t_2}}(x, y)}dxdy\\ &= \int_{-\infty}^{\infty}\int_{-\infty}^{\infty}{xy f_{X_{t_1-t_2},X_{0}}(x, y)}dxdy\\ &= E\left[X(t_1-t_2)X(0)\right] = R_X(t_1-t_2, 0) \stackrel{?}{=} R_X(\tau) \end{align}
$3^{rd}$ question: a process is SSS 
the process is WSS
From my summary (originally from websites, textbooks and so on), it seems to me that the necessary conditions of WSS requires to satisfy simultaneously the necessary condition of first-order stationrity and one of second-order stationarity(SSS). If my view is correct, WSS will always be SSS, the reverse will not always hold true, isn't it?
