Equivalent definitions of $x_{n} \downarrow a$ for $x:\mathbb N \to \mathbb R$ Let :
$$
x:\mathbb N \to \mathbb R ~,~ x_{n} \downarrow a $$
Is there any mistake with the following statement :
$$\begin{aligned}x_{n} \downarrow a &\iff  \bigwedge_{ε\in {\mathbb R}^{*+}} ~\bigvee_{N \in \mathbb N} ~\bigwedge_{n \in \mathbb N } \bigl( ~~(n > N ~\rightarrow~ x_n - a < ε) ~\land~ x_n \ge a~~\bigr) \\&\iff (x_0 \ge x_1 \ge x_2 \ge ...) \land(\inf_{n\in\mathbb N}(x_n) =a)
\end{aligned}$$
 A: I was mistaken, you're right (about being wrong). If you replace "$x_n\ge a$" with "$x_n\ge x_{n+1}$" for all $n$, then the statement is correct. Precisely, the definition of $x_n\downarrow a$ is

$(1)$ For all $n\in\Bbb N$ we have $x_n\ge x_{n+1}$, and for all $\epsilon>0$, there exists $N\in\Bbb N$ such that $n>N$ implies $|x_n-a|<\epsilon$.

I will show this is equivalent to the following:

$(2)$ $x_n\ge x_{n+1}$ for all $n\in\Bbb N$ and $\inf_n (x_n)=a$.

Suppose $(1)$ holds. We will show $a=\inf_n(x_n)$. By the definition of infimum, we need to show that $a\le x_n$ for all $n$, and if there exists some $b$ with $b\le x_n$ for all $n$, then $b\le a$. The first condition is clear, since $x_n\downarrow a$. For the second condition, suppose $b\le x_n$ for all $n$, and suppose for a contradiction that $b>a$. Let $\epsilon=b-a>0$. By assumption, there exists some $N$ such that $n>N$ implies $|x_n-a|=x_n-a<\frac{\epsilon}{2}$. But then for any such $n$, 
$$b-x_n=(b-a)-(x_n-a)>\epsilon-\frac{\epsilon}{2}=\frac{\epsilon}{2}>0$$
which contradicts our assumption $b\le x_n$. Therefore $b\le a$, and $a=\inf_n(x_n)$.
Now, suppose $(2)$ holds. Let $\epsilon>0$. Then by definition of infimum, there exists some $x_N$ such that $x_N-\epsilon<a$, i.e. $x_n-a<\epsilon$. Since $x_{n+1}\le x_n$ for all $n$, we see in fact that for all $n>N$, we have
$$|x_n-a|=x_n-a\le x_N-a<\epsilon,$$
which shows that $(1)$ holds.
