Suppose these two inequalities hold $$ \begin{cases} (1) \hspace{1cm}Pr(Y=y)\leq Pr(\epsilon\geq -a)\\ (2) \hspace{1cm}Pr(W=w)\leq Pr(\epsilon\leq -a)\\ \end{cases} $$ where $Y,W$ are discrete random variable, $\epsilon$ is a continuous random variable with cdf $F$, and $a\in \mathbb{R}$ is a scalar.
Let $G$ be the cdf of $-\epsilon$. I am confused on whether I am allowed to do the following steps without making any particular assumption on $G$ (like symmetry around zero or similar).
$$ (1) \hspace{1cm} Pr(Y=y)\leq Pr(\epsilon\geq -a) \Leftrightarrow Pr(Y=y)\leq Pr(-\epsilon\leq a) \Leftrightarrow Pr(Y=y)\leq G(a) \Leftrightarrow G^{-1}(Pr(Y=y))\leq a $$
$$ (2) \hspace{1cm} Pr(W=w)\leq Pr(\epsilon\leq -a) \Leftrightarrow Pr(W=w)\leq Pr(-\epsilon\geq a) \Leftrightarrow Pr(W=w)\leq 1-Pr(-\epsilon\leq a)\Leftrightarrow Pr(W=w)\leq 1-G(a) \Leftrightarrow 1-Pr(W=w)\geq G(a) \Leftrightarrow G^{-1}(1-Pr(W=w))\geq a $$
Hence $$ G^{-1}(Pr(Y=y))\leq a\leq G^{-1}(1-Pr(W=w)) $$
They are fine. \begin{align} (1) &\hspace{1cm} Pr(Y=y)\leq Pr(\epsilon\geq -a) \\&\Leftrightarrow Pr(Y=y)\leq Pr(-\epsilon\leq a), (\text{since } \epsilon \ge -a \iff -\epsilon \le a)\\& \Leftrightarrow Pr(Y=y)\leq G(a) ,(\text{definition of }G) \\&\Leftrightarrow G^{-1}(Pr(Y=y))\leq a , (G \text{ and } G^{-1} \text{ are non-decreasing)} \end{align}
\begin{align} (2) &\hspace{1cm} Pr(W=w)\leq Pr(\epsilon\leq -a) \\&\Leftrightarrow Pr(W=w)\leq Pr(-\epsilon\geq a), (\text{since } \epsilon \ge -a \iff -\epsilon \le a)\\& \Leftrightarrow Pr(W=w)\leq 1-Pr(-\epsilon\leq a), (\text{Since } -\epsilon \text{ is continuous})\\&\Leftrightarrow Pr(W=w)\leq 1-G(a),(\text{definition of }G) \\& \Leftrightarrow 1-Pr(W=w)\geq G(a) \text{, (just algebra)}\\& \Leftrightarrow G^{-1}(1-Pr(W=w))\geq a , (G \text{ and } G^{-1} \text{are non-decreasing)} \end{align}
Note that $G^{-1}$ refers to the generalized inverse distribution.