Let $X, A, B \in \text{Random}(\mathbb{R})$ each be a D3161: Random real number such that

(i)	\begin{equation} \mathbb{P}(A \leq B) = 1 \end{equation}

Let $E$ be an event given by $E : = \{ A > B \}$. Since $\mathbb{P}(A \leq B)$ and since \begin{equation} E = \{ A > B \} = \Omega \setminus \{ A \leq B \} \end{equation} then results

(i)	R2090: Isotonicity of probability measure
(ii)	R3719: Probability of complement event

imply \begin{equation} \mathbb{P}(E) = \mathbb{P}(A > B) = \mathbb{P}(\Omega \setminus \{ A \leq B \}) = \mathbb{P}(\Omega) - \mathbb{P}(A \leq B) = 1 - 1 = 0 \end{equation} If $\omega \in \{ X \leq A \} \setminus E$, then \begin{equation} X(\omega) \leq A(\omega) \leq B(\omega) \end{equation} Thus we have the inclusion \begin{equation} \{ X \leq A \} \setminus E \subseteq \{ X \leq B \} \setminus E \end{equation} To conclude, applying results

(i)	R2060: Probability of set difference
(ii)	R2090: Isotonicity of probability measure

we have \begin{equation} \begin{split} \mathbb{P}(X \leq A) & = \mathbb{P}(X \leq A) - 0 \\ & = \mathbb{P}(X \leq A) - \mathbb{P}(E) \\ & = \mathbb{P}(\{ X \leq A \} \setminus E) \\ & \leq \mathbb{P}(\{ X \leq B \} \setminus E) \\ & = \mathbb{P}(X \leq B) - \mathbb{P}(E) \\ & = \mathbb{P}(X \leq B) - 0 \\ & = \mathbb{P}(X \leq B) \end{split} \end{equation}