Let $E \in \mathcal{F}_{\Xi}$ be an event in $\Xi$. Since $\mathbb{P}(X = Y) = 1$ and since $\{ X \neq Y \} = \{ X = Y \}^{\complement} = \Omega \setminus \{ X = Y \}$, then result R3719: Probability of complement event implies \begin{equation} \mathbb{P}(X \neq Y) = 1 - \mathbb{P}(X = Y) = 1 - 1 = 0 \end{equation} Thus, applying results

(i)	R2060: Probability of set difference
(ii)	R3958: Intersecting subtrahend with minuend preserves set difference
(iii)	R3934: Set intersection distributes over set difference
(iv)	R4637:

yields \begin{equation} \begin{split} \mathbb{P}(X \in E) & = \mathbb{P}(X \in E) - 0 \\ & = \mathbb{P}(X \in E) - \mathbb{P}(X \neq Y) \\ & = \mathbb{P}( \{ X \in E \} \setminus \{ X \neq Y \} ) \\ & = \mathbb{P}[ \{ X \in E \} \setminus (\{ X \in E \} \cap \{ X \neq Y \}) ] \\ & = \mathbb{P}[ (\Omega \cap \{ X \in E \}) \setminus (\{ X \in E \} \cap \{ X \neq Y \}) ] \\ & = \mathbb{P}[ \{ X \in E \} \cap (\Omega \setminus \{ X \neq Y \}) ] \\ & = \mathbb{P}( \{ X \in E \} \cap \{ X = Y \} ) \\ & = \mathbb{P}( \{ Y \in E \} \cap \{ X = Y \} ) \\ & = \mathbb{P}( \{ Y \in E \} \cap \Omega \setminus \{ X \neq Y \} ) \\ & = \mathbb{P}[ (\Omega \cap \{ Y \in E \}) \setminus (\{ Y \in E \} \cap \{ X \neq Y \}) ] \\ & = \mathbb{P}[ \{ Y \in E \} \setminus (\{ Y \in E \} \cap \{ X \neq Y \}) ] \\ & = \mathbb{P}( \{ Y \in E \} \setminus \{ X \neq Y \} ) \\ & = \mathbb{P}(Y \in E) - \mathbb{P}(X \neq Y) \\ & = \mathbb{P}(Y \in E) - 0 \\ & = \mathbb{P}(Y \in E) \end{split} \end{equation} We may also opt to write the above in the equivalent form \begin{equation} \begin{split} \mathbb{P}(X \in E) & = \mathbb{P}(X \in E) - 0 \\ & = \mathbb{P}(X \in E) - \mathbb{P}(X \neq Y) \\ & = \mathbb{P}(X \in E \text{ and not } X \neq Y) \\ & = \mathbb{P}(X \in E \text{ and } X = Y) \\ & = \mathbb{P}(Y \in E \text{ and } X = Y) \\ & = \mathbb{P}(Y \in E \text{ and not } X \neq Y) \\ & = \mathbb{P}(Y \in E) - \mathbb{P}(X \neq Y) \\ & = \mathbb{P}(Y \in E) - 0 \\ & = \mathbb{P}(Y \in E) \end{split} \end{equation} Since $E \in \mathcal{F}_{\Xi}$ was arbitrary, we may conclude $X \overset{d}{=} Y$. $\square$