Let $X$ and $Y$ each be a D11: Set.

Then
\begin{equation}
X \setminus Y \subseteq X
\end{equation}

Result R47
on D70: Set difference

Set difference is subset of the minuend

Formulation 0

Let $X$ and $Y$ each be a D11: Set.

Then
\begin{equation}
X \setminus Y \subseteq X
\end{equation}

Proofs

Let $X$ and $Y$ each be a D11: Set.

If $X \setminus Y$ is empty, then it is a subset of $X$ due to R7: Empty set is subset of every set. If $X \setminus Y$ is not empty, then $x$ being an element of $X \setminus Y$ implies that $x$ is an element of $X$ by the definition of set difference. Since this is true for every $x \in X \setminus Y$, the claim follows. $\square$