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

Then
\begin{equation}
X \subseteq Y \quad \implies \quad (X \setminus Z) \subseteq (Y \setminus Z)
\end{equation}

Result R49
on D70: Set difference

Isotonicity of subtracting same set

Formulation 0

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

Then
\begin{equation}
X \subseteq Y \quad \implies \quad (X \setminus Z) \subseteq (Y \setminus Z)
\end{equation}

Formulation 1

Let $X$, $Y$, and $Z$ each be a D11: Set such that

(i) | $X \subseteq Y$ |

Then
\begin{equation}
(X \setminus Z) \subseteq (Y \setminus Z)
\end{equation}

Proofs

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

Suppose that $X$ is contained in $Y$. If $X \setminus Z$ is empty, then R7: Empty set is subset of every set guarantees that $X \setminus Z$ is a subset of $Y \setminus Z$.

Suppose then that $X \setminus Z$ is not empty and let $x \in X \setminus Z$. Now, first and foremost, $x$ belongs to $X$. Since $X$ is contained in $Y$, then $x$ belongs also to $Y$. Because we also assumed that $x \not\in Z$, it follows that $x \in Y \setminus Z$. Since $x \in X \setminus Z$ was arbitrary, the claim follows. $\square$

Suppose then that $X \setminus Z$ is not empty and let $x \in X \setminus Z$. Now, first and foremost, $x$ belongs to $X$. Since $X$ is contained in $Y$, then $x$ belongs also to $Y$. Because we also assumed that $x \not\in Z$, it follows that $x \in Y \setminus Z$. Since $x \in X \setminus Z$ was arbitrary, the claim follows. $\square$