Proof simplification I am  tasked with proving the following:
$$\varnothing   - A =   \varnothing  $$
My Attempt :
Assume there exist $x  \in $$\varnothing   - A $ then 
$$ x \in \varnothing   - A \Rightarrow x \in \varnothing \land  x \notin  A$$
However  by definition of the  empty set,  the cannot exist $x$ such that $x \in \varnothing $
and hence there doesn't exist a $x$ s.t $x \in \varnothing - A$ 
Is this correct? If so is there a more concise/rigorous way to represent this?
 A: Yes you are right, another way can be : If $A$ and $B$ are such that $A\subseteq B$, then $A\setminus B=\phi$. Since $\phi$ is a subset of every set, your question follows.
A: It looks correct. One could also say that $\varnothing - A$ must be a subset of $\varnothing$, and hence the empty set itself.
A: With bother with elements? 
Another solution might be
$$
B-A \subseteq B
\\
\emptyset\subseteq A
$$for every $A$ and $B$.
It gives you 
$$
\emptyset-A \subseteq \emptyset
,\,\,\,
\emptyset\subseteq \emptyset-A\implies \emptyset=\emptyset-A $$
A: $
\newcommand{\calc}{\begin{align} \quad &}
\newcommand{\calcop}[2]{\\ #1 \quad & \quad \text{"#2"} \\ \quad & }
\newcommand{\endcalc}{\end{align}}
$Here is just another way to write your (correct) proof.
For any $\;x\;$,
$$\calc
x \in \varnothing - A
\calcop{\equiv}{definition of $\;-\;$}
x \in \varnothing \land x \not\in A
\calcop{\equiv}{definition of $\;\varnothing\;$}
\text{false} \land x \not\in A
\calcop{\equiv}{logic: simplify}
\text{false}
\endcalc$$
By the definition of $\;\varnothing\;$, this is equivalent to $\;\varnothing - A = \varnothing\;$.
