Proof of Empty Set Subset of Every Set with Absorption Law; Valid?

Question

I know there's a proof by contradiction and a vacuous proof for $\forall S (\varnothing \subseteq S)$,

But I was practising proofs and was wondering if this proof I made up is valid,

$ \begin{array}{ll} \forall x (\neg (x \in \varnothing))\\ \forall x (\neg (x \in \varnothing) \wedge (\neg (x \in \varnothing) \vee (x \in A))) & \texttt{Absorption Law}\\ \forall x (\neg (x \in \varnothing)) \wedge \forall x ((\neg (x \in \varnothing) \vee (x \in A))) & \texttt{Universal Quantifier Distributes Over Conjunction}\\ \forall x ((\neg (x \in \varnothing) \vee (x \in A))) & \texttt{Conjunction Elimination}\\ \forall x (((x \in \varnothing) \rightarrow (x \in A))) & \texttt{Definition of Implication}\\ \varnothing \subseteq A & \texttt{Definition of Subset}\\ \hline \therefore \forall S (\varnothing \subseteq S) & \texttt{Universal Generalization}\\ \end{array} $

Answer 1

Yes, your proof is certainly correct: nice!

You can also do:

$ \begin{array}{ll} \forall x (\neg (x \in \varnothing))\\ \neg (a \in \varnothing) & \texttt{Universal Elimination}\\ \neg (a \in \varnothing) \vee (a \in A) & \texttt{Disjunction Introduction}\\ (a \in \varnothing) \rightarrow (a \in A) & \texttt{Definition of Implication}\\ \forall x ((x \in \varnothing) \rightarrow (x \in A)) & \texttt{Universal Generalization}\\ \varnothing \subseteq A & \texttt{Definition of Subset}\\ \hline \therefore \forall S (\varnothing \subseteq S) & \texttt{Universal Generalization}\\ \end{array} $