A Set is a Basis iff it is a Minimal Spanning Set
# Statement
Let $V$ be a Vector Space over Field $F$. Then $S \subset V$ is a Vector Space Basis If and Only If it is a Minimal spanning subset of $V$. That is, there does not exist any spanning $R \subset V$ so that $R \subsetneq S$.
# Proof
($\Leftarrow$) We know $S$ is a Minimal spanning subset of $V$. Suppose it is not a Vector Space Basis of $V$. Since $\text{span} S = V$, we must have that $S$ is Linearly Dependent. Then, because A Linearly Dependent Set contains a vector in the Span of the others, $\exists \mathbf{v} \in S$ so that $\mathbf{v} \in \text{span} R$ for $R = S \setminus \mathbf{v}$. Then we have that $\text{span} R \supset S$, so, because The Span of subset of a Span is also a subset of that Span $$V \supset \text{span} R \supset \text{span} S = V$$ and $\text{span} R = V$. But that would mean $S$ is not Minimal $\unicode{x21af}$. Therefore $S$ is a Vector Space Basis for $V$. $\checkmark$
($\Rightarrow$) We know $S$ is a Vector Space Basis. If $S = \emptyset$, then it is Minimal. Otherwise, suppose $S \neq \emptyset$. Let $\mathbf{v} \in S$ and consider $R \subset S \setminus {\mathbf{v}} \subsetneq S$. Suppose $R$ was a Vector Space Basis for $V$ as well. Then $\mathbf{v} \in \text{span} R = V$, so because The Subspace Span is the Set of all Linear Combinations, there exists $c_{1}, \dots, c_{n} \in F$, $\mathbf{a}{1}, \dots, \mathbf{a}{n} \in R$ for some $n \in \mathbb{Z}{\geq 0}$ so that $$\begin{align*} &\mathbf{v} = c{1} \mathbf{a}{1} + \cdots + c{n} \mathbf{a}{n}\\ \Rightarrow &\mathbf{0} = c{1} \mathbf{a}{1} + \cdots + c{n} \mathbf{a}_{n} - \mathbf{v}, \end{align*}$$ so $R$ is Linearly Dependent. Since $\mathbf{v}$ was arbitrary, $S$ must be Minimal. $\checkmark$ $\blacksquare$