Showing that $\{\overrightarrow x, \overrightarrow y, \overrightarrow z\}$ is linearly independent

Question

If the field $\Bbb{F}_2$ is a set with 2 elements $\{0,1\}$ and the addition and multiplication operations are defined by $$0+0=1+1=0 , 1+0=0+1=1 , 0*0=0*1=1*0=0 , 1*1=1$$ and each element is its own additive inverse. If we suppose that $\overrightarrow x, \overrightarrow y, \overrightarrow z \in V$ for some vector space $V$ and $\overrightarrow z \neq \overrightarrow 0$ and $\{\overrightarrow x, \overrightarrow y\}$ is linearly independent; how would I show that $\{\overrightarrow x, \overrightarrow y, \overrightarrow z\}$ is linearly independent if and only if $\overrightarrow z \notin span \{\overrightarrow x, \overrightarrow y\}$

Answer 1

If $\vec{z} \in span{\vec{x},\vec{y}} $ then it is linearly dependent, by definition. If it is linearly dependent, $\exists (\lambda,\mu), \vec{z} = \lambda \vec{x} + \mu\vec{y}$, which means it is in the span. By taking the inverse of the proposition, you find the said formula.