Vectors are Right Cancellable

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Theorem

Let $\struct {\mathbf V, +, \circ}$ be a vector space over $\GF$, as defined by the vector space axioms.

Then every $\mathbf v \in \struct {\mathbf V, +}$ is right cancellable:

$\forall \mathbf a, \mathbf b, \mathbf c \in \mathbf V: \mathbf a + \mathbf c = \mathbf b + \mathbf c \implies \mathbf a = \mathbf b$


Proof

Utilizing the vector space axioms:

\(\ds \mathbf a + \mathbf c\) \(=\) \(\ds \mathbf b + \mathbf c\)
\(\ds \leadsto \ \ \) \(\ds \paren {\mathbf a + \mathbf c} - \mathbf c\) \(=\) \(\ds \paren {\mathbf b + \mathbf c} - \mathbf c\)
\(\ds \leadsto \ \ \) \(\ds \mathbf a + \paren {\mathbf c - \mathbf c}\) \(=\) \(\ds \mathbf b + \paren {\mathbf c - \mathbf c}\)
\(\ds \leadsto \ \ \) \(\ds \mathbf a + \bszero\) \(=\) \(\ds \mathbf b + \bszero\)
\(\ds \leadsto \ \ \) \(\ds \mathbf a\) \(=\) \(\ds \mathbf b\)

$\blacksquare$


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