State/Every lax basis is contained in exactly two lax superbases

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Definition of Every lax basis is contained in exactly two lax superbases: Suppose that $\{ \pm \vec e, \pm \vec f \}$ is a lax basis. Then, there exist precisely two lax superbases containing $\{ \pm \vec e, \pm \vec f \}$. Equivalently, there exist precisely two primitive lax vectors $\pm \vec g,\pm \vec h$ such that: $$\{ \pm \vec e,\pm \vec f,\pm \vec g \} \mbox{ and } \{\pm \vec e,\pm \vec f,\pm \vec h \} \mbox{ are lax superbases}.$$

Logical Connections

This definition logically relies on the following definitions and statements: Def/Lax basis, Def/Lax superbasis, Def/Lax vector

The following statements and definitions logically rely on the material of this page: Def/Domain topograph

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State/Every lax basis is contained in exactly two lax superbases

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	Suppose that $\{ \pm \vec e, \pm \vec f \}$ is a lax basis.
	Suppose that $\pm \vec g$ is a primitive lax vector. Then $\pm \vec g$ is adjacent to $\pm \vec e$ if and only if there exists an integer $m$ such that: $$\vec g = \pm \vec f + m \vec e.$$ Similarly, $\pm \vec g$ is adjacent to $\pm \vec f$ if and only if there exists an integer $n$ such that: $$\vec g = \pm \vec e + n \vec f.$$ It follows that $\{ \pm \vec e, \pm \vec f, \pm \vec g \}$ is a superbasis if and only if there exist integers $m,n$ such that: $$\vec g = \pm \vec f + m \vec e = \pm \vec e + n \vec f.$$ When $m = n = \pm 1$, the above identities are clearly satisfied, from which we find two superbases containing $\{ \pm \vec e, \pm \vec f \}$. Namely, we find the following two "endpoints": $$\{ \pm \vec e, \pm \vec f, \pm (\vec e + \vec f) \}, \mbox{ and } \{ \pm \vec e, \pm \vec f, \pm (\vec e - \vec f) \}.$$ Furthermore, these are the only possible superbases containing the given basis. Indeed, the equality $\pm \vec f + m \vec e = \pm \vec e + n \vec f$ can be simplified to yield: $$(n \pm 1) \vec f = (m \pm 1) \vec e.$$ If $m,n \neq \pm 1$, then $n \pm 1$ and $m \pm 1$ are nonzero. It follows that $\vec f$ is a nonzero rational multiple of $\vec e$. This contradicts the assumption that $\{ \pm \vec e, \pm \vec f \}$ is a lax basis.
	Hence there are precisely two lax superbases containing the given lax basis.