What's the name of this algebraic property? [Complementary Subgroup Test]

I think this comes from the fact that if you have a group $G$ and $H$ a subgroup of $G$ then if $h\in H$ and $x\not\in H$ we get $xh\not\in H$.

The proof is by contradiction, suppose $xh=l$ with $l\in H$. Then postmultiplying by $h^{-1}$ gives $x=lh^{-1}$ which is in $H$ since $H$ is a subgroup of $G$.

I call it the Complementary Subgroup Test, since the composition law arises via the following complementary view of the Subgroup Test ($\rm\color{#c00}{ST} $), $ $ cf. below from one of my old sci.math posts.

Theorem $ $ Let $\rm\,G\,$ be a nonempty subset of an abelian group $\rm\,H,\,$ with complement set $\rm\,\bar G = H\backslash G.\,$ Then $\rm\,G\,$ is a subgroup of $\rm\,H\iff G + \bar G\, =\, \bar G. $

Proof $\ $ $\rm\,G\,$ is a subgroup of $\rm\,H\!\!\overset{\ \large \color{#c00}{\rm ST}}\iff\! G\,$ is closed under subtraction, so, complementing

$\begin{eqnarray} & &\ \ \rm G\text{ is a subgroup of }\, H\ fails\\[.2em] &\iff&\ \rm\ G\ -\ G\ \subseteq\, G\,\ \ fails\\ &\iff&\ \rm\ g_1\, -\ g_2 =\,\ \bar g\ \ \ for\ some\ \ g_i\in G,\ \ \bar g\in \bar G\\ &\iff&\ \rm\ g_2\, +\ \bar g\ \ =\,\ g_1\ for\ some\ \ g_I\in G,\ \ \bar g\in \bar G\\[.2em] &\iff&\ \rm\ G\ +\ \bar G\ \subseteq\ \bar G\ \ fails\qquad\ {\bf QED} \end{eqnarray}$

Instances of this law are ubiquitous in concrete number systems, e.g. below. For many further examples see some of my prior posts here (and also on sci.math).