Let $A \subseteq \{0,1\}^n$ be a set of vectors that satisfy some (in)equality properties, e.g.,
$$ \begin{aligned} A &= \left\{ x \in \{0,1\}^n : x_1 = x_2 \right\} \\ A &= \left\{ x \in \{0,1\}^n : x_1 < x_2, x_3 + x_4 = 1 \right\} \end{aligned} $$
More abstractly, in logical language,
$$ A = \left\{ x \in \{0,1\}^n : \varphi_1(x),\dots, \varphi_m(x) \right\} $$
where $\varphi_i$ is a property of the $x_i$'s only involving "$=$", "$+$", "$<$" and numerical constants. Sorry that I can't be more precise.
I am curious if it's true that the convex hull of $A$ (assuming $A$ non-empty in light of the counter-example that @daw gave below) contains all and only vectors in $[0,1]^n$ that satisfy the same properties. The convex hull of $A$ would be a subset of these vectors, because these properties are preserved under linear combination. So the hard part is to prove the converse. This seems to be true in simple cases, e.g. when $A=\{x\in\{0,1\}^n: x_1=x_2\}$, then $\mathrm{conv}(A)=\{x\in [0,1]^n: x_1=x_2\}$. Any thought/reference pointer would be greatly appreciated.
This is not true. Take $$ A =\{ x \in \{0,1\} :\ 2x_1=1\}, $$ then $A= \emptyset$ and $\{ x \in [0,1] :\ 2x_1=1\}=\{1/2\}$.