Let $\langle a,x\rangle=b$ be a linear constraint where $x\in\mathbb R^n$ and every entry in $a=(a_1,\dots,a_n)$ and the entry $b$ are of $m=poly(n)$ bitlength and are in $\mathbb Z_{\geq0}^{n}$ and $\mathbb Z_{\geq0}$ (non-negative) respectively. Is there an universal polyhedron $By\leq c$ (depending on $m,n$) satisfying the properties
$B\in\mathbb Z^{q\times(n+1)},c\in\mathbb Z^{q}$ where $q=poly(n)$
$\log_2\max_{i,j}|B_{i,j}c_i|=poly(n)$
$\exists x\in\mathbb R_{\geq0}^n:\langle a,x\rangle=b\iff B[a,b]'\leq c$?
At least is there a $B,c$ depending on $n$ alone if $\exists x\in\mathbb R_{\geq0}^n:\langle a,x\rangle=b$ satisfies $\exists!x\in\mathbb R_{\geq0}^n:\langle a,x\rangle=b$ (uniqueness property)?
By continuity uniqueness on positive region implies uniqueness on entire $\mathbb R^n$. But we require non-negative region and so the $\langle a,x\rangle=b$ might not be unique valued in $x\not\in\mathbb R_{\geq0}^n$ ($x\in\mathbb R_{\geq0}^n:\langle a,x\rangle=b$ might be a corner point for the polyhedron $\langle a,x\rangle=b$).