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Nonempty disjoint linear polyhedra $P=\{x:Ax\le b\}$ and $Q=\{y:Cy\le d\}$ admit $h$ with a strictly positive separation gap. A phase-I linear program minimizing common constraint violation has positive attained optimum. Its Lagrangian dual problem gives nonnegative $\lambda ,\mu$ with $A^T\lambda +C^T\mu =0$ and ${\lambda }^{T}b+{\mu }^{T}d<0$. Then $h=A^T\lambda$ satisfies $h^{T}x\le {\lambda }^{T}b<-{\mu }^{T}d\le h^{T}y$. The polyhedral hypothesis matters; arbitrary disjoint closed convex sets need not have a positive separation gap.

A system of linear inequalities requires several linear inequalities to hold simultaneously. The notation $Ax\le b$ compares coordinates componentwise, giving one inequality per row of the matrix $A$. Its solution set is the intersection of the corresponding half-spaces. For non-strict inequalities it is a linear polyhedron, which may be empty. For example, $x\ge 1$, $y\ge 1$, $x+y\le 1$ are inconsistent: adding their equivalent forms $-x\le -1$, $-y\le -1$, $x+y\le 1$ gives $0\le -1$. A Farkas certificate for linear inequalities generalizes this cancellation.

For a linear polyhedron $P=\{x\in {\mathbb{R}}^n:Ax\le b\}$, a vertex of a linear polyhedron is a feasible point $p$ at which the normals of the active constraints span ${\mathbb{R}}^n$. Equivalently, $n$ linearly independent active constraints uniquely determine $p$. A linear polyhedron may be unbounded or lower-dimensional; the definition applies in those cases as well as to a convex polytope.

Such a point is an extreme point. Indeed, if $p=\lambda u+(1-\lambda )v$ with $u,v\in P$ and $0<\lambda <1$, every active constraint at $p$ must also be an equality at $u$ and $v$. Selecting $n$ independent active rows forces $u=v=p$. Conversely, if the active normals do not span ${\mathbb{R}}^n$, take a nonzero vector $h$ annihilated by all active rows. The finitely many inactive inequalities have positive slack, so both $p+\epsilon h$ and $p-\epsilon h$ remain feasible for sufficiently small $\epsilon >0$. Their midpoint is $p$, so $p$ is not an extreme point. Thus these notions coincide for a linear polyhedron.