# Let A and B represent two linear inequalities: A : a 1 </msub> x

Let A and B represent two linear inequalities:
$A:{a}_{1}{x}_{1}+...+{a}_{n}{x}_{n}\ge {k}_{1}$
$B:{b}_{1}{x}_{1}+...+{b}_{n}{x}_{n}\ge {k}_{2}$
If A and B is unsatisfiable (does not have solution), does the following hold in general (the conjunction of two inequalities implies the summation of them )? If so, I am looking for a formal proof?
$A\wedge B\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}A+B$
${𝑎}_{1}{𝑥}_{1}+...+{𝑎}_{n}{𝑥}_{n}\ge 𝑘1\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\wedge \phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}{𝑏}_{1}{𝑥}_{1}+...{𝑏}_{n}{𝑥}_{n}\ge {𝑘}_{2}\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}{𝑎}_{1}{𝑥}_{1}+...+{𝑎}_{n}{𝑥}_{n}+{𝑏}_{1}{𝑥}_{1}+...{𝑏}_{n}{𝑥}_{n}\ge {𝑘}_{1}+{𝑘}_{2}$
and then I would like to generalize the above theorem to summation of several inequalities.
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Carolyn Farmer
It's still false, even with the unsatisfiability assumption.
Consider the inequalities
$\begin{array}{rl}-2x& >2\\ x& >3\end{array}$
Their sum is
$-x>5$
i.e., $x<-5$. But $x<-5$ does not imply that $x>3$.
###### Not exactly what you’re looking for?
quorums15lep
I don't see how the linear combination part is relevant. $A\ge {k}_{1},B\ge {k}_{2}\to A+B\ge {k}_{1}+{k}_{2}$ regardless of where $A$ and $B$come from. This can be seem by
$A\ge {k}_{2}$
$A-{k}_{1}\ge 0$
$B+\left(A-{k}_{1}\right)\ge B\ge {k}_{2}$
$B+A\ge {k}_{1}+{k}_{2}$