Let A and B represent two linear inequalities: A : a 1 x 1 + . . . + a n x n ≥ k 1 B : b 1 x 1 + . . . + b n x n ≥ 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 ∧ B ⟹ A + B 𝑎 1 𝑥 1 + . . . + 𝑎 n 𝑥 n ≥ 𝑘 1 ∧ 𝑏 1 𝑥 1 + . . . 𝑏 n 𝑥 n ≥ 𝑘 2 ⟹ 𝑎 1 𝑥 1 + . . . + 𝑎 n 𝑥 n + 𝑏 1 𝑥 1 + . . . 𝑏 n 𝑥 n ≥ 𝑘 1 + 𝑘 2 and then I would like to generalize the above theorem to summation of several inequalities.

Question

Let A and B represent two linear inequalities:  A  :      a    1        x    1    +  .  .  .  +      a    n        x    n    ≥      k    1    B  :      b    1        x    1    +  .  .  .  +      b    n        x    n    ≥      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  ∧  B    ⟹    A  +  B        𝑎    1          𝑥    1    +  .  .  .  +        𝑎    n          𝑥    n    ≥  𝑘  1      ∧            𝑏    1          𝑥    1    +  .  .  .        𝑏    n          𝑥    n    ≥        𝑘    2      ⟹          𝑎    1          𝑥    1    +  .  .  .  +        𝑎    n          𝑥    n    +        𝑏    1          𝑥    1    +  .  .  .        𝑏    n          𝑥    n    ≥        𝑘    1    +        𝑘    2  and then I would like to generalize the above theorem to summation of several inequalities.

Carolyn Farmer · Accepted Answer

It&#039;s still false, even with the unsatisfiability assumption.Consider the inequalities                    −        2        x                            &amp;gt;        2                            x                            &amp;gt;        3            Their sum is  −  x  &amp;gt;  5i.e.,   x  &amp;lt;  −  5. But   x  &amp;lt;  −  5 does not imply that   x  &amp;gt;  3.

quorums15lep · Accepted Answer

I don&#039;t see how the linear combination part is relevant.   A  ≥      k    1    ,  B  ≥      k    2    →  A  +  B  ≥      k    1    +      k    2   regardless of where   A and   Bcome from. This can be seem by  A  ≥      k    2    A  −      k    1    ≥  0  B  +  (  A  −      k    1    )  ≥  B  ≥      k    2    B  +  A  ≥      k    1    +      k    2

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

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