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

measgachyx5q9 2022-05-15 Answered
Let A and B represent two linear inequalities:
A : a 1 x 1 + . . . + a n x n β‰₯ k 1
B : b 1 x 1 + . . . + b n x n β‰₯ 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
π‘Ž 1 π‘₯ 1 + . . . + π‘Ž n π‘₯ n β‰₯ π‘˜ 1 ∧ 𝑏 1 π‘₯ 1 + . . . 𝑏 n π‘₯ n β‰₯ π‘˜ 2 ⟹ π‘Ž 1 π‘₯ 1 + . . . + π‘Ž n π‘₯ n + 𝑏 1 π‘₯ 1 + . . . 𝑏 n π‘₯ n β‰₯ π‘˜ 1 + π‘˜ 2
and then I would like to generalize the above theorem to summation of several inequalities.
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Answers (2)

Carolyn Farmer
Answered 2022-05-16 Author has 16 answers
It's still false, even with the unsatisfiability assumption.
Consider the inequalities
βˆ’ 2 x > 2 x > 3
Their sum is
βˆ’ x > 5
i.e., x < βˆ’ 5. But x < βˆ’ 5 does not imply that x > 3.
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quorums15lep
Answered 2022-05-17 Author has 4 answers
I don't see how the linear combination part is relevant. A β‰₯ k 1 , B β‰₯ k 2 β†’ A + B β‰₯ k 1 + k 2 regardless of where A and Bcome from. This can be seem by
A β‰₯ k 2
A βˆ’ k 1 β‰₯ 0
B + ( A βˆ’ k 1 ) β‰₯ B β‰₯ k 2
B + A β‰₯ k 1 + k 2
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