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

measgachyx5q9

measgachyx5q9

Answered question

2022-05-15

Let A and B represent two linear inequalities:
A : a 1 x 1 + . . . + a n x n โ‰ฅ k 1
B : b 1 x 1 + . . . + b n x n โ‰ฅ 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 โˆง B โŸน A + B
๐‘Ž 1 ๐‘ฅ 1 + . . . + ๐‘Ž n ๐‘ฅ n โ‰ฅ ๐‘˜ 1 โˆง ๐‘ 1 ๐‘ฅ 1 + . . . ๐‘ n ๐‘ฅ n โ‰ฅ ๐‘˜ 2 โŸน ๐‘Ž 1 ๐‘ฅ 1 + . . . + ๐‘Ž n ๐‘ฅ n + ๐‘ 1 ๐‘ฅ 1 + . . . ๐‘ n ๐‘ฅ n โ‰ฅ ๐‘˜ 1 + ๐‘˜ 2
and then I would like to generalize the above theorem to summation of several inequalities.

Answer & Explanation

Carolyn Farmer

Carolyn Farmer

Beginner2022-05-16Added 16 answers

It's still false, even with the unsatisfiability assumption.
Consider the inequalities
โˆ’ 2 x > 2 x > 3
Their sum is
โˆ’ x > 5
i.e., x < โˆ’ 5. But x < โˆ’ 5 does not imply that x > 3.
quorums15lep

quorums15lep

Beginner2022-05-17Added 4 answers

I don't see how the linear combination part is relevant. A โ‰ฅ k 1 , B โ‰ฅ k 2 โ†’ A + B โ‰ฅ k 1 + k 2 regardless of where A and Bcome from. This can be seem by
A โ‰ฅ k 2
A โˆ’ k 1 โ‰ฅ 0
B + ( A โˆ’ k 1 ) โ‰ฅ B โ‰ฅ k 2
B + A โ‰ฅ k 1 + k 2

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