MHB How Can Two Inequality Sets Be Combined into One?

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Combining two inequality sets A and B into a new set C using the symmetric difference operation can lead to non-convex solutions. The solutions to inequalities in two variables form convex semi-planes, and intersections of convex sets remain convex. However, applying the symmetric difference can result in a non-convex set, complicating the representation as linear inequalities. The discussion highlights the challenges in merging these sets while maintaining their properties. Ultimately, the resulting set C may not be expressible as a simple combination of linear inequalities.
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Hi everyone, let's stay I have two inequation set such as:

First one is A:=
$$X_1-X_2 \leq 1$$
$$X_1 \leq3$$
$$X_2 \geq 1$$
$$X_1,X_2 \geq 0$$

Second one is B:=
$$X_1+X_2 \geq 5$$
$$X_1\leq5$$
$$X_1\geq4$$
$$X_2\leq4$$
$$X_1,X_2 \geq 0$$

I had like to write it as a set $$C := A\oplus B$$, with C made of linear inequations too. I'm not so sure of how to tackle such problem, if anyone can help!
 
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re: Two inequation set into one.

What do you mean by $\oplus$?
 
re: Two inequation set into one.

the XOR operation, sorry I should have said so!
 
Re: Two inequation set into one.

The set of solutions to an inequality in two variables is a semi-plane. In particular, it is convex. Therefore, the set of solutions to several inequality is also convex as an intersection of convex sets. On the other hand, symmetric difference can act as set difference when one of the sets is inside another. Thus, it can turn two convex sets into a non-convex set. Therefore, the result is not always representable as the set of solutions of linear inequalities.
 

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