I'm having a problem with Mathematica.

I have a large and complicate system of 3 inequalities in 3 variables, let's call them x,y,z.

Now, I can't use reduce to solve it exactly.

There is a way to define the region of x,y such as the system is satisfied for all z?

I tried

ImplicitRegion[ForAll[z,z :elem: Reals, sys],{x,y}]

of course, it doesn't work.

I need this region for a minimization process.

Thanks for everyone that will help.

EDIT: Here is an example

sys = {};
AppendTo[sys, x^2*y*(1 + z^2 + x*z^4) >= 0];
AppendTo[sys, z^2*y*x^2 <= 0];
AppendTo[sys, z^4 + y^3 + x^8 >= 0];

Now I want to define the region of the plane x,y where the inequalities hold for all the z. For example, the point (x,y)=(1,1) is in this region, while (x,y)=(-1,1) is not.

Of course, this is just a toy problem, you can use reduce and find the solution, but in my real problem, I can't use it. This is just to give you some ideas.

Another thing, if it is too hard to find the region of the whole plane x,y, you can restrict to a bounded region for example [-1,1]x[-1,1].

  • 1
    $\begingroup$ Please provide a concrete example. $\endgroup$ – Bob Hanlon Jun 25 '20 at 0:41
  • $\begingroup$ I've edited as you asked. Hope it is enough. $\endgroup$ – Eso Jun 25 '20 at 7:42
sys = ForAll[z, And @@ {
     x^2*y*(1 + z^2 + x*z^4) >= 0,
     z^2*y*x^2 <= 0,
     z^4 + y^3 + x^8 >= 0

reg = ImplicitRegion[Resolve[sys, Reals], {x, y, z}]

(* example minimization *)
Minimize[x^4, {x, y, z} ∈ reg]

The system can be reduced further:

(* y == 0 || (x == 0 && x^8 + y^3 >= 0) *)
(* y == 0 || (y >= 0 && x == 0) *)
  • $\begingroup$ Thanks, but why Resolve would work while Reduce doesn't? ( There are so many things that I don't know...) $\endgroup$ – Eso Jun 25 '20 at 12:01
  • $\begingroup$ @Eso Resolve takes out the quantifiers. See the documentation attempts to resolve expr into a form that eliminates ForAll and Exists quantifiers $\endgroup$ – flinty Jun 25 '20 at 12:03
  • $\begingroup$ First of all, Thanks. I tried but it doesn't work for my system, it gets blocked at the resolve part. So I think we can't play with a toy system, so I want to post the real system but it is large, and copy and paste don't work. When I will understand how to do it I will post it. $\endgroup$ – Eso Jun 25 '20 at 13:59
  • $\begingroup$ @Eso - Reduce works. Simplify@Reduce@sys evaluates to (x == 0 && y >= 0) || (y == 0 && x != 0) which is equivalent to what @flinty obtained since the excluded point {0, 0} in the second alternative is covered by the first alternative. $\endgroup$ – Bob Hanlon Jun 25 '20 at 14:54
  • $\begingroup$ When I said that reduce doesn't work, I mean on the real system that I have and not on the toy problem that I proposed just as an example. Of course, it works for the toy problem, I said it into the post, and flinty shows us that works better after resolve. $\endgroup$ – Eso Jun 25 '20 at 15:26

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