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I am giving a really simple example that describes my problem really clear.

Example: "x + b < 50" for "3 < x < 6"

i need a solution set for my coefficient (e.g. for "b") that satisfies all possible values of the defined domain of a variable (e.g. "3 < x < 6") for the given function (e.g. "x + b < 50").

I need one simple output "b<=44" but i still have not been able to get a result without the defined domain variable intruding in the solution (e.g. (b <= 44&& 3 < x < 6)||(44 < b < 47 && 3 < x < 50-b))

The code below that should normally work produces no results:

FullSimplify[Reduce[ { x + b < 50 , x > 3 , x< 6 }, {b}, Reals, Backsubstitution -> True], x > 3 && x<6]

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  • $\begingroup$ CylindricalDecomposition[x + b < 50 && 3 < x < 6, {b, x}] almost gets there. $\endgroup$ – J. M.'s technical difficulties Sep 5 '17 at 11:52
  • $\begingroup$ Reduce[x + b < 50 && 3 < x < 6, {b,x}] does the same =/ unfortunately like everything i have tried so far they include x in their solutions $\endgroup$ – Panagiotis Kmd Sep 5 '17 at 11:58
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    $\begingroup$ When cross-posting, please link the threads together: community.wolfram.com/groups/-/m/t/1177429 $\endgroup$ – Szabolcs Sep 5 '17 at 12:38
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You can use ForAll for this kind of problem:

Reduce[ForAll[x, 3 < x < 6, x + b < 50], b]
(* b <= 44 *)
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  • $\begingroup$ That's exactly what i needed! You can't imagine how much time i spent on it, its really not that intuitive! Thanks a lot $\endgroup$ – Panagiotis Kmd Sep 5 '17 at 12:42

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