I have a set of assumptions, say, $x > 0$ and $y > 0$. So I specified the $Assumptions = x > 0 && y > 0 Now the inequality $x - y/2 > 0$ does not provide a truth value.

The function Simplify[x-y/2>0] produces output 2x>y, while the function Simplify[x+y/2>0] produces output True.

I actually want a truth value for expressions. But when a truth value doesn't exist, i.e., it may be either true or false, I want to know that the said inequality can take either.

Then I can just proceed with assuming $x-y/2>0$ and $x-y/2\leq 0$ as two cases.

Is there a way to figure out that the inequality is undetermined in Mathematica?

  • 1
    $\begingroup$ Think of the FindInstance command. $\endgroup$
    – user64494
    Sep 7, 2018 at 7:36
  • $\begingroup$ It's just telling you in the first instance that x-y/2 is true when 2x>y. It will be False otherwise. Since MMA doesn't know whether 2x>y, it can't give you a straight True or False answer. The second case is obviously True and MMA tells you that. $\endgroup$
    – Bill Watts
    Sep 7, 2018 at 7:49
  • $\begingroup$ Do you mean to ask if such values for x and y exist that make x - y/2 > 0 when x > 0 and y > 0? Simplify is not the way to go here, but using Exists and Reduce is a very obvious solution when you actually get to know its existence. Reduce[Exists[{x, y}, x > 0 && y > 0, x + y/2 > 0], Reals] returns True, which indicates that such values of x and y exist. What your Simplify query for x + y/2 > 0 with these assumptions tells is that all values under these conditions fulfil the requirement. (You can also check ForAll usage with Reduce). $\endgroup$
    – kirma
    Sep 7, 2018 at 9:56

1 Answer 1


I split the result to four alternatives: value is true for all variable values under assumptions, it's false for all such values, it is neither true nor false for all values (Both), or it is not a Boolean value (Unresolved):

eqnTrueFalseBothUnresolved[var_, eqn_, assum_, dom_:Reals] :=
 If[Reduce[ForAll[var, assum, eqn], dom], True,
  If[Reduce[ForAll[var, assum, !eqn], dom], False, Both, Unresolved],

Both outcomes are possible:

eqnTrueFalseBothUnresolved[{x, y}, x - y/2 > 0, x > 0 && y > 0]


Always true:

eqnTrueFalseBothUnresolved[{x, y}, x + y/2 > 0, x > 0 && y > 0]


When we don't know the value of a we can't be certain of the outcome (if a is positive, it'd be Both, otherwise True):

eqnTrueFalseBothUnresolved[{x, y}, x + y/2 > a, x > 0 && y > 0]


  • 1
    $\begingroup$ Thanks for the solution. I think I was missing the command ForAll. Since it checks for all instances in the domain, it can indicate whether there exists a solution to the inequality. $\endgroup$
    – drdebmath
    Sep 7, 2018 at 12:41

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