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I'm evaluating an expression and to me the result should be True but Mathematica is giving me a different result.

AllPositive = (A > 0 && B > 0 && X > 0 && P > 0 && L > 0);

NewCondition = ((B*P*L - X) > 0);
Simplify[(B*P*L - X) > 0, AllPositive && NewCondition]

NewCondition = ((B*P*L - X) > B);
Simplify[(B*P*L - X) > 0, AllPositive && NewCondition]

NewCondition = ((B*P - X) > B);
Simplify[(B*P - X) > 0, AllPositive && NewCondition]

NewCondition = (A*(B*P - X) > B);
Simplify[(B*P - X) > 0, AllPositive && NewCondition]

NewCondition = (A*(B*P*L - X) > B);
Simplify[(B*P*L - X) > 0, AllPositive && NewCondition]

Results:

True

True

True

True

B L P > X

Shouldn't the last Simplify also return True?

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  • 1
    $\begingroup$ Even NewCondition = (B*P*L - X > B/A); Simplify[{B/A > 0, (B*P*L - X) > 0}, AllPositive && NewCondition] does not work as expected, yielding {True, B L P > X} $\endgroup$ – bbgodfrey Sep 25 '15 at 18:01

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