Say I want to define a function fwith input x but also involving parameter p not yet specified:
f[x_] := p^x
Now I want to verify some inequalities with additional assumptions. A (hopefully) minimal example, showing my problem:
If, for some for input y in the function and for some paremeter value q>0, it holds that f[y] >= 2 q, then I would expect f[y] >= q to yield True.
However,
FullSimplify[f[y] >= q, {f[y] >= 2 q, q > 0}]
yields p^y >= q instead. Even if I add q \[Element] Reals as an assumption.
Do I overlook an obvious mathematical (non-Mathematica) issue, or how I can get Mathematica to "fully" simplify the expression? I have also looked at this issue with FullSimplify, but the solution provided did not help.
ForAllandResolve? – Mr Puh May 13 '20 at 10:56f, "local" assumptionf[y]>=2qwithq>0and the statement I want to verify,f[y] >= q, which I expected to turn intoTrue.ForAllandResolvedon't seem to fare any bettr. I hoped my example was minimalistic enough, I can add more if required. – Bernd May 13 '20 at 11:21f[x_] := Hold[E^(Log[p] x)]as your equation, this seems to give the desired result. – Mr Puh May 13 '20 at 12:07