# Why do none of FullSimplify, Simplify, and Refine simplify my expression?

I used three functions to simplify.

First

FullSimplify[Exp[y] > Exp[x], y > x > 0 && y ∈ Reals && x ∈ Reals]
(*Out: Exp[y] > Exp[x]  *)


Second

Simplify[Exp[y] > Exp[x],  y > x > 0 && y ∈ Reals && x ∈ Reals]
(*Out: Exp[y] > Exp[x]  *)


Third

Refine[Exp[y] > Exp[x], y > x > 0 && y ∈ Reals && x ∈ Reals]
(*Out: Exp[y] > Exp[x]  *)


I think I've given enough conditions to get True for the result just as with

Simplify[y > x, y > x > 0 && y ∈ Reals && x ∈ Reals]
(*Out: True   *)

• Related: (132158). – corey979 Nov 27 '16 at 12:11
• This works: With[{assum = y > x > 0 && y ∈ Reals && x ∈ Reals}, Simplify[Reduce[assum \[Implies] Exp[y] > Exp[x], Reals], assum]]. -- Again, I can't explain why there are these edge cases between different functions. The *Simplify functions tend to transform the expression into simpler expressions (fewer leaves); if there was a transformation that would change Exp[y] > Exp[x] into y > x, then it should work. Apparently there isn't one. (It's only true over the reals, which might be why.) Reduce tends to be more robust and exacting, but it only works on relations. – Michael E2 Nov 27 '16 at 13:19
• @MichaelE2 - you can simplify to assum = y > x > 0 since the presence of y and x in an inequality implies that they are real, – Bob Hanlon Nov 27 '16 at 13:47
• @BobHanlon Yes, I know. Thanks. (If you're curious, I often just copy & paste the OP's code and don't fiddle with it unless and until it seems worth it. It's hardly an answer, imo, though many times, the OP isn't really interested in the question Why? but actually just in workarounds.) – Michael E2 Nov 27 '16 at 13:49
• Refine seems a bit sketchy with Exp. For example Refine[Exp[y] > Pi, y > 2] returns unrefined Exp[y] > Pi but Refine[Exp[y] > Pi, y > 3] returns true. – Simon Woods Nov 27 '16 at 20:21

The following return True:
Resolve[Implies[ForAll[{x, y}, y > x > 0], Exp[y] > Exp[x]]]