# 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). Commented Nov 27, 2016 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. Commented Nov 27, 2016 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, Commented Nov 27, 2016 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.) Commented Nov 27, 2016 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. Commented Nov 27, 2016 at 20:21

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