Chris, thank you for the suggestion of:

Simplify[(a^b)^c == a^(b*c), Assumptions -> Element[b | c, Integers]]

That produced the results I was looking for in all of my test cases.

Thank you both for all your help and suggestions!

Hello, thank you for the suggestion, but I'm not sure how to use it:

PowerExpand["(a^b)^c"] == PowerExpand["a^(b*c)"]

Gives a result of "False".

Ah, I see that I have to use ToExpression:

PowerExpand[ToExpression["(a^b)^c"]] == PowerExpand[ToExpression["a^(b*c)"]]

Gives a result of "True". Thank you for the extra help!

I am using ToExpression because I plan on generating multiple different expressions programmatically, and then comparing them to see if any are mathematically/algebraically equivalent. I don't know how to do that without manipulating strings.

The following two expressions produce the correct output:

// expecting false, returns false

Reduce[ForAll[{a, b, c}, (a + b) - c == (a - b) + c]] 

// expecting true, returns true

Reduce[ForAll[{a, b, c}, (a + b) - c == (a - c) + b]]

But I just tried that on one of the original comparison expressions, and so far it is failing to produce any output in Mathematica. It's been running for about 10 minutes now.

// expecting true, hasn't returned any output yet...

Reduce[ForAll[{a, b, c}, (a^b)^c == (a^c)^b]]

And... Actually, it looks like it just stopped. Still no output. I guess there is a 10 minute time limit for online calculations.

Can you think of any way to compare expressions to get the following results:
(a-b)+c == (c-b)+a -> True
(a-b)+c == (a-c)+b -> False
(a-b)/(c-d) == (b-a)/(d-c) -> True
(a-b)/(c-d) == (b-a)/(c-d) -> False
(a-b)/((c-d)^(d-e)) == (a-b)*((c-d)^(e-d)) -> True
(a-b)/((c-d)^(d-e)) == (a-b)*((d-c)^(e-d)) -> False
(a^b)^c == (a^c)^b -> True
(a^b)^c == (b^a)^c -> False
a^(b*c) == (a^c)^b -> True
a^(b^c) == (a^b)^c -> False

One must remember that deciding if two expressions are identical is very difficult in its full generality. If you generate the expressions programmatically, you cannot expect to get the answer easily at the first try. Search "Constant problem" on Wikipedia.

Wrong syntax. Simplify[(a + b) - c == (a - b) + c] gives b == c, which is equivalent condition for the equality. If you insert b == c inside Equal you call Equal with one argument, which returns True whatever the argument is. Perhaps you meant

Reduce[ForAll[{a, b, c}, (a + b) - c == (a - b) + c]]

Why are you using ToExpression of a string?

Here are two conditions under which the equality is true:

Reduce[(a^b)^c == (a^c)^b &&
  Element[c | b, Integers] && a != 0]
Reduce[(a^b)^c == (a^c)^b && a > 0, Reals]

You can try FullSimplify instead of Reduce.

If you search the archives you will probably find earlier discussions of the topic.

When I ran your example code, I see that Mathematica returned imaginary results.

While I am not looking for "solutions" (actual values for a,b,c), I would like to restrict the symbolic comparison to the Real domain, hmmm maybe even more restricted to the positive Reals, in which case (I believe) the expressions are equivalent.

Is there a way to perform the symbolic comparison with this restriction?