Is this binary operation commutative? In a set $X$ we define a binary operation $*$ such that
$$\forall x, y \in X,\  (x*y)*y=y*(y*x)=x.$$
Is $*$ commutative and why?
 A: Ah, I think I got a counterexample:
Consider the free monoid $F_2$ on 2 letters $a,b$ (with empty word '$1$'), and consider its quotient
$$X:=F_2/(w^2\sim 1) $$ 
for all words $w\in F_2$.
Then, basically $X$ contains those words which doesn't have any substring of the form $ww$, and if such appears at a concatenation, they get just cancelled. For example $aba\cdot ab= a$. 
Update: not good,  as it is going to be a group, $xy=yx$ will follow from all $x^2=1$...
A: Yes, it is commutative. Here is why : 
$$
x * y = y*(y*(x*y)) = y * ((x * (x*y))*(x*y)) = y *x.
$$
I first use the identity to multiply by $y$ on the left twice, and then I replace the second $y$ by $x*(x*y)$. Then we use the fact that 
$$
(x * (x * y))*(x*y) = x
$$
to remove the bigger chunk.
Hope that helps,
A: Here's a proof that uses the automated theorem prover Prover9.
As a human, I find the output of these things difficult to read, but I find they can help by (a) suggesting an important in-between step, and (b) give a proof (even if it is unreadable), which means the result we're trying to prove is actually true, and it is possible there is a proof with a number of steps that could be understood by humans.  (Also, there is a "coolness factor" to automated proofs.)
Here's the input:
formulas(assumptions).

(x * y) * y = x.
y * (y * x) = x.

end_of_list.

formulas(goals).

x * y = y * x.

end_of_list.

and a trimmed version of the output:
============================== PROOF =================================

% Proof 1 at 0.01 (+ 0.00) seconds.
% Length of proof is 7.
% Level of proof is 3.
% Maximum clause weight is 7.
% Given clauses 4.

1 x * y = y * x # label(non_clause) # label(goal).  [goal].
2 (x * y) * y = x.  [assumption].
3 x * (x * y) = y.  [assumption].
4 c2 * c1 != c1 * c2.  [deny(1)].
5 x * (y * x) = y.  [para(3(a,1),2(a,1,1))].
7 x * y = y * x.  [para(2(a,1),5(a,1,2))].
8 $F.  [resolve(7,a,4,a)].

============================== end of proof ==========================

