Recently I have an interesting reading of Terrence Tao’s community-collaborative project on the equational theories. He mentioned an interesting algebra exercise as follows.
Given a magma (an algebraic strucuture with binary multiplication without commutativity or associativity law guaranteed) so that
$$
\begin{equation}
x = (y \circ x) \circ (y \circ (y \circ x))
\end{equation}
$$
holds for any \( x, y \), prove that
$$
\begin{equation}
x \circ x = (x \circ x) \circ x
\end{equation}
$$
holds for any \( x \).

It provoked my interest and I spent some time to figure it out.

1. Substitute \( x = t, y = t \) into the original equation to obtain
   $$
   t = (t \circ t) \circ (t \circ (t \circ t))
   $$
2. Substitute \( x = t \circ (t \circ t), y = t \circ t \) into the original equation to obtain
   $$
   t \circ (t \circ t) = ((t \circ t) \circ (t \circ (t \circ t))) \circ ((t \circ t) \circ ((t \circ t) \circ (t \circ (t \circ t))))
   $$
3. Utilize the equation in Step 1) to simplify the RHS of Step 2) to obtain
   $$
   t \circ (t \circ t) = t \circ ((t \circ t) \circ t)
   $$
4. Substitute \( x = t \circ t, y = t \) into the original equation to obtain
   $$
   t \circ t = (t \circ (t \circ t)) \circ (t \circ (t \circ (t \circ t)))
   $$
5. Substitute \( x = (t \circ t) \circ t, y = t \) into the original equation to obtain
   $$
   (t \circ t) \circ t = (t \circ ((t \circ t) \circ t)) \circ (t \circ (t \circ ((t \circ t) \circ t)))
   $$
6. Comparing the RHS of Step 4) and Step 5) while keeping Step 3)’s result in mind, we see they are equal. So the LHS are also equal, which is just what we need
   $$
   t \circ t = (t \circ t) \circ t
   $$

I have tried the lates OpenAI’s O1-preview on this puzzle. It gives a “Finish Thinking” without any output after several minutes’ painful struggling. 😉