Strong sequential equilibria and the existence of others I am working on the following game

and I have to find all strong sequential equilibria here.
I determined that here any belief derived from a fully mixed strategy gives a distribution (1/2, 1/2) over the nodes in Player 2’s information set. Given this, player 2 will chose r and player 1 will chose y. I believe that this is the unique sequential equilibrium. But is it strong?
And are there any other Nash equilibria and if so why they are not sequential?
Thanks!
There are three classes of equilibria of this game.

The first class is sequential:
\begin{equation}
(s_1,s_2)=(y,r)
\end{equation}
and the beliefs are
\begin{equation}
\mu_1(a)=\mu_1(b)=\mu_2(a\mid y)=\mu_2(b\mid y)=\frac12.
\end{equation}


The second class is not sequential, but weak perfect Bayesian:
\begin{equation}
(s_1,s_2)=(x,l)
\end{equation}
and the beliefs are
\begin{equation}
\mu_1(a)=\mu_1(b)=\frac12,\quad\text{but }\mu_2(a\mid y)=1-\mu_2(b\mid y)=p&gt;\frac58.
\end{equation}


The third class is actually a boundary case of the second class, which permits P2 to use mixed strategies. This equilibrium requires P2's off equilibrium belief to be
\begin{equation}
\mu_2(a\mid y)=1-\mu_2(b\mid y)=\frac58,
\end{equation}
so that P2 can play a mixed strategy that puts sufficiently high probability on $l$ ($\sigma_2(l)\ge\frac35$). It is also necessary that P1 does not play $y$ with positive probability, because if they do, Bayes rule would kick in, requiring that P2's belief be $(\frac12,\frac12)$ instead of $(\frac58,\frac38)$ which rationalizes P2's mixed strategy.