I'm reading Theorem 15.23 of Jech's Set Theory, where it is proved that there exists a generic extension in which there exists a Suslin tree.
The notion of forcing $(P,<)$ is the set of all countable normal trees, ordered by reverse inclusion. Let $G$ be generic, and define $\mathcal T=\bigcup G$ in $V[G]$, the author claims that $\mathcal T$ is a Suslin tree in $V[G]$. It is not hard to prove in $V[G]$ that $\mathcal T$ is a normal tree of height $\omega_1$ , and that $V[G]$ preserves cardinals, so we need only to show that $\mathcal T$ has only countable antichains in $V[G]$.
I have read the proof of this, however there is something I don't understand:
Let $\dot A$ be a name for a maximal antichain $A$ of $\mathcal T$. Let $T$ be a condition such that $T\Vdash(\dot{A}$ is a maximal antichain of $\mathcal T$). Let $T_0\leq T$ be arbitraty, then $T_0\Vdash(\dot A$ is a maximal antichain in $\mathcal T$ and $\mathcal T$ is an extension of $T_0$); I constructed a name $\dot{\mathcal T}$ for $\mathcal T$ and proved that $||T_0\subseteq\dot{\mathcal T}||=T_0$, and as $T_0$ forces that $\mathcal T$ is a normal tree and $|| T_0$ is a normal tree$||=1$; this is a $\Sigma_1$ statement, we get that $T_0$ forces that $\mathcal T$ is an extension of $T_0$, as it is a theorem of $\mathsf{ZFC}$ that if two normal trees are comparable, one is an extension of the other. Then the author claims, without proof, the following:
there exist for each $s\in T_0$ an extension $T_0'$ of $T_0$ and some $t_s\in T_0'$ such that $$ s\text{ and } t_s\text{ are comparable and }T_0'\Vdash t_s\in\dot A.$$
But I have no idea why this is true, specially on the existence of the $t_s$, and I've been stuck on this for a while.
Thanks for any help.
(Turning the comments into an answer):
Let $s\in T_0$ be arbitrary. For me, the most natural way to argue is in terms of generics, so: Go to a forcing extension by a generic $G'$ containing $T_0$, and let $\mathcal T'=\bigcup G'$ be the corresponding tree. In $V[G']$, the interpretation $A'$ of $\dot A$ (that is, $A'=\dot A_G$) is a maximal antichain. Maximality implies that $s$ is comparable to some member of $\dot A$. Call this member $t_s$.
By definition of $\mathcal T'$, there must be a $T''_0\in G'$ such that $t_s\in T''_0$, and we may also assume that $T''_0\le T_0$. Since $t_s\in A'$, some $T'''_0\in G$ must force this, $T'''_0\Vdash t_s\in \dot A$. Let $T'_0$ be a common extension of $T'''_0$ and $T''_0$ (that we may pick in $G'$, if so desired). Then $T'_0\le T_0$ is as wanted.
Since $s\in T_0$ was arbitrary, we are done.