While proving statements about formation trees for WFF's, I tried to prove the following Lemma:
Let $T$ be a binary tree with root $r$,and Denote the set of vertices of $T$ by $V$. assume $S \subseteq V$ is such that $r\in S$ and if $v\in S$ and $u$ is a child of $v$. so $S=V$
My proof is by induction on the height of $T$ and goes roughly like this: If the height of $T$ is $0$ than $T$ consists only of the single node $r$ and $S=V$ trivially. Now assume the claim is true for every tree of height $k$ at most, and let $T$ be of height $k+1$. For every child $u$ of $r$, the subtree rooted in $u$, say $T_{u}$, is of height $k$ at most. Moreover, $u$ is in $S$ since we assume $r$ is in $S$ and $S$ contains the children of all its' elements. now we can apply the induction hypothesis for $S_{u} = S\cap V(T_{u})$ to get $S_{u} = V(T_{u})$ and therefore $S$ contains the subtrees (or subtree) diverging from it and also contains $r$ so the proof is complete.
I think my proof is correct, but I tried to look this result up on the internet and didn't find anything. this seems pretty trivial so I'm surprised I couldn't find it online. If it is true I'd like to read more on it and would be glad to be referred to relevant sources. If it is false, I'd like to know why. Thank you!