# Trees with no vertex of degree 2 have more leaves than internal nodes

There is a question asked by portal about Tree having no vertex of degree 2 has more leaves than internal nodes

so we want to prove this claim by induction and an answer from Micheal Biro suggested to split the tree at an internal vertex. If the internal vertex has degree $k$, you get $k$ pieces, each with no internal vertex of degree 2.

So my question is "how did we know that each piece has no internal vertex of degree 2?"

What about trees like this tree for example: The vertex $u$ after removing one edge became an internal vertex with degree 2

PS: sorry could not put my question as a comment at the answer in the original question because I'm new and have not earned that privilege yet, and this question really got me thinking and I'm assuming that the answer is going to be really simple but I just can't see it You should only split the tree at the designated vertex, not remove that vertex (and its adjacent edges). So your example splits into three trees having $2$, $4$, and $6$ vertices respectively.