For every forest, $\mathrm{\mid }V\mathrm{\mid }=\mathrm{\mid }E\mathrm{\mid }+c$, where $c$ is the number of connected components

If $\mathrm{\mid }V(F)\mathrm{\mid }=0$, then $\mathrm{\mid }E(F)\mathrm{\mid }=0$ and $c(F)=0$, so the identity holds.

Let $\mathrm{\mid }V(F)\mathrm{\mid }>0$, assume the identity for forests with fewer vertices, and choose a vertex $v$ of degree at most one.

Put $F^{\mathrm{\prime }}:=F-v$. It remains a forest, and $V(F)$ is the disjoint union of $V(F^{\mathrm{\prime }})$ and $\{v\}$, so $\mathrm{\mid }V(F)\mathrm{\mid }=\mathrm{\mid }V(F^{\mathrm{\prime }})\mathrm{\mid }+1$.

If $v$ is isolated, then $E(F)=E(F^{\mathrm{\prime }})$, while the component family of $F$ is the disjoint union of the component family of $F^{\mathrm{\prime }}$ and $\{\{v\}\}$. Hence $\mathrm{\mid }E(F)\mathrm{\mid }=\mathrm{\mid }E(F^{\mathrm{\prime }})\mathrm{\mid }$ and $c(F)=c(F^{\mathrm{\prime }})+1$.

If $v$ has degree one, deleting it removes its unique incident edge, so $\mathrm{\mid }E(F)\mathrm{\mid }=\mathrm{\mid }E(F^{\mathrm{\prime }})\mathrm{\mid }+1$. It does not change the number of components: $v$ cannot be an internal vertex of a path between remaining vertices, and its neighbour remains in the same nonempty component after deletion. Thus $c(F)=c(F^{\mathrm{\prime }})$.

In the isolated case, the induction hypothesis gives $\mathrm{\mid }V(F)\mathrm{\mid }=\mathrm{\mid }V(F^{\mathrm{\prime }})\mathrm{\mid }+1=(\mathrm{\mid }E(F^{\mathrm{\prime }})\mathrm{\mid }+c(F^{\mathrm{\prime }}))+1=\mathrm{\mid }E(F)\mathrm{\mid }+c(F)$.

In the degree-one case, the induction hypothesis gives $\mathrm{\mid }V(F)\mathrm{\mid }=\mathrm{\mid }V(F^{\mathrm{\prime }})\mathrm{\mid }+1=(\mathrm{\mid }E(F^{\mathrm{\prime }})\mathrm{\mid }+c(F^{\mathrm{\prime }}))+1=(\mathrm{\mid }E(F^{\mathrm{\prime }})\mathrm{\mid }+1)+c(F^{\mathrm{\prime }})=\mathrm{\mid }E(F)\mathrm{\mid }+c(F)$.

The two possible degrees of $v$ have both been handled, so the identity holds for every forest.