The proof proceeds by complete induction. Let $G$ be a planar graph with $V$ vertices and $E$ edges. For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition: :For all planar graphs $G = \struct {V, E}$ such that $V + E = n$, the equation $V - E + F = 2$ holds. === Basis for the Induction === Let $V + E = 1$. This is the case of the edgeless graph of order $1$ which consists of a single vertex. It has exactly one face, which is the entire plane. {{begin-eqn}} {{eqn | l = V       | r = 1 }} {{eqn | l = E       | r = 0 }} {{eqn | l = F       | r = 1 }} {{eqn | ll= \leadsto       | l = V - E + F       | r = 2 }} {{end-eqn}} Thus $\map P 1$ is seen to hold. This is the basis for the induction. === Induction Hypothesis === Now it needs to be shown that if $\map P j$ is true, for all $j$ such that $1 \le j \le k$, then it logically follows that $\map P {k + 1}$ is true. So this is the induction hypothesis: :For all planar graphs $G = \struct {V, E}$ such that $V + E = k$, the equation $V - E + F = 2$ holds. from which it is to be shown that: :For all planar graphs $G = \struct {V, E}$ such that $V + E = {k + 1}$, the equation $V - E + F = 2$ holds. === Induction Step === This is the induction step: Let $G$ be a planar graph with $V$ vertices and $E$ edges such that $V + E = k + 1$. Let $V - E + F = n$ There are a number of cases to consider: ;$(1)$ -- Existence of vertex $v$ of degree $1$ Suppose $G$ has a vertex $v$ of degree $1$. Then we can remove $v$ and its incident edge. Let $G' = \struct {V', E'}$ be the (necessarily planar) graph which results from doing this, where $V'$ and $E'$ are the vertices and edges of $G'$. The number of faces of $G'$ is the same as the number of faces of $G$, as the removed edges does not separate the two faces, but instead juts out into the one face. So: :$V' = V - 1$ (as we have removed $1$ vertex) :$E' = E - 1$ (as we have removed $1$ edges). Thus we have: {{begin-eqn}} {{eqn | l = V - E + F       | r = \paren {V' + 1} - \paren {E' + 1} + F }} {{eqn | r = V' + 1 - E' - 1 + F }} {{eqn | r = V' - E' + F }} {{end-eqn}} But: :$V' + E' = V + E - 2 = k - 1$ and so by the induction hypothesis: :$V' - E' + F = 2$ It follows that: :$V - E + F = 2$ {{qed|lemma}} ;$(2)$ -- Existence of a loop Suppose $G$ has a loop $e$. Then we can remove $e$. Let $G' = \struct {V', E'}$ be the (necessarily planar) graph which results from doing this, where $V'$ and $E'$ are the vertices and edges of $G'$. Let $F'$ be the number of faces of $G'$. Then $F'$ one less than the number of faces of $G$, as the removal of $e$ has caused the face enclosed by $e$ to be merged with the face on the other side of $e$. So: :$V' = V$ (as we have not changed any vertices) :$E' = E - 1$ (as we have removed $1$ edge). :$F' = F - 1$ (as we have merged $2$ faces into $1$). Thus we have: {{begin-eqn}} {{eqn | l = V - E + F       | r = V' - \paren {E' + 1} + \paren {F' + 1} }} {{eqn | r = V' - E' - 1 + F + 1 }} {{eqn | r = V' - E' + F' }} {{end-eqn}} But: :$V' + E' = V + E - 1 = k$ and so by the induction hypothesis: :$V' - E' + F' = 2$ It follows that: :$V - E + F = 2$ {{qed|lemma}} ;$(3)$ -- Existence of vertex $v$ of degree $2$ Suppose $G$ has: :neither a vertex of degree $1$ :or a loop but does have a vertex $v$ of degree $2$ such that $v$ is not either end of a loop. Then we can remove $v$ and replace its two incident edges with a single edge. Let $G' = \struct {V', E'}$ be the (necessarily planar) graph which results from doing this, where $V'$ and $E'$ are the vertices and edges of $G'$. The number of faces of $G'$ is the same as the number of faces of $G$ as the two edges that were replaced with a single edge separate the same two faces. So: :$V' = V - 1$ (as we have removed $1$ vertex) :$E' = E - 1$ (as we have replaced $2$ edges with $1$ edge). Thus we have: {{begin-eqn}} {{eqn | l = V - E + F       | r = \paren {V' + 1} - \paren {E' + 1} + F }} {{eqn | r = V' + 1 - E' - 1 + F }} {{eqn | r = V' - E' + F }} {{end-eqn}} But: :$V' + E' = V + E - 2 = k - 1$ and so by the induction hypothesis: :$V' - E' + F = 2$ It follows that: :$V - E + F = 2$ {{qed|lemma}} ;$(3)$ -- Existence of vertex $v$ of degree $m$ where $m > 2$ Suppose $G$ has: :no vertex of degree $1$ :no loop :no vertex of degree $2$ Then all vertices of $G$ will be of degree $m$ where $m > 2$. Let $u$ and $v$ be two adjacent vertices of $G$ which is not a bridge. Let $e$ be the edge which is incident to both $u$ and $v$. Let $G' = \struct {V', E'}$ be the (necessarily planar) graph which results from removing $e$, where $V'$ and $E'$ are the vertices and edges of $G'$. Let $F'$ be the number of faces of $G'$. Then $F'$ one less than the number of faces of $G$, as the removal of $e$ has caused the faces separated by $e$ to be merged with the face on the other side of $e$. By construction, $E' = E - 1$. Removing $e$ will not affect the number of vertices of $G$, so $V' = V$. So: :$V' = V$ (as we have removed $1$ vertex) :$E' = E - 1$ (as we have removed $1$ edges). :$F' = F - 1$ (as we have merged $2$ faces into $1$). Thus we have: {{begin-eqn}} {{eqn | l = V - E + F       | r = V' - \paren {E' + 1} + \paren {F' + 1} }} {{eqn | r = V' - E' - 1 + F + 1 }} {{eqn | r = V' - E' + F' }} {{end-eqn}} But: :$V' + E' = V + E - 1 = k$ and so by the induction hypothesis: :$V' - E' + F' = 2$ It follows that: :$V - E + F = 2$ {{qed|lemma}} All cases have been covered. So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction. {{qed}} {{improve|clumsy and handwavey}} {{Namedfor|Leonhard Paul Euler|cat = Euler}}