How do you prove that $L=I-V+1$ in $\lambda\phi^4$ theory? It is known that the number of loops in $\lambda\phi^4$ theory is given by the formula
$$L=I-V+1$$
where $L$ is the number of loops, $I$ the number of internal lines and $V$ the number of vertices. I would like to know the proof of this statement.
 A: This formula is actually Euler's formula for planar graphs, and holds for all Feynman diagrams regardless of what theory we are in.
The proof proceeds by induction and is easy if we first disregard the case of crossing lines: 


*

*Observe that a one-loop graph has two vertices, one loop, and two internal lines, so the formula holds.

*Observe that a $(n+1)$-loop graph is produced from a $n$-loop graph by either drawing one additional line between two already existing vertices, which doesn't change $L-I$, or by adding a new vertex and connecting it to two other vertices, which doesn't change $L-I+V$.

*By induction, the formula holds for all graphs with finitely many loops.
More formally, we can say that

A Feynman diagram is called planar if the adjoint graph obtained by connecting all external lines to a single vertex is planar.

and then we have proven up to now that the formula holds for all planar Feynman graphs. Interestingly, not even all $\phi^4$ graphs are planar. Consider $2\to 2$ (or $1\to 3$)-scattering with a box diagram, where each external line is connected to its own vertex, and each vertex is connected with each other vertex. The adjoint graph is the complete graph on five vertices, which is known to be not planar.
Nevertheless, the "Feynman-Euler formula"
$$ L-I+V = 1$$ 
still holds because of the way loops are formally counted. By the general Euler formula,
$$ \#\{\mathrm{vertices}\} - \#\{\mathrm{edges}\} + \#\{\mathrm{faces}\} = 2 - 2g$$ 
where $g$ is the genus of the surface on which the graph can be drawn without intersections, and "faces" are all regions bounded by edges. A "face" does not have to have a vertex at every corner, so when you get two crossing lines in a Feynman graph, you get two additional faces that you do not count as loops - the above boxy $\phi^4 $ diagram has four faces inside the box, but only two loops.
Since every crossing of lines that cannot be eliminated by deforming the graph (and is hence a "true crossing" and not just us being too dumb to draw the graph properly) increases the genus on which you could draw the graph without crossings by $1$, the "Feynman-Euler formula" for all graphs follows from the general Euler formula.
A: Page 140 of Srednicki's QFT textbook provides a much simpler proof:

This can be seen by counting the number of internal momenta and the constraints among them. Specifically, assign an unfixed momentum to each internal line; there are [$I$] of these momenta. Then the $V$ vertices provide $V$ constraints. One linear combination of these constraints gives overall momentum conservation, and so does not constrain the internal momenta. Therefore, the number of internal momenta left unfixed by the vertex constraints is $[I] − (V−1)$, and the number of unfixed momenta is the same as the number of loops L.

