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The final paragraph states Tutte's theorem on perfect matchings in terms of the number of components of odd integer order after deleting a vertex subset. The worked perfect matching theorem for a cubic graph with at most two bridges uses this criterion directly.

Use Tutte's theorem on perfect matchings: a finite graph has a perfect matching if and only if, for every $S\subseteq V$, the number $q(S)$ of connected components of a graph of odd integer order in $G-S$ is at most $|S|$.

First, the handshaking lemma gives $3|V|=2|E|$, so $|V|$ is an even integer. Fix $S\ne \varnothing$. For each connected component of a graph $C$ of odd integer order in $G-S$, let $d(C)$ count its edges to $S$. Counting vertex degrees inside $C$ givesThus $d(C)$ is an odd integer; it is positive because $G$ is a connected graph. If $d(C)=1$, its unique outgoing edge is a bridge. Distinct such components yield distinct bridges, so there are at most two of them. Every other such component has $d(C)\ge 3$. ThereforeThe last inequality holds because each vertex of $S$ has vertex degree three. Furthermore,since components of even integer order do not affect parity, and $|V|$ is an even integer. If $q(S)>|S|$, this parity forces $q(S)\ge |S|+2$, contradicting the bound $3q(S)\le 3|S|+4$.

For $S=\varnothing$, the only component is the whole connected graph, of even integer order, so $q(\varnothing )=0$. Thus every $S$ satisfies the criterion, and $G$ has a perfect matching. This proves the perfect matching theorem for a cubic graph with at most two bridges directly from Tutte's theorem on perfect matchings.

A finite simple graph that is a cubic graph and has a perfect matching admits a partition of a set of its edges into graph paths of length three.

Proof. Delete a perfect matching $M$. The remaining spanning subgraph has vertex degree two at every vertex, so it consists of disjoint graph cycles. Orient each cyclically, giving a successor bijection $s$. Replace each $uv\in M$ by the graph path $s(u),u,v,s(v)$. The four vertices are distinct: the successor edges are not in $M$, neither is a self-loop, and injectivity of $s$ excludes $s(u)=s(v)$. Every edge of $M$ appears once in the middle, and every other edge appears once at its unique tail. Thus the graph paths cover every edge exactly once. In particular, the perfect matching theorem for a cubic graph with at most two bridges guarantees this decomposition under its hypotheses.

A finite graph has a perfect matching if and only if the displayed inequalities hold, where $o(H)$ counts the connected components of a graph $H$ with an odd integer number of vertices. The theorem is associated with William Thomas Tutte.^[ref]

Necessity has a direct counting proof: each such component of $G-S$ must send at least one edge of a perfect matching to $S$, and different components must use different vertices of $S$. The sufficiency direction is the substantive content of the theorem. The criterion is useful because vertex degree and parity estimates can prove existence without constructing the perfect matching, as in the perfect matching theorem for a cubic graph with at most two bridges.