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Let $p$ be prime and set $s=2p-2$. If a bipartite graph has edge-disjoint perfect matchings $M_0,\dots ,M_{\sigma -1}$ on nested nonempty vertex sets, and $|V(M_{i+s})|>s|V(M_i)|/(s+1)$, then the union of $s+1$ consecutive matchings in a graph has maximum degree at most $2p-1$ and more than $(p-1)|V(M_i)|$ edges. The prime-regular subgraph from Boolean polynomial constraints gives a $p$-regular subgraph. Its perfect matching decomposition gives a $k$-regular subgraph for every $k\le p$. If none exists, every $s$ steps shrink the vertex set by factor at most $s/(s+1)$, bounding $\sigma$ by a constant times $\log n$.

Choose a fixed prime $p\ge k$, and put $s=2p-2$. We first derive the polynomial step: a graph $H$ on $N$ vertices and $M>(p-1)N$ edges, with maximum degree at most $2p-1$, contains a nonempty $p$-regular subgraph. Use a Boolean variable $x_e$ for each edge and work over ${\mathbb{F}}_p$. DefineThe first product has degree at most $(p-1)N<M$, while the full squarefree monomial ${\prod }_e{x}_e$ has coefficient $(-1{)}^{M+1}\ne 0$ in the second term with its minus sign. The Combinatorial Nullstellensatz, with every grid $\{0,1\}$, gives a nonzero value of $R$. Since $R(0)=0$, this is a nonempty edge selection. The second product is then zero, so the first is nonzero. Fermat's little theorem makes each selected vertex degree zero modulo $p$. The maximum-degree bound implies that every positive selected degree is precisely $p$. This proves the prime-regular subgraph from Boolean polynomial constraints.

We next turn density into an almost-regular bipartite subgraph, keeping the logarithmic loss explicit. Let the original graph have $n\ge 2$ vertices and $m>0$ edges. Some bipartition retains at least $m/2$ edges: choose the sides independently with probability $1/2$ and average the number crossing. Its average degree is at least $m/n$. Repeatedly delete a vertex of degree at most half the current average degree. Such a deletion cannot lower that average, sinceThe process cannot delete the last positive-density graph, because the maintained average is positive. At termination the minimum degree is an integer $\sigma >m/(2n)$.

Choose the larger side $A$ of this remaining bipartite graph, with opposite side $B$. Keep exactly $\sigma$ edges at each vertex of $A$; choices at different vertices of $A$ do not conflict. Remove any now-isolated vertices of $B$. Thus $|A|\ge |B|$, and every vertex of $A$ has degree $\sigma$.

Such a half-regular graph contains a half-regular restriction with a perfect matching. Choose a minimal nonempty $X\subseteq A$ satisfying $|N(X)|\le |X|$, which exists because the whole side $A$ satisfies this inequality. In fact equality holds: if it were strict, deleting a vertex of $X$ would retain the inequality, contradicting minimality; the singleton case has at least one neighbor. Every proper subset $Y\subset X$ satisfies $|N(Y)|\ge |Y|$. Hall's marriage theorem therefore supplies a perfect matching on $X\cup N(X)$. All $\sigma$ edges incident with each vertex of $X$ remain inside that restriction.

Delete this perfect matching and repeat the restriction argument. The left degrees decrease by one at each step. We obtain $\sigma$ edge-disjoint perfect matchings $M_0,\dots ,M_{\sigma -1}$ on nested nonempty vertex sets, with sizesFor any $i$ with $i+s\le \sigma -1$, the union of $s+1=2p-1$ consecutive matchings in a graph has maximum degree at most $2p-1$ and at least $(s+1)N_{i+s}/2$ edges on $N_i$ vertices. Ifit has more than $(p-1)N_i$ edges, so the polynomial step yields a $p$-regular bipartite subgraph. Every regular bipartite graph decomposes into perfect matchings: for a set in one part, counting incident edges gives Hall's inequality, so a perfect matching exists; remove it and repeat. Choosing $k$ of these $p$ matchings in a graph gives a nonempty $k$-regular subgraph. This is the nested matching criterion for a regular subgraph.

If no $k$-regular subgraph exists, every displayed size ratio must therefore satisfy the reverse inequality. Iterating at indices $0,s,2s,\dots$ gives, with $r=\lfloor (\sigma -1)/s\rfloor$,Consequently $r\le \log n/\log ((s+1)/s)$ andCombining with $\sigma >m/(2n)$ proves the required edge bound. For example the finite constantworks for every $n\ge 2$, and the case $n=1$ has no edges. ThusThe logarithm comes from how often a nested matching in a graph vertex set can shrink by a fixed factor before becoming empty; it is not supplied by the polynomial step alone.