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  "body": "\n## Proof\n\nA [[graph|0JD5C82QMB]] is a special [[multigraph|01M5T0KDG7]], so we work with a multigraph $G = (V, E, r)$, where $V$ and $E$ are finite. We show necessity by counting, and sufficiency with a longest trail.\n\n1. Conventions. A [[walk|0VJZM0F9YM]] is a sequence $(w_0, e_1, w_1, \\dots, e_k, w_k)$ of vertices $w_t$ and edges $e_t$. We require $r(e_t) = \\set{w_{t-1}, w_t}$ for each $t$. It is a [[trail|062MQGP8M5]] if $e_1, \\dots, e_k$ are pairwise distinct.\n2. In a graph an edge is determined by its ends, so these are the usual walks and trails. $G$ is [[connected|0FFAYNMAJZ]] if $V \\ne \\varnothing$ and any two vertices are joined by a walk.\n3. Let $x$ be a vertex and $e$ an edge. Let $m_x(e) = 2$ if $e$ is a loop at $x$, and $m_x(e) = 1$ if $e$ is not a loop and $x \\in r(e)$. Otherwise $m_x(e) = 0$. The [[degree|0T0F6CCS9W]] is $\\deg(x) = \\sum_{e \\in E} m_x(e)$.\n4. For a condition $\\varphi$, let $[\\varphi]$ be $1$ if $\\varphi$ holds and $0$ otherwise. For each $t$, $m_x(e_t) = [w_{t-1} = x] + [w_t = x]$. Both cases of a loop and a non-loop agree with step 3.\n5. Let $W$ be a trail as in step 1, and let $c_W(x) = \\sum_{t=1}^{k} m_x(e_t)$. By step 4, $c_W(x) = 2 \\sum_{t=1}^{k-1} [w_t = x] + [w_0 = x] + [w_k = x]$ if $k \\ge 1$.\n6. Hence $c_W(x) \\equiv [w_0 = x] + [w_k = x] \\pmod 2$. This also holds for $k = 0$, where both sides are even.\n7. Necessity. Let $W$ be an [[Eulerian circuit|0NNYDD1AVR]]. Each edge of $G$ is exactly one $e_t$, so $c_W(x) = \\deg(x)$ for every $x$.\n8. Since $w_0 = w_k$, step 6 gives $\\deg(x) \\equiv 2[w_0 = x] \\equiv 0 \\pmod 2$. So every vertex has even degree.\n9. Sufficiency. Assume every vertex has even degree. A trail has length at most $\\abs{E}$, so there is a trail $W = (w_0, e_1, \\dots, e_k, w_k)$ of largest length.\n10. $G$ has an edge $e$, with $r(e) = \\set{a, b}$ or $\\set{a}$. Then $(a, e, b)$ or $(a, e, a)$ is a trail of length $1$. So $k \\ge 1$.\n11. $W$ is closed. Suppose $w_0 \\ne w_k$, and let $x = w_k$. By step 6, $c_W(x)$ is odd, while $\\deg(x)$ is even.\n12. So $c_W(x) \\ne \\deg(x)$. Since $c_W(x)$ sums $m_x$ over the edges of $W$, some edge $f$ not in $W$ has $m_x(f) \u003e 0$, so $x \\in r(f)$.\n13. Let $y$ be the other end of $f$, or $y = x$ if $f$ is a loop. Then $(w_0, e_1, \\dots, e_k, w_k, f, y)$ is a trail of length $k + 1$. This contradicts step 9, so $w_0 = w_k$.\n14. Let $S = \\set{w_0, \\dots, w_k}$, and suppose some edge of $G$ is not in $W$. We claim that some edge not in $W$ has an end in $S$.\n15. Suppose not, and take an edge $g$ not in $W$ with an end $z$. Then $z \\notin S$.\n16. Since $G$ is connected, there is a walk from $w_0 \\in S$ to $z \\notin S$. Some edge $h$ of this walk has an end in $S$ and an end outside $S$.\n17. By the assumption in step 15, $h$ lies in $W$. But every edge of $W$ has both ends in $S$, a contradiction.\n18. So there is an edge $g$ not in $W$ with an end $w_s \\in S$, where $0 \\le s \\le k$. Since $w_k = w_0$, we may take $s \\le k - 1$.\n19. Then $(w_s, e_{s+1}, \\dots, e_k, w_k, e_1, w_1, \\dots, e_s, w_s)$ is a trail. It uses the same $k$ edges as $W$, because $w_k = w_0$. For $s = 0$ it is $W$ itself.\n20. Appending $g$ and its other end to this trail, as in step 13, gives a trail of length $k + 1$. This contradicts step 9.\n21. So the assumption of step 14 fails, and every edge of $G$ lies in $W$. Since $W$ is a trail, each edge appears exactly once, and $w_0 = w_k$ by step 13.\n22. So $W$ is an Eulerian circuit of $G$.\n\n## Idea\n\nA trail passes through a vertex by using two of its edges, so only its ends can have an unpaired edge. For necessity this shows all degrees are even.\nFor sufficiency take a longest trail. Even degrees leave a free edge at an open end, so the trail is closed. Connectedness gives a free edge at some vertex of the closed trail, and starting the trail there would make it longer.\n",
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