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  "body": "\n## Proof\n\nWe argue by [[strong induction|0KMVFSHFGZ]] on the length $k$ of the walk.\n\n1. Claim for each $k \\in \\N$: let $W$ be a [[walk|0VJZM0F9YM]] of length $k$ from a vertex $u$ to a vertex $v$. Then there is a [[path|0JDE56HWQ8]] from $u$ to $v$ whose vertices and edges all belong to $W$.\n2. Let $W = (w_0, w_1, \\dots, w_k)$ be a walk from $u$ to $v$, and assume the claim for all lengths smaller than $k$.\n3. If the vertices $w_0, \\dots, w_k$ are pairwise distinct, then $W$ itself is a path, and we are done.\n4. Otherwise there are indices $0 \\le i \u003c j \\le k$ with $w_i = w_j$.\n5. Let $W' = (w_0, \\dots, w_i, w_{j+1}, \\dots, w_k)$. If $j = k$, this is $W' = (w_0, \\dots, w_i)$.\n6. $W'$ is a walk: consecutive vertices inside the two blocks are adjacent in $W$, and $w_i = w_j \\sim w_{j+1}$ when $j \u003c k$.\n7. $W'$ starts at $w_0 = u$. It ends at $w_k = v$ if $j \u003c k$, and at $w_i = w_j = w_k = v$ if $j = k$.\n8. The vertices of $W'$ are vertices of $W$. Its edges are edges of $W$, because the only new edge $w_i w_{j+1}$ equals the edge $w_j w_{j+1}$ of $W$.\n9. The length of $W'$ is $k - (j - i) \u003c k$. By the induction hypothesis, $W'$ contains a path from $u$ to $v$ as in the claim.\n10. By step 8, the vertices and edges of that path are vertices and edges of $W$. This proves the claim for $k$, and hence the statement.\n\n## Idea\n\nIf a walk visits some vertex twice, the part between the two visits is a detour. Cutting it out gives a shorter walk with the same ends, and this cannot go on forever.\n",
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        "locator": "Walk, trail, and path",
        "url": "https://en.wikipedia.org/wiki/Path_(graph_theory)"
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