= Solution
First take $x\ne y$. Sample a <uniform spanning tree> using <Wilson's algorithm> rooted at $y$, with $x$ processed first. The first walk runs from $x$ until its first hit of $y$, so its <loop erasure> is exactly the unique $x$-to-$y$ <graph path> of the resulting <tree>. Later attachments cannot alter that <graph path>. This is the <uniform-spanning-tree path representation of loop-erased random walk>.
Instead root <Wilson's algorithm> at $x$ and process $y$ first. The resulting <tree> still has the same uniform law, while its first attachment has the law of <loop-erased random walk> from $y$ to $x$. A <tree>'s <graph path> between two fixed <graph vertices> has the same unoriented <edge> set in both directions. Pushing the common <tree> law through this <graph path> map therefore gives
$$
\boxed{\mathcal E(\operatorname{LERW}_{x\to y})\overset d=\mathcal E(\operatorname{LERW}_{y\to x}).}
$$
Here $\mathcal E$ means the set of unoriented <edges>. In fact, reversing the complete <graph vertex> <sequence> from the first loop-erased walk has the law of the second, giving <reversibility of loop-erased random walk>. If $x=y$ and first hitting allows time zero, both <graph paths> are empty and the equality is immediate.
This is a distributional assertion, not a pathwise identity between chronological <loop erasure> and reversal. For example, the valid walk trajectory $x,a,b,a,c,b,y$ in a <graph> containing the triangle on $a,b,c$ and the <edges> $x a,b y$ erases forward to $x,a,c,b,y$. Erasing the reversed trajectory and then reversing the result gives $x,a,b,y$, a different <graph path>. Wilson's root-independent <tree> law is what proves the stated equality of <probability distributions>.
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