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Grammar and small rewording
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ccodel committed Jun 9, 2024
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\begin{proof}[Proof Sketch]
The proof proceeds in 6 steps, illustrated in~\Cref{fig:symmetry-breaking}.
In each of the steps, we will construct a new list of points that is $\sigma$-equivalent to the previous one,
and the last one will be in canonical position.
\footnote{Even though we defined $\sigma$-equivalence for sets of points,
In each of the steps, we construct a new list of points that is $\sigma$-equivalent to the previous one,
with the last one being in canonical position.\footnote{
Even though we defined $\sigma$-equivalence for sets of points,
our formalization goes back and forth between sets and lists.
Given that symmetry breaking distinguishes between the order of the points
e.g., $x$-order, this proof proceeds over lists.
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\textbf{Step 1}: we transform the list of points so that no two points share the same $x$-coordinate.
This can be done by applying a rotation to the list of points, which corresponds to multiplying by a rotation matrix.
Rotations always have determinant $1$.
\textbf{Step 2}: we translate all points by a constant vector $t$, by multiplying by a translation matrix,
so that the left most point gets position $(0, 0)$, and naturally every other point will have a positive $x$-coordinate.
\textbf{Step 2}: we translate all points by a constant vector $t$, which corresponds to multiplying by a translation matrix,
to bring the leftmost point~$p_1$ to position $(0, 0)$.
As a result, every other point has a positive $x$-coordinate.

Let $L_2$ be the list of points after Step 2, excluding $(0,0)$ which we will denote by $p_1$.
Let $L_2$ be the list of points excluding $p_1$ after Step 2.
\textbf{Step 3}: we apply the projective transformation $f: (x, y) \mapsto (y/x, 1/x)$ to every point in $L_2$,
showing that this preserves orientations within $L_2$.
To see that this mapping is a $\sigma$-equivalence consider that
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\)
We then show that $\sigma((0, 0), q, r) = \sigma_{(0, \infty)}(f(q), f(r))$ for all points $q, r \in L_2$.

\textbf{Step 4}: we sort the list $L_2$ by $x$-coordinate in increasing order,
thus obtaining a list $L_3$.
\textbf{Step 4}: we sort the list~$L_2$ by $x$-coordinate in increasing order,
thus obtaining a list~$L_3$.
This can be done while preserving $\sigma$-equivalence because sorting corresponds to a permutation,
and all permutations of a list are $\sigma$-equivalent by definition.
\textbf{Step 5}: we check whether the \textsf{Lex order} condition above is satisfied in $L_3$,
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