Tableau row reduction returns the deterministic Pauli sign

If $n=0$, the row list and its label list are empty, so the unique coefficient vector is $c=()$ and the row product is the empty product $G=I$. By [F12], $M=\pm I$, so there is a unique bit $b$ with $G=(-1{)}^{b}M$. The stabilizer ray has a unit representative $\psi$ and $M\psi =(-1{)}^b\psi$. Therefore $P_b\psi =\psi$ and $P_{1-b}\psi =0$. By [F11], [F8], and [F9], the branch probabilities are one and zero, so the deterministic measurement bit is $b$.

Assume now $n\ge 1$. Since $M$ commutes with every stabilizer row, [F4] gives $\lambda (M)\in L^{\perp }=L$ by [F3]. The row labels form a basis of $L$ by [F2], so the linear system with those rows and target $\lambda (M)$ is consistent and has a unique coefficient vector $c$. Gaussian elimination over ${\mathbb{F}}_2$ returns it by pivoting on the independent rows.

Multiplying the selected commuting rows in increasing order using [F7] produces the group element $G$ with label ${\sum }_j{c}_j\lambda (g_j)=\lambda (M)$. By [F10], $G$ and $M$ differ by a scalar; as both are Hermitian, [F5] makes that scalar $+1$ or $-1$. The exact product phase therefore determines the unique bit $b$ with $G=(-1{)}^{b}M$.

Since $G$ is a product of stabilizer rows (the empty product when $n=0$), [F6] gives $G\psi =\psi$. Thus $M\psi =(-1{)}^b\psi$. By [F11], the unit ray projector is a density operator, and [F8] and [F9] give that the corresponding branch has probability one and the opposite branch has probability zero. Hence the phase returned by the row reduction is exactly the deterministic measurement bit.