Deterministic Pauli measurement on a pure stabilizer state

If $n=0$, then $S=\{I\}$ by [F11] and the exclusion of $-I$. Write $M=(-1{)}^{b}I$ for its unique sign bit $b$. The empty row product is $g=I=(-1{)}^{b}M$, so the matching group element and bit are unique. On $H_0=\mathbb{C}$, the given stabilizer ray has a unit representative $\psi$. The projector formula gives $P_b\psi =\psi$ and $P_{1-b}\psi =0$. By [F8]–[F10] the probabilities are $1$ and $0$, respectively; the nonzero branch is the original ray and the other branch has no normalized state by Zero-probability Pauli measurement branch.

Assume now $n\ge 1$. Since $M$ commutes with $S$, [F2] gives $\lambda (M)\in L^{\perp }$. By [F3], this is $L$, so there is $g\in S$ with $\lambda (g)=\lambda (M)$. Injectivity in [F1] makes this $g$ unique.

Both $g$ and $M$ are Hermitian and have the same binary label, so [F4] and [F5] give $g=(-1{)}^{b}M$ for exactly one bit $b$. Since $g\psi =\psi$ by [F6], $M\psi =(-1{)}^b\psi$.

For $n\ge 1$, the relation $M\psi =(-1{)}^b\psi$ gives $P_b\psi =\psi$ and $P_{1-b}\psi =0$. By [F8], the unit ray has its rank-one density projector; [F9] and [F10] then give branch probabilities $1$ and $0$. The nonzero branch is the original ray, while the zero-probability branch has no normalized state by Zero-probability Pauli measurement branch.

Step 1.1 proves the full claim for $n=0$, and step 1.2 and step 2.1 and step 3.1 prove it for $n\ge 1$. Thus the statement holds for every $n\ge 0$.