Proposition: Three-qubit symmetric witness

Assume $(\mathrm{C}{)}^{\star },(\mathrm{N}{)}^{\star }$ and (S5). For

the first-qubit law is $(1/3,2/3)$, and the first-qubit reduction is $\mathrm{diag}(1/3,2/3)$.

Proof. Nullity restricts the full law to the displayed strings. Tensor-factor permutations preserve $\Omega$ and act transitively on its support. Equivariance forces equal weights, hence $1/3$ each. Composability gives the first-coordinate marginal. The accompanying last-two-qubit strings are distinct, so the first-qubit reduction has no off-diagonal entry and has the stated diagonal.