Proof, [Proposition 7.2.4, AP]. Let $\Gamma(x) = \bracsn{(\xi, x\xi)|\xi \in D(x)}$ and $\Gamma(y) = \bracsn{(\xi, y\xi)|\xi \in D(y)}$ be the graphs of $x$ and $y$, respectively. Since $x$ and $y$ are closed, $\Gamma(x)$ and $\Gamma(y)$ are closed subspaces of $H \oplus H$. Let $P_{x}, P_{y} \in B(H \oplus H)$ be the orthogonal projections of $H \oplus H$ onto $\Gamma(x)$ and $\Gamma(y)$, respectively.

Let $M_{2}(M) = M_{2}(\complex) \otimes M \subset B(H \oplus H)$, then

For any $u \in U(M')$, $ux = xu$, so for any $\xi \in D(x)$, $(u\xi, ux\xi) = (u\xi, xu\xi) \in \Gamma(x)$. On the other hand, if $(\xi, \eta) \in \Gamma(x)^{\perp}$, then for any $\theta \in D(x)$,

As $(u^{*}\theta, xu^{*}\theta) \in \Gamma(x)$, $(u\xi, u\eta) \in \Gamma(x)^{\perp}$. In other words, $\Gamma(x)$ and $\Gamma(x)^{\perp}$ are both invariant under $I_{M_2(\complex)}\otimes u$. As such, $\Gamma(x)$ is reducing for $I_{M_2(\complex)}\otimes u$, and $u$ commutes with $P_{x}$. Thus $P_{x} \in M_{2}(M)$. Similarly, $P_{y} \in M_{2}(M)$ as well.

Let $p = \begin{bmatrix}1 & 0 \\ 0 & 0\end{bmatrix} \in M_{2}(M)$, then

and

By Lemma 43.9.11, $P_{x} \sim p \sim P_{y}$. Since $M$ is tracial, $M_{2}(M)$ admits a faithful tracial state $\tau_{2} := \frac{1}{2}\text{Tr}_{2} \otimes \tau$. Thus $P_{x} \le P_{y}$ and $\tau_{2}(P_{x}) = \tau_{2}(P_{y})$ imply that $P_{x} = P_{y}$. Therefore $\Gamma(x) = \Gamma(y)$, and $x = y$.$\square$