Proof, [Theorem 7.2.8, AP]. (1): By Proposition 43.8.10, $x^{*}$ is closed and densely defined.

Let $u \in U(M')$, $\xi \in D(x)$, and $\eta \in D(x^{*})$, then

so $u^{*}\eta \in D(x^{*})$ and $\eta \in D(x^{*}u^{*})$ with $x^{*}u^{*}\eta = u^{*}x^{*}\eta$. Conversely, if $\eta \in D(x^{*}u^{*})$, then

so $\eta \in D(x^{*})$ with $x^{*}u^{*}\eta = u^{*}x^{*}\eta$. Therefore $u^{*}x^{*} = x^{*}u^{*}$ for all $u \in U(M')$, and $x^{*}$ is affiliated with $M$.

(2): For each $\eps > 0$, by Lemma 44.3.4, there exists $p, q \in \text{Proj}(M)$ with $pH \subset D(x)$, $qH \subset D(y)$, and $\tau(p), \tau(q) \ge 1 - \eps$. As such, $(p \wedge q)H \subset D(x) \cap D(y) = D(x + y)$ with $\tau(p \wedge q) \ge 1 - 2\eps$. By Lemma 44.3.5, $D(x + y)$ is dense in $H$, so $x + y$ is densely defined.

By (1) and the above case, $x^{*} + y^{*}$ is also densely defined, so $(x^{*} + y^{*})^{*}$ is a closed extension of $x + y$, and $x + y$ is closable. By Lemma 44.3.7, the closure of $x + y$ lies in $\td M$.

(3): For each $\eps > 0$, by Lemma 44.3.4, there exists $p, q \in \text{Proj}(M)$ with $pH \subset D(x)$, $qH \subset D(y)$, $yq \in M$, and $\tau(p), \tau(q) \ge 1 - \eps$.

By Lemma 44.3.6, there exists $r \in \text{Proj}(M)$ with $rH \subset D(xy)$ and

As such an $r$ exists for all $\eps > 0$, Lemma 44.3.5 implies that $D(xy)$ is dense in $H$, and $xy$ is densely defined.

By (1) and the above case, $y^{*}x^{*}$ is also densely defined, so $(y^{*}x^{*})^{*}$ is a closed extension of $xy$, and $xy$ is closable. By Lemma 44.3.7, the closure of $xy$ lies in $\td M$.

(Algebra): For each $x, y, z \in \td M$,

and

Let $\ol{xy + xz}$ denote the closure of $xy + xz$, then

By (2) and (3), $\ol{xy + xz}\in \td M$, so Proposition 44.3.3 implies that

Similarly, $(x \ol{\cdot}y)^{*} = (xy)^{*} \supset y^{*}x^{*}$ and $y^{*} \ol{\cdot}x^{*} \supset y^{*}x^{*}$. If $\ol{y^*x^*}$ is the closure of $y^{*}x^{*}$, then by (1), (3), and Proposition 44.3.3, $(x \ol{\cdot}y)^{*} = \ol{y^*x^*}= y^{*} \ol{\cdot}x^{*}$.$\square$