[proofplan] The forward implication is the trace calculation built into Murray-von Neumann subequivalence: a partial isometry identifies $p$ with a subprojection of $q$, and traciality gives equality of traces for equivalent projections. For the converse, the projection comparison theorem for factors says that either $p\precsim q$ or $q\precsim p$. If the first alternative fails, we embed $q$ into $p$, compare traces, and use faithfulness to force a strict inequality $\tau(p)>\tau(q)$, contradicting the assumed trace inequality. [/proofplan] [step:Use a partial isometry implementing subequivalence to prove the trace inequality] Assume $p\precsim q$. By definition, there are a projection $r\in M$ and a partial isometry $v\in M$ such that $r\le q$, $v^*v=p$, and $vv^*=r$. Since $\tau$ is a faithful normal tracial state, it is a faithful normal semifinite trace. By [citetheorem:9317] applied to the Murray-von Neumann equivalent projections $p=v^*v$ and $r=vv^*$, \begin{align*} \tau(p)=\tau(r). \end{align*} Since $r\le q$, the element $q-r$ is a projection in $M$. Positivity and additivity of $\tau$ give \begin{align*} \tau(q)=\tau(r)+\tau(q-r)\ge \tau(r). \end{align*} Therefore $\tau(p)\le \tau(q)$. [guided] Assume $p\precsim q$. The definition of Murray-von Neumann subequivalence gives the concrete data needed for a trace comparison: there are a projection $r\in M$ and a partial isometry $v\in M$ such that \begin{align*} r\le q,\qquad v^*v=p,\qquad vv^*=r. \end{align*} Thus $p$ is Murray-von Neumann equivalent to the subprojection $r$ of $q$. We now compare traces. The functional $\tau$ is a faithful normal tracial state, hence in particular a faithful normal semifinite trace. The hypotheses of [citetheorem:9317] are therefore satisfied for the equivalent projections $p=v^*v$ and $r=vv^*$. It follows that \begin{align*} \tau(p)=\tau(r). \end{align*} It remains to use that $r$ sits under $q$. Since $r\le q$ and both are projections, $q-r$ is again a projection in $M$. Because $\tau$ is positive and linear, \begin{align*} \tau(q)=\tau(r)+\tau(q-r). \end{align*} The projection $q-r$ is positive, so $\tau(q-r)\ge 0$. Hence \begin{align*} \tau(q)\ge \tau(r)=\tau(p). \end{align*} This proves the forward implication. [/guided] [/step] [step:Apply total comparison of projections in a factor] Assume now that $\tau(p)\le \tau(q)$. Since $M$ is a factor, the comparison theorem for projections in factors, [citetheorem:9281], applies to $p$ and $q$. Hence either $p\precsim q$ or $q\precsim p$. If $p\precsim q$, the desired conclusion already holds. It remains to rule out the alternative under the additional assumption that $p\nprecsim q$. Suppose, for contradiction, that $p\nprecsim q$. Then total comparison gives $q\precsim p$. [/step] [step:Convert the opposite comparison into a subprojection of $p$ equivalent to $q$] From $q\precsim p$, there are a projection $q_0\in M$ and a partial isometry $w\in M$ such that $q_0\le p$, $w^*w=q$, and $ww^*=q_0$. Thus $q_0\sim q$. Since $\tau$ is a faithful normal semifinite trace, [citetheorem:9317] gives \begin{align*} \tau(q_0)=\tau(q). \end{align*} [/step] [step:Use faithfulness to contradict the assumed trace inequality] If $p=q_0$, then $p\sim q$, and hence $p\precsim q$, contradicting the assumption $p\nprecsim q$. Therefore $p-q_0$ is a nonzero projection in $M$. Because $\tau$ is faithful and $p-q_0$ is a nonzero positive element of $M$, \begin{align*} \tau(p-q_0)>0. \end{align*} Since $q_0\le p$, linearity gives \begin{align*} \tau(p)=\tau(q_0)+\tau(p-q_0). \end{align*} Using $\tau(q_0)=\tau(q)$, we obtain \begin{align*} \tau(p)=\tau(q)+\tau(p-q_0)>\tau(q). \end{align*} This contradicts the hypothesis $\tau(p)\le \tau(q)$. Hence the assumption $p\nprecsim q$ is impossible, so $p\precsim q$. Combining this with the first step proves the equivalence. [/step]