[proofplan] We construct a maximal partial matching between subprojections of $p$ and subprojections of $q$ by [Zorn's lemma](/theorems/1226). The unmatched residual projections $r \le p$ and $s \le q$ must have orthogonal central carriers; otherwise a nonzero corner of $rMs$ supplies a further partial isometry and contradicts maximality. Taking $z=1-c(r)$ then kills the residual part of $p$ on the $z$-summand and kills the residual part of $q$ on the complementary summand, giving the two required Murray-von Neumann subequivalences. [/proofplan]

[step:Choose a maximal orthogonal family of partial isometries matching $p$ into $q$]Let $\mathcal{F}$ be the set of all families $(v_\alpha)_{\alpha\in A}$ of partial isometries in $M$ such that, for every $\alpha,\beta\in A$ with $\alpha\ne\beta$, \begin{align*} v_\alpha^*v_\alpha \le p,\qquad v_\alpha v_\alpha^* \le q, \end{align*} and \begin{align*} v_\alpha^*v_\alpha \perp v_\beta^*v_\beta,\qquad v_\alpha v_\alpha^* \perp v_\beta v_\beta^*. \end{align*} We order $\mathcal{F}$ by inclusion of families. Let $\mathcal{C}\subseteq \mathcal{F}$ be a chain. The union of the families in $\mathcal{C}$ is again a family satisfying the same orthogonality conditions, because every finite subfamily is contained in one member of the chain. Hence every chain has an upper bound in $\mathcal{F}$. By Zorn's lemma, there is a maximal family $(v_\alpha)_{\alpha\in A}$ in $\mathcal{F}$. Define projections $e,f\in M$ by the strong operator sums \begin{align*} e=\sum_{\alpha\in A} v_\alpha^*v_\alpha,\qquad f=\sum_{\alpha\in A} v_\alpha v_\alpha^*. \end{align*} These sums are the suprema of pairwise orthogonal families of projections in the complete projection lattice of $M$, so $e,f\in M$, with $e\le p$ and $f\le q$ by [citetheorem:9267]. Define $v\in M$ to be the strong operator sum \begin{align*} v=\sum_{\alpha\in A} v_\alpha. \end{align*} The orthogonality of the initial and final projections gives \begin{align*} v^*v=e,\qquad vv^*=f. \end{align*} Thus $v$ is a partial isometry in $M$ matching $e$ with $f$.[/step]

[guided]The goal of this step is to match as much of $p$ as possible with a subprojection of $q$. A single match is a partial isometry $w\in M$ with $w^*w\le p$ and $ww^*\le q$. A family of such matches can be added only when the initial projections and final projections are pairwise orthogonal; this is exactly why we impose \begin{align*} v_\alpha^*v_\alpha \perp v_\beta^*v_\beta,\qquad v_\alpha v_\alpha^* \perp v_\beta v_\beta^* \end{align*} for distinct indices $\alpha,\beta\in A$. Let $\mathcal{F}$ denote the collection of all such orthogonal matching families. We order it by inclusion. If $\mathcal{C}$ is a chain in $\mathcal{F}$, then the union of the families in $\mathcal{C}$ is still admissible: any finite set of its members already appears together in one element of the chain, and therefore satisfies the required orthogonality and domination conditions. Hence the union is an upper bound for the chain. Zorn's lemma gives a maximal admissible family $(v_\alpha)_{\alpha\in A}$. Now we convert the maximal family into one partial isometry. Let \begin{align*} e=\sum_{\alpha\in A} v_\alpha^*v_\alpha,\qquad f=\sum_{\alpha\in A} v_\alpha v_\alpha^* \end{align*} be the strong operator sums of the pairwise orthogonal initial and final projections. By the completeness of the projection lattice in a von Neumann algebra, supplied by [citetheorem:9267], these sums are projections in $M$. Since every $v_\alpha^*v_\alpha$ is dominated by $p$, their supremum satisfies $e\le p$; similarly $f\le q$. Define \begin{align*} v=\sum_{\alpha\in A} v_\alpha \end{align*} as the strong operator sum. Orthogonality makes the cross terms vanish, so the finite partial sums satisfy \begin{align*} \left(\sum_{\alpha\in F}v_\alpha\right)^*\left(\sum_{\alpha\in F}v_\alpha\right)=\sum_{\alpha\in F}v_\alpha^*v_\alpha \end{align*} and \begin{align*} \left(\sum_{\alpha\in F}v_\alpha\right)\left(\sum_{\alpha\in F}v_\alpha\right)^*=\sum_{\alpha\in F}v_\alpha v_\alpha^* \end{align*} for every finite subset $F\subseteq A$. Passing to strong limits gives \begin{align*} v^*v=e,\qquad vv^*=f. \end{align*} Thus $v$ is a partial isometry in $M$, and it records the maximal matching obtained from Zorn's lemma.[/guided]

[step:Show that the residual projections have orthogonal central carriers] Define the residual projections \begin{align*} r=p-e,\qquad s=q-f. \end{align*} Since $e\le p$ and $f\le q$, these are projections in $M$. We claim that \begin{align*} c(r)c(s)=0, \end{align*} where $c(r)$ and $c(s)$ denote the central carriers of $r$ and $s$. Suppose instead that $c(r)c(s)\ne 0$. If $rMs=\{0\}$, then the standard central-carrier annihilation criterion gives $c(r)c(s)=0$, a contradiction. Hence there exists $a\in M$ such that \begin{align*} ras\ne 0. \end{align*} Set $x=ras\in M$. Then $x\ne 0$, and \begin{align*} x=rxs. \end{align*} By the [polar decomposition in a von Neumann algebra](/theorems/9270) [citetheorem:9270], there is a nonzero partial isometry $u\in M$ such that \begin{align*} x=u|x|. \end{align*} Since $x=rxs$, the initial and final projections of $u$ satisfy \begin{align*} u^*u\le s,\qquad uu^*\le r. \end{align*} Thus $u^*$ is a nonzero partial isometry with \begin{align*} (u^*)^*u^*=uu^*\le r\le p \end{align*} and \begin{align*} u^*(u^*)^*=u^*u\le s\le q. \end{align*} Moreover these projections are orthogonal to every $v_\alpha^*v_\alpha$ and every $v_\alpha v_\alpha^*$ because they lie under $r=p-e$ and $s=q-f$. Therefore adjoining $u^*$ to the family $(v_\alpha)_{\alpha\in A}$ gives a strictly larger admissible family, contradicting maximality. Hence $c(r)c(s)=0$. [/step]

[step:Set the central projection and verify the comparison on the first summand] Define \begin{align*} z=1-c(r). \end{align*} Then $z\in Z(M)$ is a central projection. Since $c(r)$ dominates $r$, we have \begin{align*} zr=0. \end{align*} Using $p=e+r$, we obtain \begin{align*} zp=ze+zr=ze. \end{align*} The operator $zv\in M$ is a partial isometry because $z$ is central and $v$ is a partial isometry. Its initial and final projections are \begin{align*} (zv)^*(zv)=zv^*v=ze=zp \end{align*} and \begin{align*} (zv)(zv)^*=zvv^*=zf\le zq. \end{align*} Therefore \begin{align*} zp\precsim zq. \end{align*} [/step]

[step:Use orthogonality of central carriers to verify the comparison on the complementary summand] Since $1-z=c(r)$ and $c(r)c(s)=0$, we have \begin{align*} (1-z)c(s)=0. \end{align*} Because $c(s)$ dominates $s$, it follows that \begin{align*} (1-z)s=0. \end{align*} Using $q=f+s$, we obtain \begin{align*} (1-z)q=(1-z)f. \end{align*} The partial isometry $(1-z)v^*\in M$ has initial projection \begin{align*} \bigl((1-z)v^*\bigr)^*((1-z)v^*)=(1-z)vv^*=(1-z)f=(1-z)q \end{align*} and final projection \begin{align*} ((1-z)v^*)\bigl((1-z)v^*\bigr)^*=(1-z)v^*v=(1-z)e\le (1-z)p. \end{align*} Hence \begin{align*} (1-z)q\precsim (1-z)p. \end{align*} The central projection $z$ therefore satisfies both required comparison relations. [/step]