[proofplan] We prove the equivalence by comparing finiteness with cancellation among subprojections. First, assuming $p$ is finite, we show that a proper equivalence $e<f\leq p$ would allow us to extend the implementing partial isometry to one from a proper subprojection of $p$ onto all of $p$, contradicting finiteness. The converse is obtained by applying this cancellation criterion to a subprojection equivalent to $p$. Finally, we identify projections and partial isometries in the corner $pMp$ with the corresponding objects in $M$ dominated by $p$, which shows that finiteness is unchanged after passing to the corner. [/proofplan] [step:Convert a proper equivalence below $p$ into an equivalence with $p$] Assume that $p$ is finite in $M$. Let $e,f\in M$ be projections such that $e\leq f\leq p$ and $e\sim f$ in $M$. Choose a partial isometry $v\in M$ implementing this Murray-von Neumann equivalence. Since $e\leq f$ are projections, the standard order identities give $fe=ef=e$ and $f-e$ is a projection. In particular, \begin{align*} v^*v=e. \end{align*} \begin{align*} vv^*=f. \end{align*} Suppose, toward a contradiction, that $e\neq f$. Since $e\leq f$, the difference $f-e$ is a nonzero projection. Define \begin{align*} w:=v+(p-f)\in M. \end{align*} The projection $p-f$ is orthogonal to $f$, and since $e\leq f$, it is also orthogonal to $e$. Moreover $fv=v$ and $v^*f=v^*$ because $vv^*=f$. Hence \begin{align*} (p-f)v=0,\qquad v^*(p-f)=0. \end{align*} Therefore the cross terms vanish in the products defining the initial and final projections of $w$: \begin{align*} w^*w=(v^*+p-f)(v+p-f)=v^*v+(p-f)=e+(p-f). \end{align*} Similarly, \begin{align*} ww^*=(v+p-f)(v^*+p-f)=vv^*+(p-f)=f+(p-f)=p. \end{align*} Thus $w$ is a partial isometry in $M$ whose initial projection is \begin{align*} r:=e+(p-f) \end{align*} and whose final projection is $p$. Since \begin{align*} p-r=p-e-(p-f)=f-e, \end{align*} and $f-e\neq 0$, the projection $r$ is a proper subprojection of $p$. Hence $r\sim p$ with $r<p$, contradicting that $p$ is finite in $M$. Therefore $e=f$. [guided] Assume that $p$ is finite in $M$. We want to prove that equivalence cannot occur strictly inside $p$: if $e\leq f\leq p$ and $e\sim f$, then equality must hold. Let $e,f\in M$ be projections such that $e\leq f\leq p$ and $e\sim f$ in $M$. By the definition of Murray-von Neumann equivalence, there exists a partial isometry $v\in M$ such that \begin{align*} v^*v=e. \end{align*} \begin{align*} vv^*=f. \end{align*} Suppose for contradiction that $e\neq f$. Because $e\leq f$, the difference $f-e$ is a projection. Since $e\neq f$, this projection is nonzero. The idea is to enlarge the equivalence $e\sim f$ by adding the identity map on the remaining part $p-f$. Define \begin{align*} w:=v+(p-f)\in M. \end{align*} We now verify directly that $w$ is a partial isometry from a proper subprojection of $p$ onto $p$. First, $p-f$ is orthogonal to $f$ because $f\leq p$. Since $e\leq f$, it is also orthogonal to $e$. Also $vv^*=f$, so $fv=vv^*v=v$, and therefore $(p-f)v=0$. Taking adjoints gives $v^*(p-f)=0$. These two identities are exactly what makes the cross terms vanish. Compute the initial projection of $w$: \begin{align*} w^*w=(v^*+p-f)(v+p-f)=v^*v+(p-f)=e+(p-f). \end{align*} The equality uses $(p-f)v=0$ and $v^*(p-f)=0$. Compute the final projection: \begin{align*} ww^*=(v+p-f)(v^*+p-f)=vv^*+(p-f)=f+(p-f)=p. \end{align*} Thus $w$ is a partial isometry whose initial projection is \begin{align*} r:=e+(p-f) \end{align*} and whose final projection is $p$. This means $r\sim p$ in $M$. It remains to check that $r$ is properly smaller than $p$. Since $e\leq f\leq p$, we have \begin{align*} p-r=p-e-(p-f)=f-e. \end{align*} The projection $f-e$ is nonzero by the assumption $e\neq f$. Hence $r<p$. We have therefore found a proper subprojection $r$ of $p$ with $r\sim p$, which contradicts the definition that $p$ is finite in $M$. The contradiction proves $e=f$. [/guided] [/step] [step:Recover finiteness of $p$ from cancellation below $p$] Assume condition $2$. To prove that $p$ is finite in $M$, let $q\in M$ be a projection such that $q\leq p$ and $q\sim p$ in $M$. Apply condition $2$ with \begin{align*} e:=q,\qquad f:=p. \end{align*} Since $q\leq p\leq p$ and $q\sim p$, condition $2$ gives $q=p$. Thus no proper subprojection of $p$ is Murray-von Neumann equivalent to $p$, so $p$ is finite in $M$. [/step] [step:Identify projections and equivalences inside the corner $pMp$] Let \begin{align*} N:=pMp \end{align*} denote the corner von Neumann algebra, whose identity projection is $p$. A projection $a\in N$ is exactly a projection $a\in M$ satisfying $a\leq p$. Indeed, if $a\in N$ is a projection, then $a=pap$, so $pa=a=ap$, which is equivalent to $a\leq p$ for projections. Conversely, if $a\in M$ is a projection and $a\leq p$, then $pa=a=ap$, hence $a=pap\in pMp=N$. Next, Murray-von Neumann equivalence between projections below $p$ is the same whether computed in $M$ or in $N$. If $u\in N$ is a partial isometry, then $u^*u,uu^*\in N$, so its initial and final projections are dominated by $p$. Conversely, suppose $u\in M$ is a partial isometry and \begin{align*} u^*u\leq p,\qquad uu^*\leq p. \end{align*} Since $u=uu^*u$ and $u=u(u^*u)$, the inequalities give \begin{align*} pu=p(uu^*)u=uu^*u=u \end{align*} and \begin{align*} up=u(u^*u)p=u(u^*u)=u. \end{align*} Therefore $u=pup\in pMp=N$. Thus a partial isometry in $M$ implementing equivalence between projections dominated by $p$ actually lies in the corner. [guided] Let \begin{align*} N:=pMp. \end{align*} This is the corner von Neumann algebra, and its identity projection is $p$, not the identity of $M$ unless $p=1$. We first identify the projections of $N$. Suppose $a\in N$ is a projection. Since $a\in pMp$, we have $a=pap$. Multiplying by $p$ on either side gives $pa=a$ and $ap=a$. For projections, the relations $pa=a=ap$ are equivalent to $a\leq p$. Hence every projection of $N$ is a projection of $M$ dominated by $p$. Conversely, suppose $a\in M$ is a projection satisfying $a\leq p$. The order relation between projections means $pa=a=ap$. Therefore \begin{align*} a=pap, \end{align*} so $a\in pMp=N$. Thus the projections of the corner are precisely the projections of $M$ lying below $p$. We next check that Murray-von Neumann equivalence is also unchanged for such projections. If $u\in N$ is a partial isometry, then $u^*u$ and $uu^*$ belong to $N$, so by the projection identification just proved, both are dominated by $p$. Conversely, suppose $u\in M$ is a partial isometry such that \begin{align*} u^*u\leq p,\qquad uu^*\leq p. \end{align*} We prove that $u$ actually belongs to the corner. Since $u$ is a partial isometry, it satisfies \begin{align*} u=uu^*u \end{align*} and \begin{align*} u=u(u^*u). \end{align*} The inequality $uu^*\leq p$ gives $p(uu^*)=uu^*$, so \begin{align*} pu=p(uu^*)u=uu^*u=u. \end{align*} The inequality $u^*u\leq p$ gives $(u^*u)p=u^*u$, so \begin{align*} up=u(u^*u)p=u(u^*u)=u. \end{align*} Therefore \begin{align*} u=pup, \end{align*} and hence $u\in pMp=N$. This proves that a partial isometry in $M$ implementing equivalence between projections below $p$ is exactly the same object as a partial isometry in the corner implementing the same equivalence. [/guided] [/step] [step:Conclude that finiteness is unchanged in the corner] Using the identifications above, condition that $p$ be finite in $M$ says: there is no projection $q\in M$ with $q<p$ and $q\sim p$ in $M$. Since the projections $q\in M$ with $q\leq p$ are exactly the projections of $pMp$, and equivalence between such projections is the same in $M$ and in $pMp$, this is equivalent to saying that the identity projection $p$ has no proper subprojection in $pMp$ equivalent to $p$. That is exactly condition $3$. We have proved $1\implies 2$, $2\implies 1$, and $1\iff 3$. Therefore the three stated conditions are equivalent. [/step]