An Equivalent Formulation of the Relation $\precsim$ on $\mathcal{P}_{\infty}(A)$ For a C$^{*}$-Algebra $A$ This is from Exercise 4.7 in Rordam's book. Let $A$ be a C$^{*}$-algebra. By $\mathcal{P}_{\infty}(A)$, we mean $\bigcup_{n\in\mathbb{N}}\mathcal{P}(M_{n}(A))$, where $\mathcal{P}(M_{n}(A))$ denotes the set of projections in the C$^{*}$-algebra $M_{n}(A)$. We define a new relation $\precsim$ on $\mathcal{P}_{\infty}(A)$ as follows: for $p\in\mathcal{P}(M_{n}(A))$ and $q\in\mathcal{P}(M_{m}(A))$, we write $p\precsim q$ if there is a projection $q_{0}\in\mathcal{P}(M_{m}(A))$ such that $p\sim_{0}q_{0}\leq q$, where $\sim_{0}$ denotes relation stemming from the usual Murray von-Neumann equivalence of projections. In am trying to prove the following:

$p\precsim q$ if and only if $q\sim_{0} p\oplus p_{0}$ for some projection $p_{0}\in\mathcal{P}_{\infty}(A)$, where $p\oplus p_{0}=\operatorname{diag}(p,p_{0})$.

I have succeeded in proving the $\implies$ direction. We take $v\in M_{m,n}(A)$ such that $v^{*}v=p$ and $vv^{*}=q_{0}$. Then, letting $w=\begin{pmatrix}v & (q-q_{0})^{1/2}\end{pmatrix}\in M_{m,m+n}(A)$, we have 
$$
w^{*}w=\begin{pmatrix}p & 0\\ 0 & q-q_{0}\end{pmatrix} \qquad\text{and}\qquad ww^{*}=q.
$$
The fact that $q-q_{0}=:p_{0}$ is a projection follows from the fact that $q_{0}\leq q$ as in here. Thus, we get that $p\oplus p_{0}\sim_{0} q$.
I have been trying for a long time to prove the reverse direction $\impliedby$, but to no avail. I would really appreciate some help with the $\impliedby$ direction.
Thank you very much.
 A: Assume $p \oplus r \sim q$ via $v^*v = p \oplus r$ and $vv^* = q$ for $p,q,r \in M_n(A)$. Write
$$
 v = \begin{pmatrix} A & B \\ C & D \end{pmatrix}.
$$
Then
$$
 \begin{pmatrix} q & 0 \\ 0 & 0 \end{pmatrix} = q = vv^* = \begin{pmatrix} AA^* + BB^* & AC^* + BD^* \\ CA^* + DB^* & CC^* + DD^* \end{pmatrix}
$$
It follows that $q = AA^* + BB^*$ and $C = D = 0$.
Doing the same for $p \oplus r$ gives $p = A^*A, r = B^*B$ and $0 = A^*B = B^*A$.
Let $p_0 := q-BB^*$. Then 
$$
p = A^*A \sim AA^*  = (AA^* + BB^*) - BB^* = q -BB^* = p_0 \leq q.
$$
So $p \sim p_0 \leq p$.
For a better understanding of this relation:
The result says, that the algebraic order on $V(A)$ coincides with the order induced by $\lesssim$.
A: I came up with another proof which seems more general than the one already provided. 
Suppose $p\in\mathcal P_n(A)$, $q\in\mathcal P_m(A)$, and $p\oplus p_0\sim_0 q$ for some $p_0\in\mathcal P_k(A)$.  Then $p\oplus p_0=u^*u$ and $q=uu^*$ for some $u\in M_{n+k,m}(A)$.  Put $q_0=u(p\oplus 0_k)u^*$.  Then $q_0$ is a projection in $\mathcal P_m(A)$ with $q_0\leq u(p\oplus p_0)u^*=q$.  Writing $v=u(p\oplus 0_k)$, we have $v^*v=p\oplus 0_k\sim_0 p$ and $vv^*=q_0$, so $p\sim_0q_0\leq q$, and therefore $p\precsim q_0$.
