Projections: Ordering Given a unital C*-algebra $1\in\mathcal{A}$.
Consider projections:
$$P^2=P=P^*\quad P'^2=P'=P'^*$$
Order them by:
$$P\leq P':\iff\sigma(\Delta P)\geq0\quad(\Delta P:=P'-P)$$

Then equivalently:
  $$P\leq P'\iff P=PP'=P'P\iff\Delta P^2=\Delta P=\Delta P^*$$

How can I check this?
(Operator algebraic proof?)
 A: The assertion $P\leq Q$ means $Q-P\geq0$. Then
$$
0\leq P(Q-P)P=PQP-P\leq P^2-P=P-P=0.
$$
So $P=PQP$. Now we can write this equality as $$0=P-PQP=P(I-Q)P=[(I-Q)P]^*[(I-Q)P],$$
so $(I-Q)P=0$, i.e. $P=QP$. Taking adjoints, $P=PQ$. 
The converse also holds: if $P=PQ=QP $, then $$Q-P=Q^2-QPQ=Q (I-P)Q\geq0. $$
Edit: for the equivalence $Q-P\geq0$ iff $Q-P$ is a projection:
I think it is easier to show that $Q-P$ is a projection iff $P=PQ=QP$ (the latter being equivalent to $Q-P\geq0$ by the above). 
If $Q-P$ is a projection, then $$Q-P=(Q-P)^2=Q+P-PQ-QP,$$ so $$\tag{1}2P=PQ+QP.$$ Multiplying by $I-P$ on the left, we get $(I-P)QP=0$, or $QP=PQP$. Taking adjoints, we obtain $QP=PQ$. Now $(1)$ is $2P=2PQ$, i.e. $P=PQ=QP$. 
Conversely, if $P=QP=PQ$, then 
$$
(Q-P)^2=Q+P-PQ-QP=Q+P-2P=Q-P.
$$
A: If $Q \ge P$, then
\begin{align}
      \|Px\|^{2} & =\|QPx\|^{2}+\|(I-Q)Px\|^{2} \\
            & = (PQPx,x)+\|(I-Q)Px\|^{2} \\
            & \ge (PPPx,x)+\|(I-Q)Px\|^{2} \\
            & = \|Px\|^{2}+\|(I-Q)Px\|^{2} \\
         \implies & (I-Q)Px=0.
\end{align}
