For a C*-algebra $A$ and open projections $p,q\in A^{**}$, consider the following statements.
$\overline{p}\leq q$
$p\leq q$ and there exists open $r\in A^{**}$ with $rp=0$ and $r\vee q=1$
$p\leq q$ and there exists open $r\in A^{**}$ with $r\wedge p=0$ and $r\vee q=1$.
$p\leq q$ and there exists open $r\in A^{**}$ with $(r\wedge p)^\circ=0$ and $r\vee q=1$.
In general 1.$\Rightarrow$2.$\Rightarrow$3.$\Rightarrow$4. and they are all equivalent if $A$ is commutative. Originally I wanted to know if some of these equivalences hold even for non-commutative $A$, at least when $A$ is unital, but eventually came up with the following counterexamples using just continuous functions to $M_2$ (although I am not entirely sure about 4.$\nRightarrow$3.).
For the first two counterexamples, let $A=C(X,M_2)$, where $X=\{0\}\cup\{1/n:n\in\mathbb{N}\}$. Identify $A^{**}$ with all bounded functions from $X$ to $M_2$. Take rank $1$ projections $P,Q\in M_2$ and define open projections $p,q\in A^{**}$ by $p(0)=P$, $p(1/n)=1$, $q(0)=0$ and $q(1/n)=Q$, for all $n\in\mathbb{N}$. Then $\overline{q}(0)=Q$ and hence $\overline{q}\nleq p$ as long as $P\neq Q$, i.e. 1. fails. However, defining $r\in A$ by $r(x)=P^\perp$, for all $x\in X$, we see that $2.$ holds, as long as $P^\perp\neq Q$, i.e. 2.$\nRightarrow$1.. In fact, $r=p^{\perp\circ}$, which shows that 2. fails when $P^\perp=Q$. But in this case, defining $r\in A$ by $r(x)=R$, for any rank $1$ projection $R\neq P,Q$, we see that 3. holds, i.e. 3.$\nRightarrow$2..
For 4.$\nRightarrow$3., I need a topological space $X$ and a function $p:X\rightarrow\mathcal{P}$, where $\mathcal{P}$ denotes the rank $1$ projections in $M_2$, such that $p(x_0)=0$, for some $x_0\in X$, $p$ is continuous everywhere except at $x_0$, $p$ is not constant on any open subset of $X$, and every $r\in C(X,\mathcal{P})$ coincides with $p$ at some point. I would guess this holds if $X$ is the unit ball of $\mathbb{C}\times\mathbb{C}$ and we define $p(x)$ to be the projection onto $\mathbb{C}x$, in which case $x_0=(0,0)$. Then we let $A=C(X,M_2)$ and identify $A''$ in the atomic representation (rather than the universal representation, which is valid as the canonical $\pi:A^{**}\rightarrow A''$ is faithful on open and closed projections - see Theorem 4.3.15 of Pedersen's "C*-algebras and their Automorphism Groups") with all bounded functions from $X$ to $M_2$. Then $p$ is open and so is $q\in A''$ defined by $q(x_0)=Q$, for some $Q\in\mathcal{P}$, and $q(x)=1$ otherwise. Defining $r\in A$ by $r(x)=R$, for all $x\in X$ and any fixed $R\in\mathcal{P}\setminus\{Q\}$, shows that 4. holds. But $r(x_0)\neq0$ for any open $r\in A''$ with $r\vee q=1$ and hence, by our assumptions on $p$, $p(x)\leq r(x)$, for some $x\in X$, so $p\wedge r\neq0$ i.e. 3. fails.
