The following appears (with slightly different notation) as problem 1.26 in Bell's Set Theory: Boolean-Valued Models and Independence Proofs, 3rd edition:
Let $B$ be a complete Boolean algebra and $\{b_i:i\in I\}$ a partition of unity. Let $\{x_i:i\in I\}$ be a family of sets with $i\neq j\implies x_i\neq x_j$. Find a name $\tau$ such that, for every $i\in I$, $\|\tau=\check{x_i}\|=b_i$.
If we let $\tau$ be the mixture defined by $\text{dom}(\tau)=\{\check y:y\in x\}$ (where $x:=\bigcup_{i\in I} x_i$) and $\tau(\check y):=\bigvee_{i\in I}b_i\wedge \|\check y\in \check x_i\|$, then the Mixing Lemma guarantees $b_i\le \| \tau=\check{x_i}\|$ for every $i\in I$, so it suffices to establish the other equality.
Through a series of manipulations, I'm able to show that $$ \|\tau=\check x_i\|=\bigwedge_{y\in x\setminus x_i}\bigwedge_{z\in x_i} \bigvee_{j\in I_z\setminus I_y} b_j $$ where $I_u:=\{j\in I:u\in x_j\}$. Obviously, if $y\in x\setminus x_i$ and $z\in x_i$, then $j\in I_z\setminus I_y$, but this only says that $b_i$ is bounded above by the inner sup. More worryingly, it is immediate from the above expression that $\|\tau=x_i\|\wedge b_i=b_i$, which yields the inequality we already had, so it would strike me as odd if the form above would result in the opposite inequality.
Is there a different mixture I should be considering? I tried a few different variations with no luck. If $\tau$ any name is such that $\text{dom}(\tau)=\{\check e:e\in E\}$ for some set $E$, then one can show that $$ \|\tau=\check{x_i}\|=\bigwedge_{e\in E\cap x_i} \bigwedge_{y\in E\setminus x_i} (\tau(\check e))'\wedge\tau(\check y) $$ but there's no clear choice of $E$ and the values $\tau(\check e)$ so that the LHS is $b_i$.
Fix $i.$ We want to show $\| \tau=\check{x_i}\|\leq b_i.$ Equivalently, we want to show $1-b_i\leq \| \tau\neq\check{x_i}\|.$ You know $b_j\leq \| \tau=\check{x_j}\|$ and $\| \tau=\check{x_j}\|\leq \|\tau\neq\check{x_i}\|$ for $j\neq i.$ Also, $1-b_i=\bigvee_{j\neq i}b_j$ because $b_i\wedge\bigvee_{j\neq i}b_j=0$ and $b_i\vee \bigvee_{j\neq i}b_j=1.$ So
$$1-b_i=\bigvee_{j\neq i}b_j\leq \bigvee_{j\neq i}\| \tau=\check{x_j}\|\leq\|\tau\neq\check{x_i}\|.$$
To try and make this more general: if you want to verify $t=0$ for a given $t\in B,$ it suffices to check on a dense set of conditions, i.e. check $t\wedge b_i=0$ for each $i\in I$ where $\bigvee_i b_i=1.$ In more algebraic terms, a partition of unity $\langle b_i:i\in I\rangle$ gives you an isomorphism $\pi:B\to \prod_{i\in I} B/(1-b_i),$ and you can check a ring element is zero by checking that its image is zero.