I am trying to prove the following: for any Boolean algebra $B$, if $R$ is a $(\le)$ relation on $B$ such that $x \le y$ iff $x+y=y$, then $R$ is a partial ordering. Thus far I have shown that $R$ is reflexive and anti-symmetric. However, I am having difficulty showing that $R$ is transitive. Here is what I have so far:
$1$. Assume $(x,y)\in R$ and $(y,z)\in R$.
$2$. This implies $x+y=y$ and $y+z=z$ by definition of $R$.
$3$. I'm having trouble deducing $x+z=z$. I know that if $y=0$, then $x+0=x=0$ and I can justify using $x$ as zero element to conclude $x+z=z$; however, I do not believe I can I assume $y=0$ because $0$ is not necessarily an element included in the relation.
Any help is appreciated. Thank you!
With your notation: $x+z = x+(y+z) = (x+y)+z = y+z = z$.