Let $S$ be a set. We say that $\perp\subseteq S\times S$ is an orthogonality relation, if
- $a\perp b\Rightarrow b\perp a$
- $a\perp a\Rightarrow ((\forall b)(b\perp a))$
- $((\forall c)(c\perp a\Leftrightarrow c\perp b))\Rightarrow a=b$
Now, it's easy to see that we can define an order on $S$ by $a\le b\Longleftrightarrow(\forall d)(d\perp b\Rightarrow d\perp a)$.
We say that $(P,\le,',0,1)$ is an orthocomplementary poset, if
- $\le$ is an order
- $(\forall a)(0\le a\le 1)$
- $a\le b\Rightarrow b'\le a'$
- $a''=a$
- $a\wedge a'=0$
Now, the other way around we can define an orthogonality relation on $P$ by $a\perp b\Longleftrightarrow a\le b'$. The first and the second can be checked easily. But I do not know how to prove the last one. I must miss something easy...
If $(P,\leq,',0,1)$ is an orthocomplementary poset where you define $a \perp b$ iff $a \leq b'$, then, if $$x \perp a \Longleftrightarrow x \perp b,$$ for all $x \in P$, then $$a \perp a',$$ since $a \leq a'' = a$, whence $$a' \perp a,$$ by the first condition you claim having proven, yielding $$a' \perp b,$$ by hypothesis. Now it follows that $a' \leq b'$, by definition of $\perp$, whence $b \leq a$.
By symmetry, $a \leq b$, and therefore $a=b$.