Let $P$ be a set, $\leq$ a binary relation on $P$, reflexive, antisymmetric and transitive. Let $\wedge$ and $\vee$ be two binary operations, both commutative and associative and distributive one to each other. Let $0$ be the minimum element of $P$, $1$ the maximum. Suppose that for every $a\in P$ there exists $a'\in P$ such that $a\wedge a'=0$ and $a\vee a'=1$.
Now take $a,b\in P$ and assume $a\leq b$. Can i prove that $b'\leq a'$? How?
No, you haven't assumed enough connections between the $\land$ and $\lor$ operations on the one hand and the partial order $\leq$ on the other. Consider, for example, the set $P$ of all subsets of $\{p,q,r\}$, let $\land$ and $\lor$ be intersection and union, and let $0$ be the empty set and $1$ be $\{p,q,r\}$, so $a'$ is the complement of $a$. Now you can define $\leq$ to be, for example, the linear order $$ \varnothing<\{p\}<\{q\}<\{r\}<\{q,r\}<\{p,r\}<\{p,q\}<\{p,q,r\}, $$ and find that complementation does not reverse this order relation.