Consider two element in a lattice $L$ says a and b. we say that a covers b and write $a \succ b$ if $a > b$ and if there is no x in L such that $a > x > b$.
A lattice is said be Semmimodular if for all a,b in L, $a \succ a \wedge b$ and $b \succ a \wedge b$ $\Rightarrow a \vee b \succ a$ and $a \vee b \succ b$
we know that $ \rho \wedge \sigma = \rho \cap \sigma $ for any $\rho, \sigma \in E(X)$.
Let $\rho , \sigma$ be equivalences on $X$ such that $\rho \succ \rho \wedge \sigma$ and $\sigma \succ \rho \wedge \sigma$
therefore $\rho \succ \rho \cap \sigma$ and $\sigma \succ \rho \cap \sigma$
Suppose that $\rho \cap \sigma $ classes are $A_1, A_2 , \cdots $
I want to prove that exactly one $\rho -$ class that is the union of two $\rho \cap \sigma$ classes and other $\rho$ and $\rho \cap \sigma$ classes are same .
I mean that $\rho$ classes of the form $A_1 \cup A_2 , A_3 , A_4 , \cdots $
Any help would be appreciated . Thank you
It might be easier if you use the following definition of semi-modular lattice:
$\mathbf{L}$ is semi-modular if for every $a, b, c \in L$, $$a \prec b \Rightarrow a \vee c \preceq b \vee c.$$ Then, I suppose you know that in the lattice of partitions of a set, $\rho \prec \sigma$ if $\sigma$ results from $\rho$ by uniting two of its blocks.
Now, given that characterization of the cover relations, if $\rho, \sigma, \pi$ are partitions (equivalent relations) of a set, if $\rho \prec \sigma$, then either $\rho \vee \pi = \sigma \vee \pi$ (if in $\pi$ there are elements of both of those blocks of $\rho$, so that those blocks are also united in $\rho \vee \pi$) or $\rho \vee \pi \prec \sigma \vee \pi$, so that $a \vee c \preceq b \vee c$.