Boolean algebra-Modular lattice

Let $L=\{a,b,c,d,e,f\}$; $P(L)$ is the set of all partitions of $L$, and $\le$ is the order relation on $P(L)$ defined as:

if $r$ and $t$ are relations, then $r\le t$ iff every block in $r$ is a subset of some block in $t$.

Show that the lattice $(P(L),\le)$ is not modular.

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Do you mean for $r$ and $t$ to be equivalence relations, and for the underlying set to be $P(L\times L)$? Otherwise block makes no sense to me. – Brian M. Scott Dec 6 '11 at 20:38
I don't understand your definition. "If $r$ and $t$ are relations"... relation on what? If they are relations on $L$, then they are elements of $P(L\times L)$, not of $P(L)$. If they are relations on $P(L)$, then they are not elements of $P(L)$. And what is a "block" of a relation? – Arturo Magidin Dec 6 '11 at 20:39
No, I think it's not on LxL. I found all subsets of L, and I noticed r1={{a,b}, {c}, {d}, {e}, {f}} r2={{c,d,e}, {a}, {b}, {f}} r3={{a,b,c,f}, {d}, {e}} r4={{a,b,c,d,e}, {f}} r5={{a,b,c,d,e,f}} – Ggg Dec 6 '11 at 20:55
do those 5 subsets make N5? – Ggg Dec 6 '11 at 20:56
You’re looking at the lattice of partitions of $L$. The set of partitions of $L$ is not $P(L)$; it’s a subset of $P(P(L))$. – Brian M. Scott Dec 6 '11 at 21:04

Your comments indicate that you’re really looking at the lattice $(P,\le)$ of partitions of $L$, where for $r,s\in P$ we define $r\le s$ iff for each $x\in r$ there is a $y\in s$ such that $x\subseteq y$. (That is, each piece of $r$ is a subset of some piece of $s$.) Note that $P$ is not $\wp(L)$, or even a subset of $\wp(L)$: it’s a subset of $\wp(\wp(L))$.
HINT: Let $1$ be the trivial partition whose only member is $\{a,b,c,d,e,f\}$, and let $0$ be the partition $\{\{a\},\{b\},\{c\},\{d\},\{e\},\{f\}\}$. Let $$r=\{\{a,b,c\},\{d,e,f\}\}$$ and $$s=\{\{a,d\},\{b,e\},\{c,f\}\}\;.$$
1. Can you show that $r\land s=0$ and $r\lor s=1$? That is, can you show that $0$ is the only partition $\le$ both $r$ and $s$, and $1$ is the only partition $\ge$ both $r$ and $s$?
2. Can you find a partition $x$ of $L$ such that $x=x\le s$ and $$x\lor(r\land s)\ne(x\lor r)\land s=1\land s=s\;?$$
@Ggg: Yes, $P(L)$ is the set of all subsets of $L$, but $\{\{a,b,c,d,e\},\{f\}\}$, for instance, is not a subset of $L$, so it’s not a member of $P(L)$. It’s a set of subsets of $L$, so it’s a member of $P(P(L))$. – Brian M. Scott Dec 6 '11 at 21:16