Does uniqueness of irredundant representation as meet of meet-irreducibles in a lattice of finite length imply semi-modularity?

Question

Question

It is "Donnellan - Lattice Theory" exercise 119.

Prove that a lattice of finite length in which every element has a unique representation as meet of meet-irreducible elements without redundancy is semi-modular.

Notation

A lattice $L$ is said to be of finite length if $L$ has a minimum $0$ and a maximum $1$ and there is a natural $n$ such that all totally ordered subsets have cardinality at most $n$.

An element $a\in L$ is said to be meet-irreducible if for all elements $b,c\in L$, then $a=b\wedge c$ implies $a=b$ or $a=c$.

A representation of an element $a\in L$ as meet of meet-irreducible elements is a finite set $S$ of meet-irreducible elements such that $\bigwedge S=a$.

A representation $S$ is said to be without redundancy if for all $s\in S$ we have $\bigwedge(S\setminus\{s\})\nleq s$.

A lattice is said to be semi-modular if for any $a, b, c$ such that $b \wedge c < a < c < b \vee c$, there is an element $d$ such that $b \wedge c < d \leq b$ and $a = (a \vee d) \wedge c$.

Answer 1

I found an answer in the article "Dilworth - Lattices with Unique Irreducible Decompositions".

For every $a\in L$ let $S_a$ be its unique representation without redundancy. Also let us say $a\succ b$ if $a$ covers $b$.

It is known that a lattice $L$ of finite length is semi-modular if and only if it is Birkhoff, that is:

(B) For $a,b\in L$, if $a\succ a\wedge b$, then $a\vee b\succ b$.

So it suffices to prove (B). Let $a\succ a\wedge b$ and suppose that $a\vee b > b' > b$. Then, since $\bigwedge S_b=b \ngeq b'$, there is a $q \in S_b$ such that $q \ngeq b'$, so $q \ngeq a$. Hence $a>a\wedge q\geq a\wedge b$, so $a\wedge b = a\wedge q = \bigwedge S_a\wedge q$. Moreover, $a\vee b'\geq a\vee b>b'$, so $\bigwedge S_{b'}=b'\ngeq a$, so there is a $q'\in S_{b'}$ such that $q' \ngeq a$. Then $a>a\wedge q'\geq a\wedge b$, so $a\wedge b = a\wedge q' = \bigwedge S_a\wedge q'$. But $q \neq q'$ and, by uniqueness of representation of $a\wedge b$, we have $S_{a\wedge b}\subseteq S_a\cup\{q\}$ and $S_{a\wedge b}\subseteq S_a\cup\{q'\}$, so $S_{a\wedge b}\subseteq S_a$, so $\bigwedge S_{a\wedge b}\geq\bigwedge S_a$, so $a\wedge b\geq a$, a contradiction.