Proof Cancellation property in lattice implies distributivity

The proof on page 16 of the linked paper is incorrect.

One easy way to show that ``cancellativity'' implies distributivity is to show:

(1) any lattice with a sublattice isomorphic to either $M_3$ or $N_5$ must fail cancellation, and
(2) any lattice that has no sublattice isomorphic to $M_3$ or $N_5$ is distributive.

But one can argue with elements alone, without referring to sublattices, or to $M_3$, or $N_5$, as I will show.

I will type $a+b$ instead of $a\vee b$, and $ab$ instead of $a\wedge b$, since this is easier for me to type and to read. The cancellation rule is now: $a+b=a+c$ and $ab=ac$ imply $b=c$. This is a universally quantified statement. Also, it is invariant under interchanging join and meet, so if we can derive any consequences from cancellation, then we automatically have the dual statement as well.

For example, we prove the first statement of the following lemma, and the second follows automatically by duality.

Lemma 1. (Modular law) Assume that $L$ satisfies cancellation. Then $L$ satisfies the laws
(1) $a(b+ac) = ab+ac$, and
(2) $a+(b(a+c)) = (a+b)(a+c)$.

Proof of (1). Let $U=a(b+ac)$ and $V=ab+ac$. We have $ab \leq V \leq U\leq a$. Meeting throughout with $b$ yields $ab\leq Vb\leq Ub\leq ab$, so $Ub=Vb$. We also have $ac \leq V \leq U\leq b+ac$. Joining throughout with $b$ yields $b+ac\leq b+V \leq b+U\leq b+ac$, so $b+V=b+U$. By cancellation we have $U=V$, which is what (1) claims. \\\

Lemma 2. Assume that $L$ satisfies cancellation. Then $L$ satisfies $ab+(a+b)c=ac+(a+c)b$.

Proof. Let $U=ab+(a+b)c$ and $V=ac+(a+c)b$. Note that $V$ is obtained from $U$ by interchanging the roles of $b$ and $c$ By Lemma 1(1), with $B=(a+c)b$,
$$ aV=a(B+ac)=aB+ac=a(a+c)b+ac=ab+ac. $$ Now if we do the same calculation interchanging the roles of $b$ and $c$ we obtain $aU=ac+ab$, so $aU = aV$.

We now argue that $a+U=a+V$. $$ a+V=a+ac+(a+c)b=a+(a+c)b=(a+c)(a+b), $$ where for the last equality we used Lemma 1(2). Now, interchanging the roles of $b$ and $c$ we obtain $a+U=(a+b)(a+c)=a+V$. By cancellation, $U=V$, which is what the lemma claims. \\\

At this point we have deduced some equational consequences of cancellation, which together are sufficient to derive distributivity. We argue that, for $U=ab+(a+b)c$ and $V=ac+(a+c)b$,

(1) $a$ meet $(b+c)$ is the same as $a$ meet $(a+b)(a+c)(b+c)$;
(2) $(a+b)(a+c)(b+c)=U+V$;
(3) but since $U=V$, by Lemma 2, we get $a(b+c)=a(U+V)=aU$;
(4) finally $aU=ab+ac$.

(1) is clear, since $a(a+b)(a+c)(b+c)=a(a+c)(b+c)=a(b+c)$.

For (2), $$ U+V=ab+(a+b)c+ac+(a+c)b=(a+b)c+(a+c)b=(a+c)(b+(a+b)c)=(a+c)(a+b)(b+c). $$ This uses Lemma 1(2) for each of the last 2 equalities.

(3) just summarizes the progress so far.

For (4), $aU=a(ab+(a+b)c)=ab+a(a+b)c=ab+ac$. The first equality uses Lemma 1(1).