Proof Cancellation property in lattice implies distributivity A lattice is distributive iff it has the cancellation property (for all $a,b,c$ in the lattice $L$, if $a\lor b=a\lor c$ and $a \land b=a \land c$ then $a=b$). Showing that distribution implies cancellation is easy and a proof can be found here:
https://core.ac.uk/download/pdf/55337078.pdf
at page 15. The proof for the converse seem wrong since it considers the case in which $a\lor b=a\lor c$ but doesn't seem to show the distributive property holds for when that equality doesn't hold. Is the proof of the book correct and am I missing something? If so, what is a proper proof to the theorem?
 A: The property is true, but the given proof seems indeed to be incorrect. One way to prove that cancellation implies distributivity is to use the following characterisation of distributive lattices:

A lattice is distributive if and only if none of its sublattices is
  isomorphic to the pentagon lattice $N_5$ or to the diamond lattice $M_3$.

Now, it is easy to see that neither $N_5$ nor $M_3$ are cancellative lattices: with the notation of Wikipedia, one has $x \vee y = z \vee y = 1$ and $x \wedge y = z \wedge y = 0$ in both cases.
A: 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).
