Show that every finite distributive lattice is a reduct of a Heyting algebra.
I was thinking the following claim could be helpful but I haven't figured out how to prove this either:
The class of bounded distributive lattices (L, ∨, ∧, 0, 1) such that for
each a, b ∈ L there is a largest c ∈ L with a∧c ≤ b is precisely the class of reducts
of Heyting algebras (to {∨, ∧, 0, 1}).
Best Answer
If $H$ is a Heyting algebra, then for every $a,b\in H$, $a\rightarrow b$ is the greatest element $c\in H$ such that $a\land c \leq b$.
To turn a distributive lattice $(L,\lor,\land,\bot,\top)$ into a Heyting algebra $(L,\lor,\land,\bot,\top, \rightarrow)$, you just need to define the $\rightarrow$ operation. And the observation above says that for all $a,b\in L$, there is at most one choice of how to define $a\rightarrow b$.
This proves your claim: $L$ is the reduct of a Heyting algebra if and only if for all $a,b\in L$, there is a greatest element $c$ such that $a\land c\leq b$. Now...
Hint: The obvious way to try to find the greatest element with a property is to take the join of all elements with that property. Of course, you can only take the join of finitely many elements, and you need to check that the property is preserved by joins...