Is there something like a Heyting Ring? I would like to know whether a Heyting algebra gives rise to ring in a similar way that a Boolean algebra gives rise to a Boolean ring. In a Boolean algebra $(B,\lor,\land,\lnot,0,1)$ I can define multiplication and addition as follows:
$$a * b = a \land b$$
$$a + b = (a \land \lnot b) \lor (b \land \lnot a)$$
And I get a Boolean ring $(B,+,*,0,1)$. From the Boolean ring we can reconstruct the Boolean algebra again via the following definitions (right?):
$$a \land b = a * b$$
$$a \lor b = (a + 1) * (b + 1) + 1$$
In a Heyting algebra $(H,\rightarrow,\land,0,1)$ we do only have a pseudo complement. Can we nevertheless define a ring, that in turn would allow us to reconstruct the algebra?
Best Regards
 A: Though not directly analogous to your example of a boolean ring, a 2010 paper in Algebra Universalis approaches the problem of classifying linearly ordered Heyting Algebras via the concept of a Godel Ring. The idea is that the semiring of ideals of a suitable ring should form a Heyting Algebra.
For example, the semiring of ideals of every Von Neumann Regular Ring form a Heyting Algebra; furthermore, if the ring in question is commutative, this Heyting Algebra is in fact a Boolean Algebra.
A: There are no terms that would define a group structure (let alone a ring) in all Heyting algebras.
The first observation is that if $f(x)$ is a term-definable involution, then $f(x)=x$: by inspection of the Rieger–Nishimura lattice, we see that all other unary term-definable $f$ are either constant in Boolean algebras, or satisfy $f(x)=f(\neg\neg x)$ ($x$ is $p$ in the picture):

Then, assume that $+,-,0$ are terms defining a group in any Heyting algebra (where the $0$ is not necessarily the bottom of the Heyting algebra, which I will denote as $\bot$, and the group is not assumed commutative, despite the additive notation). Since $f(x)=-x$ is an involution, we must have $-x=x$, i.e., the group is of exponent $2$. Thus, $g(x)=\bot+x$ and $h(x)=\top+x$ are also involutions, hence $\bot+x=x=\top+x$, which implies $\top=\bot$, a contradiction.
