# Fibers of the morphism from the free Heyting algebra to the free Boolean algebra

For $$k\in\mathbb{N}$$, let $$H_k$$ be the free Heyting algebra on $$k$$ variables $$p_1,\ldots,p_k$$ and $$B_k$$ be the free Boolean algebra on the same $$k$$ variables. Thus, $$B_k$$ has $$2^{2^k}$$ elements (corresponding to all possible truth-value tables with $$k$$ entries), while $$H_k$$ is infinite as soon as $$k\geq 1$$.

Since Boolean algebras are, in particular, Heyting algebras, there is a natural morphism $$\psi \colon H_k \to B_k$$ (taking each $$p_i$$ to the corresponding Boolean variable).

Example: For $$k=1$$ (and writing $$p:=p_1$$), the algebra $$H_1$$ is the (infinite) Rieger-Nishimura lattice (see here), while $$B_1$$ has four elements namely $$\bot,p,\neg p,\top$$. The morphism $$\psi$$ takes the single element $$\bot$$ to $$\bot$$, the two $$p$$ and $$\neg\neg p$$ to $$p$$, the single $$\neg p$$ to $$\neg p$$, and every other element of $$H_1$$ to $$\top$$. So three of its fibers are finite while the last is infinite.

Question: Which elements $$u \in B_k$$ have finite fiber $$\psi^{-1}(u)$$, and how can we describe their cardinalities or, better, their elements?

## 1 Answer

$$\let\eq\leftrightarrow$$Notice that $$\psi(A)=u$$ iff $$\vdash_\mathrm{CPC}A\eq u$$ iff $$\vdash_\mathrm{IPC}\neg\neg(A\eq u)$$. (I will write just $$\vdash$$ for $$\vdash_\mathrm{IPC}$$.) Thus:

• $$\bot$$ has a one-element fiber consisting of $$\bot$$: if $$\vdash\neg\neg(A\eq\bot)$$, then $$\vdash\neg A$$.

• For each $$i$$, $$u_i:=p_i\land\bigwedge_{j\ne i}\neg p_j$$ has a two-element fiber consisting of $$p_i\land\bigwedge_{j\ne i}\neg p_j$$ and $$\neg\neg p_i\land\bigwedge_{j\ne i}\neg p_j$$: if $$\vdash\neg\neg(A\eq u_i)$$, then $$\vdash A\to\neg p_j$$ for all $$j\ne i$$, thus $$A$$ is equivalent to $$\bigwedge_{j\ne i}\neg p_j\land A(\bot,\dots,\bot,p_i,\bot,\dots)$$. The formula $$A':=A(\bot,\dots,\bot,p_i,\bot,\dots)$$ of one variable also has to imply $$\neg\neg p_i$$, hence by your analysis of the Rieger–Nishimura lattice, it must be equivalent to $$\neg\neg p_i$$ or to $$p_i$$. (It can’t be $$\bot$$.)

• Likewise, $$u=\bigwedge_i\neg p_i$$ has a one-element fibre consisting of $$\bigwedge_i\neg p_i$$.

In the remaining cases, the fiber is infinite:

• If $$u=\bigwedge_{i\in I}p_i\land\bigwedge_{i\notin I}\neg p_i$$ where $$|I|\ge2$$, fix $$j\ne k$$ in $$I$$. Then the fiber of $$u$$ includes all formulas of the form $$\bigwedge_{\substack{i\in I\\i\ne j,k}}p_i\land\bigwedge_{i\notin I}\neg p_i\land A(p_j,p_k),$$ where $$A(p_j,p_k)$$ is a formula of two variables implying $$\neg\neg(p_j\land p_k)$$ and implied by $$p_j\land p_k$$. That there are infinitely many such formulas follows from the fact that in the universal intuitionistic frame of rank $$2$$, there are infinitely many points such that the only leaf they see is the one satisfying both variables.

• If $$u$$ has $$\ge2$$ satisfying assignments $$e,e'$$, let $$I_{a,b}$$, $$a,b=0,1$$, be the set of indices of variables assigned to $$a$$ by $$e$$, and to $$b$$ by $$e'$$. Without loss of generality, $$I_{0,1}\ne\varnothing$$, hence we may fix $$j\in I_{0,1}$$. Then $$\psi^{-1}(u)$$ includes all formulas of the form $$\neg\neg u\land A(p_j)$$ where $$A$$ is a formula in one variable such that $$\vdash\neg\neg A(p_j)$$. All these formulas are inequivalent: if $$\vdash\neg\neg u\land A(p_j)\to A'(p_j),$$ we may substitute $$\bot$$ for all $$p_i$$ such that $$i\in I_{0,0}$$, $$\top$$ for $$i\in I_{1,1}$$, $$p_j$$ for $$i\in I_{0,1}$$, and $$\neg p_j$$ for $$i\in I_{1,0}$$. This substitution turns $$\neg\neg u$$ into a theorem, and leaves $$A$$ and $$A'$$ unaffected, hence $$\vdash A(p_j)\to A'(p_j).$$

• Ah, it's the second time you make me realize how important the double negation translation can be! I hope next time I'll remember it. Mar 15 '19 at 15:44