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Let $\mathcal{L}$ be a propositional language and let $\text{Prop}(\mathcal{L})$ be the set of all the propositions of the language $\mathcal{L}$.

Let $(H,\wedge,\vee,\rightarrow,1,0)$ be an Heyting algebra. An evaluation of the propositions of $\mathcal{L}$ in $(H,\wedge,\vee,\rightarrow,1,0)$ is a map $V:\text{Prop}(\mathcal{L}) \to H$ such that $V(\top)=1$, $V(\bot)=0$, $V(P \wedge Q)=V(P) \wedge V(Q)$, $V(P \vee Q)=V(P) \vee V(Q)$ and $V(P \rightarrow Q)=V(P) \rightarrow V(Q)$.

Let $\Gamma$ be a set of propositions of the language $\mathcal{L}$. Suppose that there exist an Heyting algebra $(H,\wedge,\vee,\rightarrow,1,0)$ and an evaluation $V:\text{Prop}(\mathcal{L}) \to H$ such that $V(P)=1$ for every $P \in \Gamma$.

Can I construct a Boolean algebra $(B,\wedge,\vee,\rightarrow,\neg,1,0)$ and an evaluation $V':\text{Prop}(\mathcal{L}) \to B$ such that $V'(P)=1$ for every $P \in \Gamma$?

effezeta
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1 Answers1

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Consider a homomorphism h from the Heyting algebra H into the two elements Boolean algebra 2. Then the induced valuation $h\circ V$ witnesses the satisfiability of $\Gamma$ over 2.

For a proof which does not make use the axiom of choice you can also proceed as follows. Let $(H,\lor,\land,\to,1,0)$ be the Heyting algebra witnessing the satisfiability of $\Gamma$ under the valuation $V$. Remember that its subset of regular elements $H_\neg=\{ x\in H : x=\neg \neg x \}$ forms a Boolean algebra $(H_{\neg},\lor',\land,\to,1,0)$, where $\lor':=\neg \neg (x\lor y)$ -- see also this question.

Now consider the map $f:H\to H_\neg$ such that $f(x)=\neg \neg x$. The same proof that establishes that $H_\neg$ is a Boolean algebra also gives that $f(0)=0, f(1)=1, f(x\land y)=f(x)\land f(y)$ and $f(x\to y)=f(x)\to f(y)$ (see the link above). For disjunction, as you pointed out in the comments, it is not true in general that $\neg\neg (x\lor y)=\neg\neg x\lor \neg\neg y$. However, since Heyting algebras satisfy the De morgan rule $\neg(a\lor b)=\neg a\land \neg b$ and also simplify triple negation to single negation, we have:

$$ f(x\lor y)=\neg\neg (x\lor y) = \neg(\neg x\land \neg y) = \neg(\neg \neg \neg x \land \neg \neg \neg y) = \neg \neg (\neg \neg x \lor \neg \neg y)=f(x)\lor' f(y). $$

Which establishes that $f:H\to H_{\neg}$ is a homomorphism. Then the valuation $f\circ V$ witnesses the satisfiability of $\Gamma$ over $H_\neg$.

D.Q.
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  • How can I construct the homomorphism $h:H \to {1,0}$? – effezeta May 28 '23 at 18:43
  • It should work if you quotient H by a maximal filter. – D.Q. May 28 '23 at 18:59
  • I think I cannot construct a maximal filter on H, I can prove that it exists using Zorn's lemma but this proof is not constructive. – effezeta May 28 '23 at 19:08
  • Yes, indeed. Is that a problem? Are you working without choice? – D.Q. May 28 '23 at 19:15
  • Yes, I am working in constructive mathematics. – effezeta May 28 '23 at 19:19
  • Ok what if you send H to its Boolean algebra of regular elements $x=\neg\neg x $. It seems this should work. – D.Q. May 28 '23 at 19:33
  • Are you sure that $f(x)=(x \rightarrow 0) \rightarrow 0$ is an homomorphism? I think that the condition $f(x \vee y) = f(x) \vee f(y)$ doesn't hold. – effezeta May 28 '23 at 21:11
  • I think it should, the point is that when you verify this condition you should keep in mind that on the right hand side you have a novel join operator on the Boolean algebra of regular elements and not the original one, given by $x\lor’ y := \neg \neg (x\lor y)$. – D.Q. May 28 '23 at 21:37
  • I'm a bit confused. $(H,\wedge,\vee,\rightarrow,1,0)$ is an Heyting algebra, $\neg x$ means $x \rightarrow 0$, I define $H'={ x \in H \mid x = \neg \neg x }$, I have that $(H',\wedge',\vee',\rightarrow',\neg',1,)$ is a Boolean algebra in which every operation $'$ is defined by $x'y = \neg \neg (xy)$. I define the map $f:H \to H'$ setting $f(x) = \neg \neg x$. I have to prove that $f(xy) = \neg \neg (xy)$ is equal to $f(x) ' f(y) = (\neg \neg x)'(\neg \neg y) = \neg \neg ((\neg \neg x)(\neg \neg y))$. How can I prove this equality? – effezeta May 28 '23 at 23:53
  • I made an edit to the answer, hope it is clear. – D.Q. May 29 '23 at 09:58
  • Thank's! I checked that $f(0)=0$,$f(1)=1$,$f(x \wedge y)=f(x) \wedge f(y)$ and that $f(x \vee y)=f(x) \vee' f(y)$. But how can I check that $f(x \rightarrow y)=f(x) \rightarrow f(y)$? – effezeta May 29 '23 at 19:05
  • @effezeta: Note that $\neg\neg(x \rightarrow y) = \neg\neg x \rightarrow \neg\neg y$ holds for any $x,y \in H$ in every Heyting algebra $H$ since it is an intuitionistic tautology. – Z. A. K. Jun 01 '23 at 00:16
  • @Z.A.K.: I know that it is an intuitionistic tautology, but how can I prove it using the axioms of Heyting algebras? – effezeta Jun 02 '23 at 10:05
  • @effezeta: Use the fact that $\neg\neg(x \rightarrow y) \leq \neg\neg x \rightarrow \neg\neg y$ iff $\neg\neg(x \rightarrow y) \wedge \neg\neg x \leq \neg\neg y$. But $\neg\neg(x \rightarrow y) \wedge \neg\neg x = \neg\neg ((x \rightarrow y)\wedge x) = \neg\neg (y \wedge x)$. So you need to prove $\neg\neg (y \wedge x) \leq \neg\neg y$, which follows immediately from $y\wedge x \leq y$ and the fact $\neg$ reverses order. Inequality in other direction is easy. (NB you could use the completeness/soundness proofs for Heyting semantics to transform the IPC deduction into a direct proof as well). – Z. A. K. Jun 02 '23 at 11:42