An Application of Rooted Kripke Models in Intuitionistic Logic

Question

Consider the intuitionistic propositional logic (IPL).

• Assume that $A$ is a proposition such that $\vdash_{\mathsf{IPL}} (\neg\neg A \to A) \to A \lor \neg A$. Show that $\vdash_{\mathsf{IPL}} \neg A$ or $\vdash_{\mathsf{IPL}} \neg\neg A$.
• Suppose $A$ is a proposition such that $\nvdash_{\mathsf{IPL}} \neg A$ and $\nvdash \neg\neg A$. Show that there is a proposition $B$ such that $\nvdash_{\mathsf{IPL}} ((A \to B) \to B) \land ((B \to A) \to A) \to A \lor B.$

For the first part, my idea is to go with the same line of thought used to prove the disjunction property in IPL. So, I tried to prove the contrapositive form combined with the soundness and completeness theorems. Let us assume that $\nvDash \neg A$ and $\nvDash \neg\neg A$. Then, there are Kripke models $k_1 = (W_1, R_1, L_1)$ and $k_2=(W_2, R_2, L_2)$ with roots $w_1$ and $w_2$ such that $k_1, w_1 \nvDash \neg A$ and $k_2, w_2 \nvDash \neg\neg A$. Here $W_i$ refer to the worlds of the models, $R_i \subset W_i \times W_i$ are reflexive and transitive relations and $L_i$ are labeling functions with the monotone property. We construct a new Kripke model $k=(W, R, L)$ with a the root $w$ such that $W = W_1 \cup W_2$, $R=R_1 \cup R_2 \cup \{(w, v) : v = w \, \text{or} \,v \in W_1 \, \text{or} \, v \in W_2\}$ and $L|_{W_1} = L_1$, $L|_{W_2} = L_2$ and $L(w) = \emptyset$. Now, in the root of the new model, we have $k, w \nvDash \neg A$ and $k, w \nvDash \neg\neg A$. Now, what would be the next step for proving that $k, w \nvDash (\neg\neg A \to A) \to A \lor \neg A$?

For the second part, it seems that we should use the result from the first part. So, I should choose a $B$ such that the aforementioned expression becomes equivalent with $(\neg\neg A \to A) \to A \lor \neg A$. Does $B = \neg A$ work?

Answer 1

First off, I have appended an addendum, it might be better to check it first.

I denote a Kripke model for intuitionistic logic by the triple $\mathcal{M}=\langle W, \leq, V\rangle$ where $W$ is the set of possible worlds, $\leq$ is the accessibility relation on $W$ and $V$ is the variable valuation function. For the propositions employed without argument, I refer the reader to the chapter on intuitionistic logic in van Dalen's Logic and Structure (freely available at the Internet Archive).

• For a proposition $A$, we want to show that $$\underbrace{\vdash_{IPL}(\neg\neg A\rightarrow A)\rightarrow A\vee\neg A}_{antecedent\,statement}\implies\underbrace{\vdash_{IPL}\neg A\mbox{ or }\vdash_{IPL}\neg\neg A}_{consequent\,statement}$$

Hence, for any model $\mathcal{M}$ and world $w$

$$\mathcal{M},w\Vdash(\neg\neg A\to A)\rightarrow A\vee\neg A\implies\mathcal{M},w\Vdash\neg A\mbox{ or }\mathcal{M},w\Vdash\neg\neg A$$

Then, for the antecedent statement, $\forall w'\,(w\leq w')$,

$$\mathcal{M},w'\Vdash\neg\neg A\rightarrow A\implies\mathcal{M},w'\Vdash A\vee\neg A$$

Therefore, whenever either $\mathcal{M},w'\not\Vdash\neg\neg A\mbox{ or }\mathcal{M},w'\Vdash A$, we must get $\mathcal{M},w'\Vdash A\vee\neg A$

(i) In case that $\mathcal{M},w'\Vdash A$, we get $\mathcal{M},w'\Vdash A\vee\neg A$ immediately, for $A\rightarrow A\vee B$ is an axiom of IPL.

Using the theorem of IPL $A\rightarrow\neg\neg A$, we have also obtained one of the disjuncts in the consequent statement that we have set out to obtain.

(ii) In case that $\mathcal{M},w'\not\Vdash\neg\neg A$, we note that $\forall w\, (w\leq w')$ and $\neg A\leftrightarrow \neg\neg\neg A$ is a theorem of IPL. Therefore,

$$\mathcal{M},w'\not\Vdash\neg\neg A\iff\mathcal{M},w\Vdash\neg A$$

Thus, we get $\mathcal{M},w'\Vdash A\vee\neg A$ by the same axiom mentioned.

Notice that this is also the other disjunct that we have set out to obtain, and we are done.

• We suppose that $A$ is a proposition such that $\nvdash_{IPL}\neg A$ and $\nvdash_{IPL}\neg\neg A$. We want to show that there is a proposition $B$ such that

$$\nvdash_{IPL}((A\rightarrow B)\rightarrow B)\wedge((B\rightarrow A)\rightarrow A)\rightarrow A\vee B$$

Hence, in Kripke semantics formulation, there is a model $\mathcal{M}$ such that

$$\mathcal{M}, w\Vdash((A\rightarrow B)\rightarrow B)\wedge((B\rightarrow A)\rightarrow A),\mbox{ but }\mathcal{M},w\not\Vdash A\vee B$$

Since $\nvdash_{IPL}\neg\neg A$ and $\vdash_{IPL}A\rightarrow\neg\neg A$, we infer that $A$ is not a theorem of IPL. So, we may take $\neg A$ as the sought $B$ in order to be able to write $\not\Vdash A\vee\neg A$. Hence, by the same model that we have obtain $\mathcal{M}, w\not\Vdash A\vee\neg A$, if we can show that

$$\mathcal{M}, w\Vdash((A\rightarrow\neg A)\rightarrow\neg A)\wedge((\neg A\rightarrow A)\rightarrow A)$$

we are done.

For this, we take $W=\{w, w'\}$ such that $w\leq w'$ and let $V(A, w)=0$ and $V(A, w')=1$. Since $A$ is true at $w'$, $\neg A$ is false at $w$ in conformance to $\not\vdash_{IPL}\neg A$. We can check the formula against the truth condition of implication and see that $\mathcal{M}$ is indeed a model for the antecedent statement.

Addendum

It seems to be helpful to dwell on the proof methodology. Suppose we want to semantically (i.e., by models and truth-conditions) show

$$(\phi\rightarrow\psi)\rightarrow\theta$$

We can follow two paths (arranging the worlds/nodes of evaluation appropriately):

(1) Assume $\phi$. Then, for any model $\mathcal{M}$ of $\phi$, write

$$\mathcal{M}\Vdash\phi\\\vdots\\\mathcal{M}\Vdash\psi\\\mathcal{M}\Vdash\phi\rightarrow\psi\\\vdots\\\frac{\mathcal{M}\Vdash\theta}{\mathcal{M}\Vdash(\phi\rightarrow\psi)\rightarrow\theta}$$

(2) Assume $\phi\rightarrow\psi$. Then, for any model $\mathcal{M}$ of $\phi\rightarrow\psi$, write

$$\mathcal{M}\not\Vdash\phi\qquad\mathcal{M}\Vdash\psi\\\vdots\quad\qquad\vdots\\\frac{\mathcal{M}\Vdash\theta\qquad\mathcal{M}\Vdash\theta}{\mathcal{M}\Vdash(\phi\rightarrow\psi)\rightarrow\theta}$$

The path (1) is clear by the definition of logical consequence: Having assumed $\phi$, we derive $\psi$ and then $\theta$.

But why (2)? Because of the truth-conditions:

$$\mathcal{M}, w\Vdash\phi\rightarrow\psi\iff\forall w'(w\leq w')\mbox{ either }\mathcal{M}, w'\not\Vdash\phi\mbox{ or }\mathcal{M}, w'\Vdash\psi$$

Hence, assuming the left-hand side is equivalent to assuming the right-hand side.

Answer 2

Since $\vdash ((\neg \neg A \to A) \to A) \to \neg\neg A$ and $\vdash ((\neg\neg A \to A) \to \neg A) \to \neg A$ are theorems of intuitionistic propositional logic (easy exercises), it suffices to show that we have either $\vdash (\neg\neg A \to A) \to A$ or $\vdash (\neg\neg A \to A) \to \neg A$. Now we apply the same idea as for the proof of the disjunction property:

Assume we have rooted Kripke models $(V, v)$ and $(W, w)$ such that $V, v \nVdash (\neg\neg A \to A) \to A$ and $W, w \nVdash (\neg\neg A \to A) \to \neg A$. This means:

• There is a $v' \geq_V v$ with $V, v' \Vdash \neg\neg A \to A$ and $V, v' \nVdash A$; hence also $V, v' \nVdash \neg\neg A$.
• There is a $w' \geq_W w$ with $W, w' \Vdash \neg\neg A \to A$ and $W, w' \nVdash \neg A$.

We restrict our attention to the respective futures ("cones") of $v'$ and $w'$, $V' = \{v'\} \uparrow$ and $W' = \{w'\} \uparrow$, and build a new rooted model $(U, u)$ from $(V', v')$ and $(W', w')$ in the usual way, such that $u$ reaches all the worlds in $V'$ and $W'$. Then:

• $U, u \nVdash A$ and $U, u \nVdash \neg A$ by monotonicity, hence $U, u \nVdash A \lor \neg A$.
• Furthermore, we have $U, u \Vdash \neg\neg A \to A$. This is the key part of the argument, and we have to use the structure of the formula $\neg\neg A \to A$ here. Given $u' \geq_U u$, we must show that $U, u' \Vdash \neg\neg A$ implies $U, u' \Vdash A$. We have three cases:
  • If $u' \geq_U v'$ or $u' \geq_U w'$ then we already have $U, u' \Vdash \neg\neg A \to A$ by monotonicity.
  • If $u' = u$, then we have $U, u \nVdash \neg\neg A$ by monotonicity since $V, v' \nVdash \neg\neg A$, so that the implication holds vacuously.

Therefore we conclude that $U, u \nVdash (\neg\neg A \to A) \to A \lor \neg A$. By contrapositive and completeness of Kripke semantics, this shows that $\vdash (\neg\neg A \to A) \to A \lor \neg A$ implies either $\vdash (\neg\neg A \to A) \to A$ or $\vdash (\neg\neg A \to A) \to \neg A$, and thus either $\vdash \neg\neg A$ or $\vdash \neg A$ by the theorems above and modus ponens.

I expect that looking at normal forms/cut-free proofs (à la Gentzen) would yield a somewhat simpler proof.

Answer 3

Here are the outline of proofs.

• For the first part, construct the following Kripke model with root $r$ connected to the green models which is shown schematically in the below figure. Use the definition of negation in IPL semantics to see how this construction works for proving the contrapositive form of the first part of the question.

• For the second part, choose $B=\neg A$. Also, notice that $\vdash_{\mathsf{IPL}} (\neg A \to A) \leftrightarrow \neg\neg A$ and $\vdash_{\mathsf{IPL}} (A \to \neg A) \leftrightarrow \neg A$. Using these facts, it is easy to show that $$\vdash_{\mathsf{IPL}} ((A \to \neg A) \to \neg A) \land ((\neg A \to A) \to A) \leftrightarrow \neg\neg A \to A.$$ Now, use the first part to get the final result.