A not not Brief Introduction to Kripke Semantics for Propositional Intuitionistic Logic

2023-08-16T00:00:00+00:00

In this post we motivate the study of Kripke semantics for propositional intuitionistic logic by showing fundamental properties at the proof theory level using \(n+1\)-valued logics. Some non-theorems of propositional intuitionistic logic (like the excludded middle principle) are discussed using this approach.

Introduction

With the discovery of non-Euclidean geometries in the middle of the nineteenth century (Trudeau 2001), mathematicians questioned the logical foundations of mathematics. In fact, the latter culminated in the fragmentation of the mathematical community resulting in three main schools of thought:

Platonism:

The practicioners consider the subject at study to be independent of the epistemic knowledge of mathematical objects and their relations. Instead, these were regarded as abstract entities which can only be accessed through meaningful propositions.

Formalism:

This trend considers the linguistic approach an essential aspect for the discussion of the mathematical theories using axiomatic systems and precise rules to understand and communicate mathematical truth. Formalism is quite often critized by its opponents to reduce mathematical investigations to a simple manipulation of symbols. Regardless of this unintended byproduct, formalism’s main point is to provide a suitable framework to understand the consistency of mathematics, a program heavely supported by the mathematician David Hilbert at the beginning of the 20th century. It was not until logician Kurt Gödel with his Incompleteness Theorems that showed the limitations of this program.

Intuitionism:

This school was initiated by the ideas of L.E.J. Brouwer, a Dutch mathematician who worked in several mathematical subjects ranging from topology to foundations of mathematics. He was a critic of the Formalism approach. The crucial missing part was the lack of involvement of the knowledge-subject. The thinking activity is of huge importance for him. Such activity grounds on our Ur-intuition of the flow of time (Richardson 2011). Thus, mathematical work accounts to `mental constructions’ in the intuitionistic sense.

In this light, an intuitionistic assertion constitutes complete knowlege on the subject. To say that something is true in the intuitionisc sense accounts to having a proof of it. Notice that the intention here is radically different in the negation. Knowing that something is false accounts to having a proof of the impossibility of the fact, contrary to the position of not having a proof of it.

A typical example in intuitionistic mathematics is to discuss the excludded middle principle \(\neg A \lor A\). Such statement cannot be accepted intuitioniscally since it contributes more to an undecided moment rather than knowledge about the statement. In the rest of this paper we will discuss axiomatic systems and models which allows us to understand intuitionistic ideas.

A formal system \(J\) for propositional intuitionistic logic

The following axioms were extract from (Hodel 2013):

Language: The symbols of the language are:
- \(p_1, p_2, p_3, \dots\): An infinite list of propositional variables
- \(\neg, \rightarrow, \land, \lor\): negation, implication, conjunction, and disjunction symbol
- \((, )\): left and right parenthesis for punctuation
Axioms:
- J1: \(A \rightarrow (B \rightarrow A)\)
- J2: \((A \rightarrow (B \rightarrow C)) \rightarrow ((A \rightarrow B) \rightarrow (A \rightarrow C))\)
- J3: \(A \land B \rightarrow A\)
- J4: \(A \land B \rightarrow B\)
- J5: \(A \rightarrow (B \rightarrow (A \land B))\)
- J6: \(A \rightarrow (A \lor B)\)
- J7: \(B \rightarrow (A \lor B)\)
- J8: \((A \rightarrow C) \rightarrow ( ( B \rightarrow C) \rightarrow (( A \lor B) \rightarrow C))\)
- J9: \((A \rightarrow B) \rightarrow (( A \rightarrow \neg B) \rightarrow \neg A)\)
- J10: \(\neg A \rightarrow (A \rightarrow B)\)
Rules of Inference:
- Modus Ponens

As in classical logic \(A \leftrightarrow B\) is abbreviated as \((A \rightarrow B) \land (B \rightarrow A)\). It is important to mention that the other connective symbols do not share similar abbreviations as in the classical counterpart. Since \(J\) as a formal system contains \(J1\) and Modus Ponens as unique rule of inference, by the comment on Theorem 3 in Section 3.5 of (Hodel 2013) we have that \(J\) satisfies the Deduction theorem. A proof of the above is shown in the next theorem:

Theorem 1: {#theorem1} (Deduction theoreom for \(J\)) If \(\Gamma \cup \{ A \} \vdash_J B\), then \(\Gamma \vdash_J A \rightarrow B\)

Proof: We proceed by induction on the number of steps in the proof of \(B\) using \(\Gamma \cup \{A\}\). Assume \(\Gamma \cup \{A \} \vdash_J B\), to prove \(\Gamma \vdash_J A \rightarrow B\). There are three cases:

\(B\) is an axiom of \(J\) or \(B \in \Gamma\): From this we have that \(\vdash_J B\) Since \(J\) has \(J1\), we can conclude that \(\Gamma \vdash_J B \rightarrow (A \rightarrow B)\). By Modus Ponens we obtain \(\Gamma \vdash_J A \rightarrow B\)
\(B\) is in \(A\): In this case, the following is proof of \(\Gamma \vdash_J A \rightarrow A\).
1. \(A \rightarrow (A \rightarrow A)\), Axiom \(J1\)
2. \(A \rightarrow ((A \rightarrow A) \rightarrow A)\), Axiom \(J1\) 3. \((A \rightarrow ((A \rightarrow A) \rightarrow A)) \rightarrow ( (A \rightarrow (A \rightarrow A)) \rightarrow (A \rightarrow A))\), Axiom \(J2\) 4. \((A \rightarrow (A \rightarrow A)) \rightarrow (A \rightarrow A)\), Modus Ponens(2, 3)
3. \(A \rightarrow A\), Modus Ponens(1, 4)
\(B\) is obtained from \(C\) and \(C \rightarrow B\) by an application of Modus Ponens. By induction, we have from \(\Gamma \cup \{ A \} \vdash_J B\) and \(\Gamma \cup \{ A \} \vdash_J C \rightarrow B\) the following \(\Gamma \vdash_J A \rightarrow B\) and \(\Gamma \vdash_J A \rightarrow (C \rightarrow B)\). By \(J2\) we have that \(\Gamma \vdash_J (A \rightarrow (C \rightarrow B)) \rightarrow ((A \rightarrow B) \rightarrow (A \rightarrow C))\). A double application of Modus Ponens, we obtain \(\Gamma \vdash_J A \rightarrow C\).

Theorem 2: \(\neg A \lor A\) and \(\neg \neg A \rightarrow A\) are not theorems of \(J\).

Proof: Let us introduce an \(n+1\)-valued logic as follows: A truth assignment is a function \(\phi : PROP \rightarrow \{0, 1, \dots, n\}\); such an assignment extends to all formulas \(FOR(\neg, \lor, \land, \rightarrow)\) of \(J\) according to these rules:

\[\phi(A \lor B) = \min \{ \phi(A), \phi(B) \}\]
\[\phi(\neg A) = \begin{cases} 0 & \phi(A) = n \ n & \phi(A) < n \end{cases}\]
\[\phi(A \land B) = \max \{ \phi(A), \phi(B) \}\]
\[\phi(A \rightarrow B) = \begin{cases} 0 & \phi(A) \geq \phi(B) \ \phi(B) & \phi(A) < \phi(B) \end{cases}\]

We will prove that for every theorem \(A\) in \(J\) we have that \(\phi(A) = 0\) by induction on the length of the proof:

Base case: Since the base case accounts to axioms in \(J\), we need to prove that every axiom in \(J\) evaluates to 0 under \(\phi\):
- Case \(J1\): Since \(\phi(A) \geq \phi(A)\) and \(\phi(A) \geq 0\) we have that \(\phi(A) \geq \phi(B \rightarrow A)\), so \(\phi(A \rightarrow (B \rightarrow A)) = 0\).
- Case \(J2\): To prove \(\phi(( A \rightarrow (B \rightarrow C)) \rightarrow ((A \rightarrow B) \rightarrow (A \rightarrow C))) = 0\).
  - Case 1. \(\phi(A) \geq \phi(B \rightarrow C)\). This implies \(\phi(A \rightarrow (B \rightarrow C)) = 0\). Thus, it is enough to prove that \(\phi((A \rightarrow B) \rightarrow (A \rightarrow C)) = 0\), i.e. \(\phi(A \rightarrow B) \geq \phi(A \rightarrow C)\).
    - Subcase 1. \(\phi(A) \geq \phi(B)\) and \(\phi(A) \geq \phi(C)\): This holds since \(0 \geq 0\).
    - Subcase 2. \(\phi(A) \geq \phi(B)\) and \(\phi(A) < \phi(C)\): It is enough to prove \(0 \geq \phi(C)\), i.e. \(\phi(C) = 0\). Suppose by contradiction that \(\phi(C) \neq 0\). We have that \(\phi(C) > \phi(A) \geq \phi(B)\), so \(\phi(C) > \phi(B)\). This implies that \(\phi(B \rightarrow C) = \phi(C)\). Since \(\phi(A) \geq \phi(B \rightarrow C)\) we have that \(\phi(A) \geq \phi(C)\). Therefore, \(\phi(C) > \phi(A) \geq \phi(C)\), a contradiction.
    - Subcase 3. \(\phi(A) < \phi(B)\) and \(\phi(A) \geq \phi(C)\): This holds since \(\phi(B) \geq 0\).
    - Subcase 4. \(\phi(A) < \phi(B)\) and \(\phi(A) < \phi(C)\): To prove that \(\phi(B) \geq \phi(C)\). Suppose by contradiction that \(\phi(B) < \phi(C)\), thus \(\phi(B \rightarrow C) = \phi(C)\). Since \(\phi(A) \geq \phi(B \rightarrow C)\) we have that \(\phi(A) \geq \phi(C)\). Since \(\phi(C) > \phi(A) \geq \phi(C)\) we reach a contradiction.
  - Case 2. \(\phi(A) < \phi(B \rightarrow C)\). The latter implies that \(\phi(A \rightarrow (B \rightarrow C)) = \phi(B \rightarrow C)\). Hence, it is enough to prove that \(\phi(B \rightarrow C) \geq \phi((A \rightarrow B) \rightarrow (A \rightarrow C))\).
    - Subcase 1. \(\phi(B) \geq \phi(C)\): From the latter we have that \(\phi(B \rightarrow C) = 0\). Hence, \(\phi(A) < 0\), but \(\phi(A) \geq 0\), a contradiction.
    - Subcase 2. \(\phi(B) < \phi(C)\): It is enough to prove that \(\phi(C) \geq \phi((A \rightarrow B) \rightarrow (A \rightarrow C))\). Since \(\phi(B) < \phi(C)\) we have that \(\phi(B \rightarrow C) = \phi(C)\). Since \(\phi(A) < \phi(B \rightarrow C)\) we can conclude that hence \(\phi(C) > \phi(A)\). Let us prove the following <<observation>{=html}>observation: For any two formulas \(A, B \in FOR(\neg, \land, \lor, \rightarrow)\), we have that \(\phi(A) \geq \phi(B \rightarrow A)\). This happens because \(\phi(A) \geq 0\) and \(\phi(A) \geq 0\) and by definition of we have that \(\phi(B \rightarrow A)\) is either \(0\) or \(\phi(A)\). From this observation we have that \(\phi(A \rightarrow C) \leq \phi((A \rightarrow B) \rightarrow (A \rightarrow C))\). Since \(\phi(A \rightarrow C) = \phi(C)\) we can conclude that \(\phi(C) \geq \phi((A \rightarrow B) \rightarrow (A \rightarrow C))\).
- Case \(J3\): This follows from \(\max \{ \phi(A), \phi(B) \} \geq \phi(A)\).
- Case \(J4\): This follows from \(\max \{ \phi(A), \phi(B) \} \geq \phi(B)\).
- Case \(J5\): We notice the following cases:
  - Case \(\phi(A) < \phi(B) = \max \{ \phi(A), \phi(B) \}\): This reduces \(\phi(A \rightarrow (B \rightarrow A \land B)) = 0\) to check that \(\phi(A) \geq 0\), which is true.
  - Case \(\phi(B) < \phi(A) = \max \{ \phi(A), \phi(B) \}\): This reduces \(\phi(A \rightarrow (B \rightarrow A \land B)) = 0\) to check that \(\phi(A) \geq \max\{ \phi(A), \phi(B) \}\), which is true since \(\phi(A) = \max \{ \phi(A), \phi(B) \}\).
  - Case \(\phi(A) = \phi(B) = \max \{ \phi(A), \phi(B) \}\): This reduces \(\phi(A \rightarrow (B \rightarrow A \land B)) = 0\) to check that \(\phi(A) \geq 0\) which is true.
- Case \(J6\): This follows from \(\min \{ \phi(A), \phi(B) \} \leq \phi(A)\).
- Case \(J7\): This follows from \(\min \{ \phi(A), \phi(B) \} \leq \phi(A)\).
- Case \(J8\): To prove that \(\phi((A \rightarrow C) \rightarrow ((B \rightarrow C) \rightarrow ((A \lor B) \rightarrow C))) = 0\).
  - Case 1. \(\phi(A) \geq \phi(C)\): It is enough to prove that \(\phi((B \rightarrow C) \rightarrow ( ( A \lor B) \rightarrow C)) = 0\), i.e. \(\phi(B \rightarrow C) \geq \phi((A \lor B) \rightarrow C)\).
    - Subcase 1. \(\phi(B) \geq \phi(C)\). It is enough to prove that \(\phi(A \lor B) \geq \phi(C)\). Since \(\phi(A) \geq \phi(C)\) and \(\phi(B) \geq \phi(C)\) then \(\phi(A \lor B) = \min \{ \phi(A), \phi(B) \} \geq \phi(C)\).
    - Subcase 2. \(\phi(B) < \phi(C)\). Since \(\phi(A) \geq \phi(C) > \phi(B)\) we conclude that \(\phi(A \lor B) = \phi(B)\). Because \(\phi(B) < \phi(C)\), we have that \(\phi(B \rightarrow C) = \phi(C)\). Using the previous [[observation][observation]], we have that \(\phi(C) \geq \phi((A \lor B) \rightarrow C)\), thus \(\phi(B \rightarrow C) \geq \phi((A \lor B) \rightarrow C)\).
  - Case 2. \(\phi(A) < \phi(B \rightarrow C)\). We will prove the following observation <<observation2>{=html}>observation: \(\phi((A \land B) \rightarrow C) = \phi(A \rightarrow (B \rightarrow C))\).
    - Subcase 1. \(\phi(A) \geq \phi(B \rightarrow C)\): This means that \(\phi(A \rightarrow (B \rightarrow C)) = 0\), hence we need to prove that \(\phi((A \land B) \rightarrow C) = 0\). We can see that \(\phi(A \rightarrow (B \rightarrow C)) = 0\) implies that \(\phi(A) \geq \phi(B \rightarrow C)\), which means that if \(\phi(B) < \phi(C)\) we have that \(\phi(A) \geq \phi(B \rightarrow C) = \phi(C)\). Suppose by contradiction that \(\phi((A \land B) \rightarrow C) \neq 0\), so \(\phi(A \land B) < \phi(C) \neq 0\). Thus, \(\phi(C) > \phi(A)\) and \(\phi(C) > \phi(B)\). The latter entails \(\phi(A) \geq \phi(C) > \phi(A)\), a contradiction.
    - Subase 2. \(\phi(A) < \phi(B \rightarrow C)\): This implies that \(\phi(A \rightarrow (B \rightarrow C)) = \phi(B \rightarrow C)\). So we need to prove that \(\phi((A \land B) \rightarrow C) = \phi(B \rightarrow C)\). We notice that \(\phi(B) < \phi(C)\), otherwise \(\phi(B \rightarrow C) = 0\) so \(\phi(A) < 0\), a contradiction. From this, we conclude that \(\phi(B \rightarrow C) = \phi(C)\), which reduces proving \(\phi(A \rightarrow (B \rightarrow C)) = \phi(B \rightarrow C)\) to prove \(\phi(A \rightarrow (B \rightarrow C)) = \phi(C)\) instead. Since \(\phi(A) < \phi(B \rightarrow C) = \phi(C)\), we have that \(\phi(C) > \max \{\phi(A), \phi(B) \}\). Therefore, \(\phi((A \land B) \rightarrow C) = \phi(C)\) as desired.

Returning to our original problem, we have that \(\phi(A) < \phi(B \rightarrow C)\), hence it is enough to prove \(\phi(A) \geq \phi((B \rightarrow C) \rightarrow ((A \lor B) \rightarrow C))\). From our previous [[observation2][observation]], we notive that \(\phi(A) \geq \phi(((B \rightarrow C) \land (A \lor B)) \rightarrow C)\), so by our first [[observation][observation]] the latter is true.

Case \(J9\): To prove \(\phi((A \rightarrow B) \rightarrow ((A \rightarrow \neg B) \rightarrow \neg A)) = 0\), i.e \(\phi(A \rightarrow B) \geq \phi((A \rightarrow \neg B) \rightarrow \neg A)\).
- Case 1. \(\phi(A) \geq \phi(B)\): To prove \(\phi((A \rightarrow \neg B) \rightarrow \neg A) = 0\), i.e. \(\phi(A \rightarrow \neg B) \geq \phi(\neg A)\).
  - Subcase 1. \(\phi(A) \geq \phi(\neg B)\): Since \(\phi(A) \geq \phi(B)\) and \(\phi(A) \geq \phi(\neg B)\) we conclude that \(\phi(A) = n\), thus \(\phi(\neg A) = 0\), so \(\phi(A \rightarrow \neg B) \geq \phi(\neg A)\) reduces to \(0 \geq 0\) which is true.
  - Subcase 2. \(\phi(A) < \phi(\neg B)\): This reduces \(\phi(A \rightarrow \neg B) \geq \phi(\neg A)\) to prove \(\phi(\neg B) \geq \phi(\neg A)\). Since \(\phi(A) \geq \phi(B)\) we have that \(\phi(\neg B) > \phi(B)\). This implies that \(\phi(\neg B) = n\), otherwise \(\phi(\neg B) = 0\) and \(\phi(B) = n\), but it cannot be the case that \(\phi(\neg B) > n\). The latter reduces \(\phi(\neg B) \geq \phi(\neg A)\) to prove \(n \geq \phi(\neg A)\) which is true.
- Case 2. \(\phi(A) < \phi(B)\): This means that \(\phi(A \rightarrow B) = \phi(B)\). To prove that \(\phi(B) \geq \phi((A \rightarrow \neg B) \rightarrow \neg A)\).
  - Subcase 1. \(\phi(A \rightarrow \neg B) \geq \phi(\neg A)\). This means that \(\phi((A \rightarrow \neg B) \rightarrow \neg A) = 0\), which reduces \(\phi(B) \geq \phi((A \rightarrow \neg B) \rightarrow \neg A)\) to \(\phi(B) \geq 0\) which is true.
  - Subcase 2. \(\phi(A \rightarrow \neg B) < \phi(\neg A)\). The latter means that \(\phi((A \rightarrow \neg B) \rightarrow \neg A) = \phi(\neg A)\). To prove \(\phi(B) \geq \phi(\neg A)\).
Suppose by contradiction that \(\phi(B) < \phi(\neg A)\). Since \(\phi(A) \geq \phi(B)\) we have that \(\phi(\neg A) > \phi(A)\). So \(\phi(\neg A) = n\), otherwise \(\phi(A) = n\) and \(\phi(\neg A) > n\), which is not possible. The latter also entails that \(\phi(A) < n\). Additionally, \(\phi(B) < n\), otherwise \(n < \phi(\neg A)\), which is not possible. From the latter \(\phi(\neg B) = n\). Since \(\phi(A) < n = \phi(\neg B)\), we have that \(\phi(A \rightarrow \neg B) = \phi(\neg B) = n\). But this implies that \(n > n\), a contradiction.
Case \(J10\): To prove that \(\phi(\neg A \rightarrow (A \rightarrow B)) = 0\), i.e. \(\phi(\neg A) \geq \phi(A \rightarrow B)\).
- Subcase 1. \(\phi(A) \geq \phi(B)\): So \(\phi(A \rightarrow B) = 0\), so \(\phi(\neg A) \geq \phi(A \rightarrow B)\) reduces to \(\phi(\neg A) \geq 0\), which is true.
- Subcase 2. \(\phi(A) < \phi(B)\): This implies that \(\phi(A) < n\), otherwise \(n < \phi(B)\), which is not possible. Additionally, \(\phi(A \rightarrow B) = \phi(B)\). Since \(\phi(A) < n\) we have that \(\phi(\neg A) = n\), thus \(\phi(\neg A) \geq \phi(A \rightarrow B)\) reduces to \(n > \phi(A \rightarrow B)\) which is true.
Inductive case: Let \(\langle A_1, A_2, \dots, A_n, A_{n+1} \rangle\) be proof in \(J\) of size \(n+1\). We notice that the subproof \(\langle A_1, A_2, \dots, A_n \rangle\) satisfies the Inductive hypothesis, i.e. \(\phi(A_i) =\) for every \(1 \leq i \leq n\). We need to show that \(\phi(A_{n+1}) = 0\). Several cases are noticed:
- \(A_{n+1}\) is an axiom of \(J\). Then by the base case we have that \(\phi(A_{n+1}) = 0\) as desired.
- \(A_{n+1}\) was obtained using Modus Ponens using some \(A_i, A_j := A_i \rightarrow A_{n+1}\) in the proof with \(i, j \leq n\). By the inductive hypothesis, we have that \(\phi(A_i) = 0\) and \(\phi(A_i \rightarrow A_{n+1}) = 0\), which means that \(0 = \phi(A_i) \geq \phi(A_{n+1})\), thus \(\phi(A_{n+1}) = 0\).

With this invariant we conclude that \(\phi(A) = 0\) for every \(\vdash_J A\).

We notice that with an assignment \(\phi : PROP \rightarrow \{0, 1, 2\}\) such that \(\phi(A) = 1\) we have that \(\phi(\neg A \lor A) = \min \{ \phi(\neg A), \phi(A) \} = \min \{ 2 , 1 \} = 1\). Additionally, \(\phi(\neg \neg A) = 0\) since \(\phi(\neg A) = 2\), so \(\phi(\neg \neg A \rightarrow A) = 1\) since \(0 = \phi(\neg \neg A) < \phi(A) = 1\).

It is important to notice that the \(n+1\)-valued logic introduced in the previous theorem can be considered an invariant for the propositional intuitionisc formal system. However, this truth assignment does not constitute a semantics for the system \(J\). In fact, there are no finite smeantics for intuitionistic logic as we will observe with the following theorem:

Lemma 1: {#lemma1} For \(n \geq 2\), let \(D_n\) denote the formula:

\((p_1 \leftrightarrow p_2) \lor (p_1 \leftrightarrow p_3) \lor \dots \lor (p_1 \leftrightarrow p_n) \lor \dots\)
\(\lor (p_2 \leftrightarrow p_3) \lor \dots \lor (p_2 \leftrightarrow p_n) \lor \dots\)
\(\lor (p_{n-1} \leftrightarrow p_n)\)

We have that \(\not \vdash_J D_n\).

Proof: We use the \(n+1\)-valued logic previously defined in Theorem 1. We notice that \(\phi(D_n) = \min_{1 \leq i<j \leq n} \{ \phi(p_i \leftrightarrow p_j) \}\). Let us suppose by contradiction that \(\vdash_J D_n\). Thus, by Theorem 1 we have that \(\phi(D_n) = 0\), so there are \(1 \leq i<j \leq n\) such that \(\phi(p_i \leftrightarrow p_j) = 0\). Since \(p_i \leftrightarrow p_j\) stands for \((p_i \rightarrow p_j) \land (p_i \rightarrow p_j)\) we have that \(\max \{ \phi(p_i \rightarrow p_j), \phi(p_j \rightarrow p_i) \} = 0\). The latter implies that \(\phi(p_i \rightarrow p_j) = 0\) and \(\phi(p_j \rightarrow p_i) = 0\), which entail that \(\phi(p_i) \geq \phi(p_j)\) and \(\phi(p_j) \geq \phi(p_i)\). These inequalities can be combined into \(\phi(p_i) = \phi(p_j)\). So if we pick a truth assignment such that \(\phi(p_i) = i\) we notice that \(D_n\) does not hold for all truth assignments in the \(n+1\)-valued logic.

Theorem: Consider the language with connectives \(\neg, \lor, \land, \rightarrow\). A matrix for this language is a 6-tuple \(M = \langle S, S_0, H_\neg, H_\lor, H_\land, H_\rightarrow \rangle\), where \(S\) is a nonempty set whose elements are called truth values, \(S_0\) is a subset of \(S\) whose elements are called designated values, and \(H_\lor, H_\land, H_\rightarrow\), and \(H_\neg\) are truth functions for \(\lor, \land, \rightarrow\), and \(\neg\). A truth assignment for \(M\) is a function \(\phi : PROP \rightarrow S\). Such an assignment extends \(FOR(\neg, \lor, \land, \rightarrow)\) in the usual way.

There is no matrix \(M\) with \(S\) finite such that for every formula \(A\), \(\vdash_J A \Leftrightarrow \phi(A) \in S_0\) for every truth assignment \(\phi\) for \(M\).

Proof: Let us assume by contradiction that such matrix \(M\) exists with \(n\) elements. We realize that \(D_{n+1}\) is not a theorem of \(J\) from Lemma 1, so there is a truth assignment \(\phi\) for \(M\) such that \(\phi(D_{n+1}) \not \in S_0\). By the pigeonhole principle, there are \(1 \leq j < k \leq n+1\) such that \(\phi(p_i) = \phi(p_k)\), i.e. more propositional variables than truth values. Let \(E_{n+1} = D_{n+1}\) be obtained from \(D_{n+1}\) by replacing \((p_j \leftrightarrow p_k)\) with \((p_k \leftrightarrow p_k)\). Since \(\phi(p_j \leftrightarrow p_k) = H_\leftrightarrow(\phi(p_j), \phi(p_k))\) and \(H_\leftrightarrow\) is a truth function, we have that \(H_\leftrightarrow(\phi(p_j), \phi(p_k)) = H_\leftrightarrow(\phi(p_k), \phi(p_k))\) since \(\phi(p_k) = \phi(p_j)\). So \(H_\leftrightarrow(\phi(p_k), \phi(p_k)) = \phi(p_k \leftrightarrow p_k)\). Thus \(\phi(D_{n+1}) = \phi(E_{n+1})\).

Let us prove the following theorem in \(J\): \(\vdash_J p_k \leftrightarrow p_k\).

\(p_k \rightarrow (p_k \rightarrow p_k)\), Axiom \(J1\)
\(p_k \rightarrow ((p_k \rightarrow p_k) \rightarrow p_k)\), Axiom \(J1\)
\((p_k \rightarrow ((p_k \rightarrow p_k) \rightarrow p_k)) \rightarrow ( (p_k \rightarrow (p_k \rightarrow p_k)) \rightarrow (p_k \rightarrow p_k))\), Axiom \(J2\)
\((p_k \rightarrow (p_k \rightarrow p_k)) \rightarrow (p_k \rightarrow p_k)\), Modus Ponens(2, 3)
\(p_k \rightarrow p_k\), Modus Ponens(1, 4)
\((p_k \rightarrow p_k) \rightarrow ( (p_k \rightarrow p_k) \rightarrow ( (p_k \rightarrow p_k) \land (p_k \rightarrow p_k)))\), Axiom \(J5\)
\((p_k \rightarrow p_k) \rightarrow ( (p_k \rightarrow p_k) \land (p_k \rightarrow p_k))\), Modus Ponens(5, 6)
\((p_k \rightarrow p_k) \land (p_k \rightarrow p_k)\), Modus Ponens(5, 7)
\((p_k \leftrightarrow p_k)\), Definition of \(\leftrightarrow\) (8)

Using the Axiom \(J6\) (or \(J7\)) we can introduce any number of formulas to a theorem in \(J\). Hence, \(\vdash E_{n+1}\), thus \(\phi(E_{n+1}) \in S_0\) according to our assumption of the existence of a matrix \(M\). This however, entails that \(\phi(D_{n+1}) \in S_0\) but that is a contradiction.

Kripke Semantics for Propositional Intuitionistic Logic

Definition:(Dalen 2013) A Kripke model is a tuple \(\mathcal{K} = \langle K, \Sigma \rangle\) such that \(K\) ia a non-empty partially ordered set, \(\Sigma : K \rightarrow 2^{PROP}\) is a function on \(K\) that assigns a collection of propositional variables to each element \(k\) in \(K\) satisfying the following:

\(\bot \not \in \Sigma(k)\) for all \(k\) in \(K\)
If \(k \leq l\) then \(\Sigma(k) \subseteq \Sigma(l)\)

Each element in \(K\) can be interpreted as the states of the knowlege-subject, which progression over time is captured by the partially order relation \(\leq\) over \(K\). Additionally, the map \(\Sigma\) can be understood as the knowledge state of the knowlege-subject at each time. The function \(\Sigma\) tells us which atoms are `true’ in each state \(k \in K\). We can extend \(\Sigma\) to all formulas.

Lemma 2: \(\Sigma\) has a unique extension to a function on \(K\) satisfying the following:

\(\psi \lor \phi \in \Sigma(k)\) if and only if \(\psi \in \Sigma(k)\) or \(\phi \in \Sigma(k)\)
\(\psi \land \phi \in \Sigma(k)\) if and only if \(\psi \in \Sigma(k)\) and \(\phi \in \Sigma(k)\)
\(\psi \rightarrow \phi \in \Sigma(k)\) if and only if for all \(l \geq k\) if \(\psi \in \Sigma(l)\) then \(\phi \in \Sigma(l)\)

The negation connective here is abbreviated as \(\neg A := A \rightarrow \bot\). We write \(k \models \phi\) for \(\phi \in \Sigma(k)\), pronouncing it as \(k\) forces \(\phi\).

Lemma 3: (Monotonicity of \(\models\)) Let \(k, l \in K\) and \(\phi \in FORM(\neg, \lor, \land, \rightarrow)\). If \(k \leq l\), then \(k \models \phi\) implies \(l \models \phi\)

Proof: Induction on \(\phi\).

\(\phi\) is atomic: Holds by definition of Kripke structure.
\(\phi\) is \(\phi_1 \lor \phi_2\): Let \(k \models \phi_1 \lor \phi_2\) and \(k \leq l\). So \(k \models \phi_1 \lor \phi_2\) if and only if \(k \models \phi_1\) or \(k \models \phi_2\). By induction, we have that \(l \models \phi_1\) or \(l \models \phi_2\) so \(l \models \phi_1 \lor \phi_2\).
\(\phi\) is \(\phi_1 \land \phi_2\): Similar to the previous case.
\(\phi\) is \(\phi_1 \rightarrow \phi_2\): Let \(k \models \phi_1 \rightarrow \phi_2\) and \(l \geq k\). Suppose \(p \geq l\) and \(p \models \phi_1\). Since \(p \geq k\), \(p \models \phi_2\). Hence, \(l \models \phi_1 \rightarrow \phi_2\).

The last lemma captures the idea that knowlege is preserved `over time’ of the knowlege-subject and as such it can only be incremented.

References

Dalen, Dirk van. 2013. Logic and Structure. Springer London. https://doi.org/10.1007/978-1-4471-4558-5.

Hodel, R. E. 2013. An Introduction to Mathematical Logic. Dover Books on Mathematics. Dover Publications, Incorporated. https://books.google.com/books?id=SxRYdzWio84C.

Richardson, George P. 2011. “Reflections on the Foundations of System Dynamics.” System Dynamics Review 27 (3): 219–43. https://doi.org/https://doi.org/10.1002/sdr.462.

Trudeau, Richard J. 2001. “The Possibility of Non-Euclidean Geometry.” In The Non-Euclidean Revolution: With an Introduction by h.s.m Coxeter. Birkhäuser Boston. https://doi.org/10.1007/978-1-4612-2102-9_5.

Jose Abel Castellanos Joo