PHIL 452 ยท Modal Logic
For all sets of formulas \(\Gamma\) and formula \(\varphi\),
\[ \begin{array}{lll} \Gamma \nvdash \varphi & &\Gamma \not \models \varphi\\ \ \ \Updownarrow & & \ \ \Updownarrow \\ \Gamma, \neg \varphi \ \textsf{consistent} & \Rightarrow & \Gamma, \neg \varphi \ \textsf{satisfiable} \\ \end{array} \]
For all sets of formulas \(\Gamma\),
\[ \begin{array}{lll} \Gamma \ \textsf{consistent} & \Rightarrow & \Gamma \ \textsf{satisfiable} \\ \end{array} \]
Given a consistent set \(\Gamma\), we will explain how to expand it into a maximal consistent set \(\Sigma\).
\[\begin{array}{clllc} \Gamma \ \text{consistent} & & & & \phantom{\Gamma \ \text{satisfiable}}\\ \phantom{\Downarrow} & & & & \phantom{\Uparrow}\\ \phantom{\Gamma \subseteq \Sigma \ \text{and} \ \Sigma \ \text{maximal consistent}} & & \phantom{\Rightarrow} & & \phantom{\Gamma \subseteq \Sigma \ \text{and} \ \Sigma \ \text{satisfiable}}\\ \end{array}\]
\[\begin{array}{clllc} \Gamma \ \text{consistent} & & & & \phantom{\Gamma \ \text{satisfiable}}\\ \Downarrow & & & & \phantom{\Uparrow}\\ \Gamma \subseteq \Sigma \ \text{and} \ \Sigma \ \text{maximal consistent} & & \phantom{\Rightarrow} & & \phantom{\Gamma \subseteq \Sigma \ \text{and} \ \Sigma \ \text{satisfiable}}\\\\ \end{array}\]
\[\begin{array}{clllc} \Gamma \ \text{consistent} & & & & \phantom{\Gamma \ \text{satisfiable}}\\ \Downarrow & & & & \phantom{\Uparrow}\\ \Gamma \subseteq \Sigma \ \text{and} \ \Sigma \ \text{maximal consistent} & & \Rightarrow & & \Gamma \subseteq \Sigma \ \text{and} \ \Sigma \ \text{satisfiable}\\ \end{array}\]
\[\begin{array}{clllc} \Gamma \ \text{consistent} & & & & \Gamma \ \text{satisfiable}\\ \Downarrow & & & & \Uparrow\\ \Gamma \subseteq \Sigma \ \text{and} \ \Sigma \ \text{maximal consistent} & & \Rightarrow & & \Gamma \subseteq \Sigma \ \text{and} \ \Sigma \ \text{satisfiable} \end{array}\]
\[\begin{array}{clllc} \color{#7a003c}{\Gamma \ \text{consistent}} & & & & \Gamma \ \text{satisfiable}\\ \color{#7a003c}{\Downarrow} & & & & \Uparrow \\ \color{#7a003c}{\Gamma \subseteq \Sigma \ \text{and} \ \Sigma \ \text{maximal consistent}} & & \Rightarrow & & \Gamma \subseteq \Sigma \ \text{and} \ \Sigma \ \ \text{satisfiable}\\ \end{array}\]
Lindenbaum theorem. If \(\Gamma\) is consistent, then \(\Gamma\) is a subset of a maximal consistent set \(\Sigma\).
Proof. Enumerate all the formulas of propositional logic
\[ \varphi_0, \varphi_1, \varphi_2, \dots \] and build a chain
\[ \Gamma = \Sigma_0 \subseteq \Sigma_1 \subseteq \Sigma_2 \subseteq \cdots \] where \(\Sigma_{0}\) is \(\Gamma\) and \(\Sigma_{n+1}\) is \(\Sigma_n, \varphi_n\) if that set is consistent, and \(\Sigma_n\) otherwise.
Let \(\Sigma = \bigcup_n \Sigma_n\). Notice that the identity of \(\Sigma\) is very sensitive to the details of the enumeration.
Let \(\varphi_0, \varphi_1, \dots\) be an enumeration of the formulas of propositional logic.
Define a chain \(\Sigma_0 \subseteq \Sigma_1 \subseteq \Sigma_2 \subseteq \cdots\) by the recursion:
\[\begin{array}{lll} \Sigma_0 & = & \Gamma \\ \Sigma_{n+1} & = & \begin{cases} \Sigma_n, \varphi_n & \text{if consistent}\\ \Sigma_n & \text{otherwise} \end{cases} \end{array}\]
Proof. Induction on \(n\).
Base Case: \(\Sigma_0\) is consistent. \(\Sigma_0\) is just \(\Gamma\), which we have assumed to be consistent.
Inductive Step: If \(\Sigma_n\) is consistent, then so is \(\Sigma_{n+1}\). Suppose \(\Sigma_n\) is consistent. Then \(\Sigma_{n+1}\) is either \(\Sigma_n, \varphi_n\) if consistent, or \(\Sigma_n\), otherwise. Either way, \(\Sigma_{n+1}\) is consistent.
Proof. Induction on \(n\).
Base Case: for every \(m < 0\), \(\Sigma_n \subseteq \Sigma_0\). Trivial.
Inductive Step: If for every \(m < n\), \(\Sigma_m \subseteq \Sigma_n\), then for every \(m < n +1\), \(\Sigma_m \subseteq \Sigma_{n+1}\). By construction, \(\Sigma_n \subseteq \Sigma_{n+1}\). So, if \(m \leq n\), then \(\Sigma_m \subseteq \Sigma_n \subseteq \Sigma_{n+1}\).
Proof. Given a finite subset \(\Delta\) of \(\Sigma\), let \(\varphi_n\) be the member of \(\Delta\) of highest index. Then \(\Delta \subseteq \Sigma_{n+1}\).
Proof. For a reductio, suppose \(\Sigma \vdash \bot\). Then there is a finite derivation of \(\bot\) from \(\Sigma\). So, there is a derivation of \(\bot\) from a finite subset \(\Delta\) of \(\Sigma\), making \(\Delta\) inconsistent. But we showed that every finite subset of \(\Sigma\) is consistent.
Proof. If \(\varphi\) is a formula, then it has an index in the enumeration, namely, \(\varphi\) is \(\varphi_n\). So, if \(\Sigma_n, \varphi_n\) is consistent, then \(\varphi \in \Sigma_{n+1} \subseteq \Sigma\).
\[\begin{array}{clllc} \Gamma \ \text{consistent} & & & & \Gamma \ \text{satisfiable}\\ \Downarrow & & & & \Uparrow \\ \color{#7a003c}{\Gamma \subseteq \Sigma \ \text{and} \ \Sigma \ \text{maximal consistent}} & & \color{#7a003c}{\Rightarrow} & & \color{#7a003c}{\Gamma \subseteq \Sigma \ \text{and} \ \Sigma \ \text{satisfiable}}\\ \end{array}\]
Henkin lemma. If \(\Sigma\) is maximal consistent, then \(\Sigma\) is satisfiable.
Proof. Given a maximally consistent \(\Sigma\), define the canonical assignment:
\[ \begin{array}{lll} v_\Sigma(p) = 1 & \text{iff } & p \in \Sigma \end{array} \] We use an induction on the complexity of formulas to prove: \[ \begin{array}{lll} V_\Sigma(\varphi) = 1 & \text{iff } & \varphi \in \Sigma \end{array} \]
Proof. Induction on the complexity of formulas:
\[ \begin{array}{lllll} \text{Base case}: & & & & \\ &V_\Sigma(p) = 1 & \text{iff } & v_\Sigma(p) = 1 & \\ & & \text{iff } & \varphi \in \Sigma & \textsf{definition of} \ v_\Sigma \\[8pt] \text{Induction step for } \neg\textbf{:} & & & & \\ & V_\Sigma(\neg \varphi) = 1 & \text{iff } & V_\Sigma(\varphi) = 0 & \textsf{definition of} \ V_\Sigma \\ & & \text{iff } & \varphi \notin \Sigma & \textsf{inductive hypothesis} \\ & & \text{iff } & \neg \varphi \in \Sigma & \textsf{maximal consistency} \\[8pt] \text{Induction step for } \to\textbf{:} & & & & \\ & V_\Sigma(\varphi \to \psi) = 1 & \text{iff } & V_\Sigma(\varphi) = 0 \ \text{or} \ V_\Sigma(\psi) = 1 & \textsf{definition of} \ V_\Sigma \\ & & \text{iff } & \varphi \notin \Sigma \ \text{or} \ \psi \in \Sigma & \textsf{inductive hypothesis} \\ & & \text{iff } & \varphi \to \psi \in \Sigma & \textsf{maximal consistency} \\ \end{array} \]
\[\begin{array}{clllc} \Gamma \ \text{consistent} & & & & \color{#7a003c}{\Gamma \ \text{satisfiable}}\\ \Downarrow & & & & \color{#7a003c}{\Uparrow} \\ \Gamma \subseteq \Sigma \ \text{and} \ \Sigma \ \text{maximal consistent} & & \Rightarrow & & \color{#7a003c}{\Gamma \subseteq \Sigma \ \text{and} \ \Sigma \ \text{satisfiable}}\\ \end{array}\]
If a valuation \(V_\Sigma\) satisfies \(\Sigma\) and \(\Gamma\) is a subset of \(\Sigma\), then \(V\) satisfies \(\Gamma\).
\[\begin{array}{clllc} \Gamma \ \text{consistent} & & & & \Gamma \ \text{satisfiable}\\ \Downarrow \ \text{\scriptsize Lindenbaum} & & & & \Uparrow \ \text{\scriptsize monotonicity} \\ \Gamma \subseteq \Sigma \ \text{and} \ \Sigma \ \text{maximal consistent}) & & \underset{\text{\scriptsize Henkin lemma}}{\Rightarrow} & & \Gamma \subseteq \Sigma \ \text{and} \ \Sigma \ \text{satisfiable}\\ \end{array}\]
Completeness. If a set \(\Gamma\) is consistent, then \(\Gamma\) is satisfiable.