PHIL 452 ยท Modal Logic
The axiom system is sound and complete for the semantics of propositional logic.
For each set of formulas \(\Gamma\) and each formula \(\varphi\), \[ \begin{array}{lll} \Gamma \vdash \varphi & \Leftrightarrow & \Gamma \models \varphi\\ \end{array} \]
The axiom system is sound if a formula \(\varphi\) is derivable from a \(\Gamma\) only if \(\varphi\) is a logical consequence of \(\Gamma\).
Soundness For each set of formulas \(\Gamma\) and each formula \(\varphi\), \[ \begin{array}{lll} \Gamma \vdash \varphi & \Rightarrow & \Gamma \models \varphi\\ \end{array} \]
Proof. Use a complete induction on the length of derivations. Set the condition \(\Phi(n)\) on length as:
For each set of formulas \(\Gamma\) and each formula \(\varphi\), if \(\langle \chi_1, \cdots \chi_n\rangle\) is a derivation of \(\varphi\) from \(\Gamma\) of length \(n\), then \(\Gamma \models \varphi\).
The axiom system is complete if if a formula \(\varphi\) is derivable from a \(\Gamma\) whenever \(\varphi\) is a logical consequence of \(\Gamma\).
Completeness For each set of formulas \(\Gamma\) and each formula \(\varphi\), \[ \begin{array}{lll} \Gamma \models \varphi & \Rightarrow & \Gamma \vdash \varphi\\ \end{array} \]
We will set out to prove a reformulation of completeness in terms of consistency, a proof-theoretic counterpart of satisfiability.
For each set of formulas \(\Gamma\) and each 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 each set of formulas \(\Gamma\),
\[ \begin{array}{lll} \Gamma \ \textsf{consistent} & \Rightarrow & \Gamma \ \textsf{satisfiable} \\ \end{array} \]
A set of formulas \(\Gamma\) is consistent iff \(\bot\) is not derivable from \(\Gamma\). That is, iff \(\Gamma \nvdash \varphi\).
Given a set of formulas \(\Gamma\) and a formula \(\varphi\), \[ \begin{array}{lll} \Gamma \vdash \varphi & \Leftrightarrow & \Gamma, \neg \varphi \ \ \text{is inconsistent}.\\ \end{array} \]
Consistency is the proof-theoretic counterpart of satisfiability:
Given a set of formulas \(\Gamma\) and a formula \(\varphi\), \[ \begin{array}{lll} \Gamma \models \varphi & \Leftrightarrow & \Gamma, \neg \varphi \ \ \text{is unsatisfiable}.\\ \end{array} \]
Given a set of formulas \(\Gamma\) and a formula \(\varphi\), \[ \begin{array}{lll} \Gamma \vdash \varphi & \Rightarrow & \Gamma, \neg \varphi \ \ \text{is inconsistent}.\\ \end{array} \]
Proof. Assume \(\Gamma \vdash \varphi\). Here is a derivation of \(\bot\) from \(\Gamma, \neg \varphi\):
\[ \begin{array}{llll} 1 & & \neg \varphi \to (\neg \bot \to \neg \varphi) & \textsf{A1}[\neg \varphi/p, \bot/q] \\ 2 & & \neg \varphi & \textsf{premise} \\ 3 & & \neg \bot \to \neg \varphi & \textsf{MP} \ 1, 2\\ 4 & & (\neg \bot \to \neg \varphi) \to (\varphi \to \bot) & \textsf{theorem} \ (4.3)\\ 5 & & \varphi \to \bot & \textsf{MP} \ 3, 4\\ 6 & & \varphi & \Gamma \vdash \varphi\\ 7 & & \bot & \textsf{MP} \ 4, 5\\ \end{array} \]
Given a set of formulas \(\Gamma\) and a formula \(\varphi\), \[ \begin{array}{lll} \Gamma, \neg \varphi \ \ \text{is inconsistent} & \Rightarrow & \Gamma \vdash \varphi\\ \end{array} \]
Proof. Assume \(\Gamma, \neg \varphi\) is inconsistent, i.e., \(\Gamma, \neg \varphi \vdash \bot\). Here is a derivation of \(\varphi\) from \(\Gamma\):
\[ \begin{array}{llll} 1 & & \neg \varphi \to \neg \bot & \Gamma \vdash \neg \varphi \to \neg \bot \\ 2 & & (\neg \varphi \to \neg \bot) \to ((\neg \varphi \to \bot) \to \varphi) & \textsf{A3}[\varphi/p, \bot/q]\\ 3 & & (\neg \varphi \to \bot) \to \varphi & \textsf{MP} \ 1, 2\\ 4 & & \neg \varphi \to \bot & \Gamma \vdash \neg \varphi \to \bot \\ 5 & & \varphi & \textsf{MP} \ 3, 4\\ \end{array} \] Steps 1 and 4 rely on two key observations:
A set of formulas \(\Gamma\) is maximal consistent iff
\(\Gamma\) is consistent, and
for every formula \(\varphi\), \(\Gamma, \varphi\) is consistent only if \(\varphi \in \Gamma\)
Given a maximal consistent set of formulas \(\Gamma\) and two formulas \(\varphi\) and \(\psi\), \[ \begin{array}{llll} 1. &\Gamma \vdash \varphi & \Leftrightarrow & \varphi \in \Gamma\\ 2. & \neg \varphi \in \Gamma & \Leftrightarrow & \varphi \notin \Gamma\\ 3. & (\varphi \to \psi) \in \Gamma & \Leftrightarrow & \varphi \notin \Gamma \ \text{or} \ \psi \in \Gamma \end{array} \]
Proof. Given a maximal consistent set \(\Gamma\) and a formula \(\varphi\): \[ \begin{array}{lll} &\Gamma \vdash \varphi & \Leftrightarrow & \varphi \in \Gamma\\ \end{array} \]
(\(\Rightarrow\)) Assume \(\Gamma \vdash \varphi\). Suppose, for a reductio, that \(\varphi \notin \Gamma\). Since \(\Gamma\) is maximal consistent, \(\Gamma, \varphi\) is inconsistent. Since \(\Gamma \vdash \varphi\), \(\Gamma \vdash \bot\) if \(\Gamma, \varphi \vdash \bot\). So, \(\Gamma\) is inconsistent. Contradiction.
(\(\Leftarrow\)) If \(\varphi \in \Gamma\), then there is a one-line derivation of \(\varphi\) from \(\Gamma\).
Proof. Given a maximal consistent set \(\Gamma\) and a formula \(\varphi\): \[ \begin{array}{llll} &\neg \varphi \in \Gamma & \Leftrightarrow & \varphi \notin \Gamma\\ \end{array} \]
(\(\Rightarrow\)) Suppose \(\neg \varphi \in \Gamma\). By 1, \(\Gamma \vdash \neg \varphi\). Since \(\Gamma\) is consistent, \(\Gamma \nvdash \varphi\), and, by 1, \(\varphi \notin \Gamma\). The justification for \(\Gamma \nvdash \varphi\) is as follows. Since \(\vdash \neg \varphi \to (\varphi \to \bot)\) (4.7), we have that \(\Gamma \vdash \varphi\) only if \(\Gamma \vdash \bot\).
(\(\Leftarrow\)) Suppose \(\varphi \notin \Gamma\). By maximal consistency, \(\Gamma, \varphi\) is inconsistent. Given 4.4. and 4.5. \(\Gamma, \neg \neg \varphi\) is inconsistent. So, \(\Gamma \vdash \neg \varphi\).
Proof. Given a maximal consistent set \(\Gamma\) and two formulas \(\varphi\) and \(\psi\): \[ \begin{array}{llll} &\varphi \to \psi \in \Gamma & \Leftrightarrow & \varphi \notin \Gamma \ \text{or} \ \psi \in \Gamma\\ \end{array} \]
(\(\Rightarrow\)) Suppose \(\varphi \to \psi \in \Gamma\). Then \(\Gamma \vdash \varphi \to \psi\). We now argue that if \(\varphi \in \Gamma\), then \(\psi \in \Gamma\). If \(\varphi \in \Gamma\), then \(\Gamma \vdash \varphi\) and \(\Gamma \vdash \psi\). By 1 above, \(\psi \in \Gamma\).
(\(\Leftarrow\)) Suppose \(\varphi \notin \Gamma\) or \(\psi \in \Gamma\).
If \(\varphi \notin \Gamma\), then \(\Gamma \vdash \neg \varphi\). Since \(\Gamma \vdash \neg \varphi \to (\varphi \to \psi)\), we infer that \(\Gamma \vdash \varphi \to \psi\).
If \(\psi \in \Gamma\), then \(\Gamma \vdash \psi\). Since \(\Gamma \vdash \psi \to (\varphi \to \psi)\), we infer that \(\Gamma \vdash \varphi \to \psi\).
Notice that \(\psi \to (\varphi \to \psi)\) is a substitution instance of axiom 1.
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 \(\Gamma\),
\[ \begin{array}{lll} \Gamma \ \textsf{consistent} & \Rightarrow & \Gamma \ \textsf{satisfiable} \\ \end{array} \]
On one side, a syntactic condition on a set of formulas \(\Gamma\):
On the other side, a semantic condition on \(\Gamma\):
We want to bridge the two:
\[\Gamma \ \text{consistent} \ \Longrightarrow \ \Gamma \ \text{satisfiable}.\]
Given a consistent set \(\Gamma\), we will explain how to expand it into a maximal consistent set \(\Sigma\).
\[\begin{array}{clllc} \Gamma \ \text{consistent} & & & & \phantom{\Gamma \ \text{satisfiable}}\\ \phantom{\Downarrow} & & & & \phantom{\Uparrow}\\ \phantom{\Gamma \subseteq \Sigma \ \text{and} \ \Sigma \ \text{maximal consistent}} & & \phantom{\Rightarrow} & & \phantom{\Gamma \subseteq \Sigma \ \text{and} \ \Sigma \ \text{satisfiable}}\\ \end{array}\]
\[\begin{array}{clllc} \Gamma \ \text{consistent} & & & & \phantom{\Gamma \ \text{satisfiable}}\\ \Downarrow & & & & \phantom{\Uparrow}\\ \Gamma \subseteq \Sigma \ \text{and} \ \Sigma \ \text{maximal consistent} & & \phantom{\Rightarrow} & & \phantom{\Gamma \subseteq \Sigma \ \text{and} \ \Sigma \ \text{satisfiable}}\\\\ \end{array}\]
\[\begin{array}{clllc} \Gamma \ \text{consistent} & & & & \phantom{\Gamma \ \text{satisfiable}}\\ \Downarrow & & & & \phantom{\Uparrow}\\ \Gamma \subseteq \Sigma \ \text{and} \ \Sigma \ \text{maximal consistent} & & \Rightarrow & & \Gamma \subseteq \Sigma \ \text{and} \ \Sigma \ \text{satisfiable}\\ \end{array}\]
\[\begin{array}{clllc} \Gamma \ \text{consistent} & & & & \Gamma \ \text{satisfiable}\\ \Downarrow & & & & \Uparrow\\ \Gamma \subseteq \Sigma \ \text{and} \ \Sigma \ \text{maximal consistent} & & \Rightarrow & & \Gamma \subseteq \Sigma \ \text{and} \ \Sigma \ \text{satisfiable} \end{array}\]
Lindenbaum theorem. If \(\Gamma\) is consistent, then \(\Gamma\) is a subset of a maximal consistent set \(\Sigma\).
Henkin lemma. Every maximal consistent set \(\Sigma\) is satisfiable under the assignment \[ \begin{array}{lll} v(p) = 1 & \Leftrightarrow & p \in \Sigma \end{array} \]
Subsets of satisfiable sets are satisfiable.