PHIL 452 · Modal Logic
Given a formula \(\varphi\) with propositional variables \(p_1, \dots, p_n\) and formulas \(\psi_1, \dots, \psi_n\), we define
\[ \varphi[\psi_1/p_1, \dots, \psi_n/p_n] \]
by the recursion:
\[ \begin{array}{lll} p_i[\psi_1/p_1, \dots, \psi_n/p_n] & := & \psi_i \\ (\neg \chi) [\psi_1/p_1, \dots, \psi_n/p_n] & := & \neg \chi[\psi_1/p_1, \dots, \psi_n/p_n] \\ (\chi \to \rho)[\psi_1/p_1, \dots, \psi_n/p_n] & := & (\chi[\psi_1/p_1, \dots, \psi_n/p_n] \to \rho[\psi_1/p_1, \dots, \psi_n/p_n]) \end{array} \]
\[ \begin{array}{lll} p[\neg q] & := & \neg q \\ \neg (p \to \neg q)[\neg p/p, (p \to q)/q] & := & \neg (\neg p \to \neg (p \to q)) \\ (p \to (q \to p))[(q \to p)/p, (p \to q)/q] & := & ((q \to p) \to ((p \to q) \to (q \to p))) \\ \end{array} \]
Axioms are substitution instances of one of the three formulas below:
\[\begin{array}{lll} \textsf{A1} & & p \to (q \to p)\\ \textsf{A2} & & (p \to (q \to r)) \to ((p \to q) \to (p \to r))\\ \textsf{A3} & & (\neg p \to \neg q) \to ((\neg p \to q) \to p) \end{array}\]
There is one rule of inference:
\[\begin{array}{lll} \textsf{Modus Ponens} & & \varphi, \ (\varphi \to \psi)/\psi\\ \end{array}\]
A formula \(\varphi\) is derivable from a set of well-formed formulas \(\Gamma\), written \(\Gamma \vdash \varphi\), iff, there is a finite sequence of well-formed formulas \((\chi_1, \dots , \chi_n)\) such that
and for each \(m \leq n\), either
\(\chi_m\) is an axiom, or
\(\chi_m \in \Gamma\), or
\(\chi_k = (\chi_l \to \chi_m)\), for some \(k, l <m\).
Such a finite sequence \(( \chi_1, ..., \chi_n )\) is called a derivation (or a proof) of \(\varphi\) from \(\Gamma\).
A formula \(\varphi\) is a theorem, written \(\vdash \varphi\), iff there is a derivation of \(\varphi\) from the empty set.
\(\vdash p \to p\)
\[\begin{array}{lll} 1 & p \to ((p \to p) \to p) & \textsf{A1}[p/p, (p \to p)/q]\\ 2 & (p \to ((p \to p) \to p))\to & \\ & ((p \to (p \to p)) \to (p \to p)) & \textsf{A2}[p/p, (p \to p)/q, p/r]\\ 3 & (p \to (p \to p)) \to (p \to p) & \textsf{MP} \ 1, 2\\ 4 & p \to (p \to p) & \textsf{A1}[p/p, p/q]\\ 5 & p \to p & \textsf{MP} \ 3, 4\\ \end{array}\]
The argument generalizes: \(\vdash \varphi \to \varphi\)
We now develop a library of subroutines we can use in the course of more complex proofs.
For all sets of formulas \(\Gamma\) and formulas \(\varphi\), and \(\psi\), \[ \begin{array}{lll} \Gamma, \varphi \vdash \psi & \Leftrightarrow & \Gamma \vdash \varphi \to \psi\\ \end{array} \]
Notation: \[ \begin{array}{lll} \Gamma, \varphi & := & \Gamma \cup \{\varphi\} \\ \varphi \vdash \psi & := & \{\varphi\} \vdash \psi \\ \end{array} \]
Given a condition \(\Phi(n)\) on positive integers, if
if for all \(m < n\), \(m\) satisfies the condition, then \(n\) does,
then all positive integers satisfy the condition.
Set the condition \(\Phi(n)\) on length as:
For all sets of formulas \(\Gamma\), and formulas \(\chi_1\), … \(\chi_n\), \(\varphi\), and \(\psi\),
if \(\langle \chi_1, \cdots \chi_n\rangle\) is a derivation of \(\psi\) from \(\Gamma, \varphi\) of length \(n\), then \(\Gamma \vdash \varphi \to \psi\).
We argue that for each \(n\), if for all \(m < n\), \(m\) satisfies the condition, then \(n\) does.
For all sets of formulas \(\Gamma\) and formulas \(\chi_1\), … \(\chi_m\), \(\varphi\), and \(\psi\),
if \(\langle \chi_1, \cdots \chi_m\rangle\) is a derivation of \(\psi\) from \(\Gamma, \varphi\) of length \(m\), then \(\Gamma \vdash \varphi \to \psi\).
For all sets of formulas \(\Gamma\) and formulas \(\chi_1\), … \(\chi_n\), \(\varphi\), and \(\psi\),
if \(\langle \chi_1, \cdots \chi_n\rangle\) is a derivation of \(\psi\) from \(\Gamma, \varphi\) of length \(n\), then \(\Gamma \vdash \varphi \to \psi\).
Consider a derivation of \(\psi\) from \(\Gamma, \varphi\) of the form: \[( \chi_1, ..., \chi_{n}).\] Thus \(\chi_{n}= \psi\), and for each \(1 \leq i \leq n\), \(\chi_{i}\) is either:
\[\begin{array}{lll} 1 & \psi \to (\varphi \to \psi) & \textsf{A1}[\psi/p, \varphi/q]\\ 2 & \psi & \textsf{axiom}\\ 3 & \varphi \to \psi & \textsf{MP} \ 1, 2\\ \end{array}\]
\[\begin{array}{lll} 1 & \psi \to (\varphi \to \psi) & \textsf{A1}[\psi/p, \varphi/q]\\ 2 & \psi & \textsf{premise}\\ 3 & \varphi \to \psi & \textsf{MP} \ 1, 2\\ \end{array}\]
Case 3. \(\psi\) is \(\varphi\). Then \(\vdash \varphi \to \psi\).
This is because \(\varphi \to \psi\) is just \(\varphi \to \varphi\) and \(\vdash \varphi \to \varphi\). So, \(\Gamma \vdash \varphi \to \varphi\)
Case 4. \(\psi\) is the output of an application of \(\textsf{MP}\) to two earlier lines \(\chi_l\) and \(\chi_k\). That means that \(\chi_k\) is \((\chi_l \to \psi)\) for some \(k, l < n\). Because both \(k, l < n\), there are derivations of \(\chi_l\) and \(\chi_l \to \psi\), respectively, of length less than \(n\).
By inductive hypothesis:
\(\Gamma \vdash \varphi \to \chi_l\)
\(\Gamma \vdash \varphi \to (\chi_l \to \psi)\).
Combine the derivations of \(\varphi \to \chi_l\) and \(\varphi \to (\chi_l \to \psi)\) from \(\Gamma\) into a derivation of \(\varphi \to \psi\) from \(\Gamma\):
\[ \begin{array}{lll} 1 & \cdots & \cdots\\ \vdots & \vdots & \vdots\\ k & \varphi \to (\chi_l \to \psi) & \Gamma \vdash \varphi \to (\chi_l \to \psi) \\ \vdots & \vdots & \vdots \\ m & \varphi \to \chi_l & \Gamma \vdash \varphi \to \chi_l\\ %\vdots & \vdots & \vdots \\ m+1 & (\varphi \to (\chi_1 \to \psi)) \to & \textsf{A2}[\varphi/p,\chi_l/q,\psi/r]\\ & ((\varphi \to \chi_l)\to (\varphi \to \psi)) & \\ n+2 & (\varphi \to \chi_l)\to (\varphi \to \psi) & \textsf{MP} \ k, n\\ n+3 & \varphi \to \psi & \textsf{MP} \ m, n+1\\ \end{array} \]
To prove a conditional, we adjoin the antecedent to our assumptions and set out to derive the consequent from them. If we can, we are entitled to the conditional on the basis of the original assumptions.
\(\vdash (\varphi \to \psi) \to ((\psi \to \chi)\to (\varphi \to \chi))\)
Proof. Three applications of the Deduction Theorem deliver: \[\{\varphi \to \psi \} \vdash (\psi \to \chi) \to (\varphi \to \chi).\] \[\{ \varphi \to \psi, \psi \to \chi\} \vdash \varphi \to \chi,\] \[\{\varphi \to \psi, \psi \to \chi, \varphi\} \vdash \chi.\] Here is a derivation of \(\chi\) from \(\{\varphi \to \psi, \psi \to \chi, \varphi\}\): \[ \begin{array}{lll} 1 & \varphi \to \psi & \textsf{premise} \\ 2 & \psi \to \chi & \textsf{premise}\\ 3 & \varphi & \textsf{premise} \\ 4 & \psi & \textsf{MP} \ 1, 3\\ 5 & \chi & \textsf{MP} \ 2, 4\\ \end{array} \]
\(\vdash (\neg \psi \to \neg \varphi) \to (\varphi \to \psi)\)
Proof. Two applications of the Deduction Theorem deliver: \[\neg \psi \to \neg \varphi \vdash \varphi \to \psi \] \[\{\neg \psi \to \neg \varphi, \varphi\} \vdash \psi\] Here is a derivation of \(\psi\) from \(\{\neg \psi \to \neg \varphi, \varphi\}\):
\[ \begin{array}{lll} 1 & \neg \psi \to \neg \varphi & \\ 2 & \varphi \to (\neg \psi \to \varphi) & \textsf{A1}[\varphi/p,\neg \psi/q]\\ 3 & \varphi & \textsf{premise} \\ 4 & \neg \psi \to \varphi & \textsf{MP} \ 2, 3\\ 5 & (\neg \psi \to \neg \varphi) \to ((\neg \psi \to \varphi) \to \psi) & \textsf{A3}[\psi/p, \varphi/q]\\ 6 & (\neg \psi \to \varphi) \to \psi & \textsf{MP} \ 1, 5\\ 7 & \psi & \textsf{MP} \ 4, 6 \end{array} \]
\(\vdash \neg \neg \varphi \to \varphi\)
Proof. Given the Deduction Theorem, it suffices to prove: \[\neg \neg \varphi \vdash \varphi\] Here is a derivation of \(\varphi\) from \(\neg \neg \varphi\):
\[\begin{array}{lll} 1 & \neg \neg \varphi \to (\neg \varphi \to \neg \neg \varphi) & \textsf{A1}[\neg\neg\varphi/p, \neg \varphi/q]\\ 2 & \neg \neg \varphi & \textsf{premise} \\ 3 & \neg \varphi \to \neg \neg \varphi & \textsf{MP} \ 1, 2 \\ 4 & (\neg \varphi \to \neg \neg \varphi) \to ((\neg \varphi \to \neg \varphi) \to \varphi) & \textsf{A3}[\varphi/p, \neg \varphi/q]\\ 5 & (\neg \varphi \to \neg \varphi) \to \varphi & \textsf{MP} \ 3, 4\\ 6 & \neg \varphi \to \neg \varphi & \textsf{theorem} \ (4.1)\\ 7 & \varphi & \textsf{MP} \ 5, 6 \end{array}\]
\(\vdash \varphi \to \neg \neg \varphi\)
Proof. Combine two observations above:
\[\begin{array}{lll} 1 & \neg \neg \neg \varphi \to \neg \varphi & \textsf{theorem} \ (4.4)\\ 2 & (\neg \neg \neg \varphi \to \neg \varphi) \to (\varphi \to \neg \neg \varphi) & \textsf{theorem} \ (4.3) \\ 3 & \varphi \to \neg \neg \varphi & \textsf{MP} \ 1, 2 \end{array}\]
\(\vdash (\varphi \to \psi) \to (\neg \psi \to \neg \varphi)\)
Proof. Given the Deduction Theorem, it suffices to prove: \[\varphi \to \psi \vdash \neg \psi \to \neg \varphi\]
\[ \begin{array}{llll} 1 & & \varphi \to \psi & \textsf{premise} \\ 2 & & \psi \to \neg \neg \psi & \textsf{theorem} \ (4.5)\\ 3 & & \neg \neg \varphi \to \varphi & \textsf{theorem} \ (4.4)\\ 4 & & (\neg \neg \varphi \to \varphi) \to ((\varphi \to \psi) \to (\neg \neg \varphi \to \psi)) & \textsf{theorem} \ (4.2) \\ 5 & & (\varphi \to \psi) \to (\neg \neg \varphi \to \psi) & \textsf{MP} \ 3, 4 \\ 6 & & \neg \neg \varphi \to \psi & \textsf{MP} \ 1, 4 \\ 7 & & (\neg \neg \varphi \to \psi) \to ((\psi \to \neg \neg \psi) \to (\neg \neg \varphi \to \neg \neg \psi)) & \textsf{theorem} \ (4.2) \\ 8 & & (\psi \to \neg \neg \psi) \to (\neg \neg \varphi \to \neg \neg \psi) & \textsf{MP} \ 6, 7\\ 9 & & \neg \neg \varphi \to \neg \neg \psi & \textsf{MP} \ 2, 8\\ 10 & & (\neg \neg \varphi \to \neg \neg \psi) \to (\neg \psi \to \neg \varphi) & \textsf{theorem} \ (4.3)\\ 11 & & \neg \psi \to \neg \varphi & \textsf{MP} \ 9, 10 \end{array} \]
Use the Deduction Theorem to justify:
\[ (\varphi \to \neg \varphi) \to \neg \varphi \]
Proof. Given the Deduction Theorem, we target \(\{\varphi \to \neg \varphi\} \vdash \neg \varphi\).
\[\begin{array}{lll} 1 & \varphi \to \neg \varphi & \textsf{premise} \\ 2 & (\neg \neg \varphi \to \neg \varphi) \to ((\neg \neg \varphi \to \varphi) \to \neg \neg \neg \varphi) & \textsf{A3} \\ 3 & (\varphi \to \neg \varphi) \to (\neg \neg \varphi \to \neg \varphi) & \textsf{theorem} \ (4.6)\\ 4 & \neg \neg \varphi \to \neg \varphi & \textsf{MP} \ 1, 3\\ 5 & (\neg \neg \varphi \to \varphi) \to \neg \neg \neg \varphi & \textsf{MP} \ 2, 4\\ 6 & \neg \neg \varphi \to \varphi & \textsf{theorem} \ (4.4)\\ 7 & \neg \neg \neg \varphi & \textsf{MP} \ 5, 6\\ 8 & \neg \neg \neg \varphi \to \neg \varphi & \textsf{theorem} \ (4.4)\\ 9 & \neg \varphi & \textsf{MP} \ 7, 8\\ \end{array}\]
\(\vdash \neg \varphi \to (\varphi \to \psi)\)
Proof. Two applications of the Deduction Theorem deliver: \[\neg \varphi \vdash \varphi \to \psi\] \[\{\neg \varphi, \varphi\} \vdash \psi\] Here is a derivation of \(\psi\) from \(\{\neg \varphi, \varphi\}\): \[\begin{array}{lll} 1 & \neg \varphi \to (\neg \psi \to \neg \varphi) & \textsf{A1}[\neg \varphi/p, \neg \psi/q] \\ 2 & \neg \varphi & \textsf{premise} \\ 3 & \neg \psi \to \neg \varphi & \textsf{MP} \ 1, 2\\ 4 & (\neg \psi \to \neg \varphi) \to (\varphi \to \psi) & \textsf{theorem} \ (4.3)\\ 5 & \varphi \to \psi & \textsf{MP} \ 3, 4\\ 6 & \varphi & \textsf{premise} \\ 7 & \psi & \textsf{MP} \ 5, 6\\ \end{array}\]