PHIL 452 · Modal Logic
Given a condition on natural numbers, if
then all natural numbers satisfy the condition.
\[\begin{array}{llll} 0 & < & 1 & = & 2^0 \\ 1 & < & 2 & = & 2^1\\ 2 & < & 4 & = & 2^2 \\ \cdots & & \cdots & & \cdots \\ \end{array}\]
For all natural numbers \(n\), \(n < 2^n\).
Set \(\Phi(n)\) as: \(n < 2^n\), and observe:
\(\Phi(0)\)
\(0 < 1 = 2^0\).
If \(\Phi(n)\), then \(\Phi(n+1)\).
If \(n < 2^n\), then \(2^n \geq 1\) and \(n + 1 < 2^{n} + 2^{n} = 2^{n+1}\).
By induction, we conclude that for all natural numbers \(n\), \(n < 2^n\).
Given a condition on positive integers, if
then all positive integers satisfy the condition.
\[\begin{array}{lllll} 1 + 3& = & 4 & = & 2^2\\ 1 + 3 + 5 & = & 9 & = & 3^2\\ 1 + 3+ 5 +7 & = & 16 & = & 4^2\\ \cdots & & \cdots & & \cdots \\ \end{array}\]
The sum of the first \(n\) odd numbers is \(n^2\).
Set \(\Phi(n)\) as: \(\sum_{k = 1}^{n}{2k - 1} = n^2\):
\(\Phi(1)\)
\(\sum_{k = 1}^{1} 2k - 1 = 1^2\).
If \(\Phi(n)\), then \(\Phi(n+1)\).
Suppose \(\sum_{k=1}^{n} 2k-1 = n^2\), and note:
\[\begin{array}{lll} \sum_{k=1}^{n+1} 2k-1 & = & (\sum_{k=1}^{n} 2k-1) + 2 (n+1) - 1\\ & = & n^2 + 2n +1\\ & = & (n+1)^2 \\ \end{array}\]For all natural number \(n\), if both \(A\) and \(B\) have each at most \(n\) members, then \(|A| = |B|\).
Set \(\Phi(n)\) be: If two sets \(A\) and \(B\) have each at most \(n\) members, then \(|A| = |B|\).
\(\Phi(0)\)
If \(A\) and \(B\) have at most \(0\) members, then they are both empty and \(|A| = |B|\).
If \(\Phi(n)\), then \(\Phi(n+1)\)
Suppose the condition holds for \(n\) and let \(A\) and \(B\) be two sets with at most \(n+1\) members. Let \(A^-\) and \(B^-\) be obtained from \(A\) and \(B\) by substracting one member from each. By inductive hypothesis, \(|A^-| = |B^-|\), whence \(|A| = |A^-| + 1 = |B^-| +1 = |B|\).
By induction, if two sets \(A\) and \(B\) have each at most \(n\) members, then \(|A| = |B|\)
Caution
It would seem to follow that all finite sets have the same cardinality, which is false.
Where is the flaw in the argument?
Complete Induction
Given a condition on natural numbers, if
then all natural numbers satisfy the condition.
Complete Induction without a Base Case
Given a condition \(\Phi(n)\) on natural numbers, if
if for all \(m < n\), \(m\) meets the condition, then \(n\) does,
then all natural numbers satisfy the condition.
We define by recursion a set of formulas for a formal language. Suppose we start with a simple vocabulary, which consists of only three symbols:
\[@, \#, \%\]
We could define a set of formulas for that language by means of the following recursion:
\[\varphi ::= @ \ | \ \% \varphi \ | \ (\varphi \ \# \ \psi)\]
Here is how to read that notation now.
\(@\) is a formula of the language
if \(\varphi\) is a formula of the language, then \(\% \varphi\) is a formula
if \(\varphi\) and \(\psi\) are formulas of the language, then \((\varphi \ \# \ \psi)\) is a formula.
That means that the set of formulas of the language includes:
\[@, \%@, \%\% @, \dots, (@ \ \# \ @), (@ \ \# \% @), \% (@ \ \# \ @), \dots\]
Choose a stock of propositional variables or atoms:
\[ AT := p, q, r, \cdots \]
We define the formulas of a propositional language \(\mathcal{L}\) recursively:
\[ \varphi ::= AT \ | \ \neg \varphi \ | \ (\varphi \to \psi) \]
Here is how to read the last line:
All atoms are formulas.
If \(\varphi\) is a formula, then \(\neg \varphi\) is a formula.
If \(\varphi\) and \(\psi\) are formulas, then \((\varphi \to \psi)\) is a formula.
Nothing else is a formula.
\[ \begin{array}{lll} \top & := & (p \to p)\\ \bot & := & \neg \top \\ (\varphi \vee \psi) & := & (\neg \varphi \to \psi)\\ (\varphi \wedge \psi) & := & \neg (\varphi \to \neg \psi)\\ (\varphi \leftrightarrow \psi) & := & (\varphi \to \psi) \wedge (\psi \to \varphi) \end{array} \]
Given a condition on formulas of \(\mathcal{L}\), if
all atoms meet the condition,
whenever a formula \(\varphi\) meets the condition, \(\neg \varphi\) meets the condition as well, and
whenever two formulas \(\varphi\) and \(\psi\) meet the condition, \((\varphi \to \psi)\) meets the condition as well,
then all formulas of \(\mathcal{L}\) meet the condition.
Call a formula balanced iff it has the same number of right and left parentheses.
\[\begin{array}{ll} p & \text{is balanced} \\ \neg (p \to q) & \text{is balanced} \\ (p \to (\neg p \to (q \to r))) & \text{is balanced} \\ \cdots & \cdots \\ \end{array}\]
All formulas are balanced.
Set \(\Phi(\varphi)\) as: \(\varphi\) is balanced:
All atoms are balanced.
Atoms have an equal number of left and right parentheses, namely \(0\).
If \(\varphi\) is balanced, then \(\neg \varphi\) is balanced.
Suppose \(\varphi\) is balanced. \(\neg \varphi\) contains the same number of left and right parentheses as \(\varphi\). So, \(\neg \varphi\) is balanced.
If \(\varphi\) and \(\psi\) are each balanced, then \((\varphi \to \psi)\) is balanced.
If \(n\) and \(m\) are the number of left and right parentheses for \(\varphi\) and \(\psi\) respectively, then \(n+m+1\) is the number of left and right parentheses for \((\varphi \to \psi)\). So, \((\varphi \to \psi)\) is balanced.
By induction on the complexity of formulas, all formulas are balanced.
An assignment for a propositional language \(\mathcal{L}\) is a function \(v\) from \(AT\) into \(\{0, 1\}\), which maps a propositional variable to a truth value.
A valuation \(V\) based on an assignment \(v\) is a function from formulas into \(\{0, 1\}\) such that: \[ \begin{array}{lll} V(p) & = & v(p)\\ & & \\ V(\neg \varphi) & = & \begin{cases}1 \ \ \ \text{if} \ V(\varphi) =0\\ 0 \ \ \ \text{if} \ V(\varphi) = 1 \end{cases}\\ & & \\ V(\varphi \to \psi) & = & \begin{cases}1 \ \ \ \text{if} \ V(\varphi) =0 \ \text{or} \ V(\psi) = 1\\ 0 \ \ \ \text{if} \ V(\varphi) = 1 \ \text{and} \ V(\psi) = 0\end{cases}\\ \end{array} \]
Let \(v\) map all atoms into \(0\). Then: \[ \begin{array}{lll} V(p) & = & 0\\ & & \\ V(\neg p) & = & 1\\ & & \\ V(\neg p \to \neg \neg q) & = & 0\\ & & \\ V(\neg \neg q \to \neg \neg p) & = & 1\\ \end{array} \]
A valuation \(V\) satisfies a formula \(\varphi\) if, and only if, \(V(\varphi) = 1\).
A set of formulas \(\Gamma\) is satisfiable if, and only if, some valuation \(V\) satisfies all formulas in \(\Gamma\).
A formula \(\varphi\) is valid if, and only if, every valuation \(V\) satisfies \(\varphi\).
Given a set of formulas \(\Gamma\), a formula \(\varphi\) is a logical consequence of \(\Gamma\), written \(\Gamma \models \varphi\) if, and only if, for every valuation \(V\), \(V\) satisfies all formulas in \(\Gamma\) only if \(V\) satisfies \(\varphi\).