Propositional logic

Definition 2..1 (Primitive propositions)  

$ P = \{ p_1, p_2, p_3, \cdots \}$ is the set of primitive propositions.

Definition 2..2 (Proposition)  

A proposition is a subset of strings of symbols from the alphabet $ (,),\Rightarrow ,\bot ,p_1,p_2,\cdots$.

Definition 2..3 (Language)  

The language $ L$ (or $ L(P)$) is the set of propositions

  1. $ P \subset L$
  2. $ \bot \in L$
  3. if $ p,q \in L$, $ (p \Rightarrow q) \in L$

$ L_n$ is the set of propositions of length $ \le n$.

Definition 2..4 (Shorthands)  

$ \neg p $ $ (p \Rightarrow \bot )$ ``not $ p$''
$ p \vee q$ $ ((\neg p) \Rightarrow q)$ ``p or q''
$ p \wedge q$ $ \neg (p \Rightarrow (\neg q))$ ``p and q''

Definition 2..5 (Valuation)  

A valuation is a function. $ v: L \mapsto \{ 0, 1 \}$.

  1. $ v(\bot )=0$
  2. $ v(p \Rightarrow q) = \begin{cases}0&\text{if $v(p)=1,v(q)=0$}\\
1&\text{otherwise}
\end{cases}$

Theorem 2..6 (Valuations agreeing on the basic propositions are the same)  

If valuations $ v,v'$ have $ v(p) = v'(p)$ for all $ p \in P$ then $ v
\cong v'$.

Proof. $ v = v'$ on $ L_1$. If $ v(p) = v'(p)$ and $ v(q) = v'(q)$ then $ v(p
\Rightarrow q) = v'(p \Rightarrow q)$ so by induction $ \forall L_n$.

$ \qedsymbol$

Theorem 2..7 (A function defined on the primitive propositions can be extended to a valuation)  

Given $ w: P \mapsto \{0,1\}$ there exists a valuation $ v$ such that $ v(p) = w(p)$ for all $ p \in P$.

Proof. Let $ v(p) = w(p)$ for all $ p \in P$ and let $ v(\bot )=0$. Extend.

$ \qedsymbol$

Definition 2..8 (Model)  

If $ v(p)=1$, $ p$ is true in $ v$. $ v$ is a model of $ p$.

Definition 2..9 (Tautology)   If $ v(p)=1$ for all valuations $ v$ then $ p$ is a tautology. $ \models p$.

Observation 2..10 (Techniques for proving something is a tautology)  

Draw a truth table, also note that if $ v(p \Rightarrow q) = 0$ then $ p = 1$ and $ q = 0$.

Definition 2..11 (Semantic implication, entails)  

If $ S \subseteq L$, $ t \in L$ and $ v(s)=1$ for all $ s \in S$ means $ v(t) = 1$ then $ S$ entails or semantically implies $ t$ ( $ S \models
t$). That is, every model of $ S$ is a model of $ t$.

John Fremlin 2010-02-17