$\newcommand\infer[3][]{ \begin{array}[b]{c c c c} \style{border-bottom:1px solid;}{ \begin{array}[b]{c c c c} #3 \\ \end{array} } & \hspace{-1em}\raise{-0.5em}{\text{#1}} \\ #2 \end{array} }$

Lecture 39a: completeness

In this part of the lecture, we outlined the proof of completentess of our proof rules for propositional logic. Recall the definition of completeness: a proof system is complete if whenever $\phi_1, \dots, \phi_n ⊨ \psi$ , we have $\phi_1,\dots,\phi_n ⊢ φ$ .

To show completeness, we assume that $⊨ φ$ ; we want to build a valid proof tree for $⊢φ$ .

To do this, we introduce a new notation. If $I$ is an interpretation, we write $I ⊢ φ$ as shorthand for $ψ_P, ψ_Q, \dots ⊢ φ$ where $ψ_P$ is just $P$ if $I(P) = T$ and is $¬P$ if $I(P) = F$ , and similarly for all of the other base propositions in the formula $φ$ .

Our approach to constructing the proof tree for $⊢ φ$ will be to give a proof of $I ⊢ φ$ for each interpretation $I$ , and then repeatedly use law of excluded middle and $∨$ elimination to snap these separate proofs together. For example, if the variables of $φ$ are $P$ , $Q$ , and $R$ , we will build proofs of $P,Q,R ⊢ φ$ , and $P, Q, ¬R ⊢ φ$ , and $P,¬Q,R ⊢ φ$ , etc., and then we will combine them as follows:

$\infer[($∨$ elim)]{⊢ φ}{ \infer[excl. mid.]{⊢ P ∨ ¬P}{\hspace{1in}} & \infer[($∨$ elim)]{P⊢ φ}{ \infer{⊢ Q ∨ ¬Q}{\hspace{1in}} & \infer{P,Q⊢φ}{ \infer{⊢ R ∨ ¬R}{\hspace{1in}} & \infer{P,Q,R ⊢ φ}{constructed~below} & \infer{P,Q,¬R ⊢ φ}{constructed~below} } & \infer{P,¬Q⊢φ}{\hspace{.4in}\vdots\hspace{.4in}} } & \infer[($∨$ elim)]{¬P⊢ φ}{\hspace{.4in}\vdots\hspace{.4in}} }$

Therefore, our goal is to show that if $I ⊨ φ$ then $I ⊢ φ$ . We will prove this by induction on the structure of $φ$ . We need a stronger inductive hypothesis: Let $P(φ)$ be the statement "if $I ⊨ φ$ then $I ⊢ φ$ and if $I \not ⊨ φ$ then $I ⊢ ¬φ$ ." We will show $P(Q)$ , $P(¬φ)$ , $P(φ∧ψ)$ , etc. As above, we will leave some of these as review exercises.

$P(Q)$ . Assume $I ⊨ Q$ . This means that $Q[I] = T$ , or in other words, $I(Q) = T$ . This means that $Q$ is one of the assumptions in $I ⊢ Q$ , so we can use the assumption rule $\cdots,Q ⊢ Q$ . On the other hand, if $I \not⊨ Q$ , then $¬Q$ is one of the assumptions in $I ⊢ Q$ , and again, we can use the assumption rule.
$P(¬φ)$ . We inductively assume $P(φ)$ . If $I ⊨ ¬φ$ , then by the way $⊨$ is defined, we see that $I \not⊨ φ$ . Therefore, by $P(φ)$ , we see that $I ⊢ ¬φ$ as required. If, on the other hand, $I \not ⊨ ¬φ$ , then we know that $I ⊨ φ$ . Therefore $I ⊢ φ$ . We want to show $I ⊢ ¬¬φ$ . We can build the following proof:

$\infer[($∨$ elim)]{I ⊢ ¬¬φ}{ \infer[(excl. mid.)]{I ⊢ (¬φ) ∨ (¬¬φ)}{\hspace{1in}} & \infer[(absurd)]{I, ¬φ ⊢ ¬¬φ}{ I ⊢ φ & \infer[(assum)]{I,¬φ⊢¬φ}{\hspace{1in}} } & \infer[(assum)]{I, ¬¬φ ⊢ ¬¬φ}{\hspace{1in}} }$

$P(φ∧ψ)$ . We inductively assume $P(φ)$ and $P(ψ)$ . If $I ⊨ φ∧ψ$ then we see that $I ⊨ φ$ and $I ⊨ ψ$ (using the definition of $eval$ ), so by induction we have $I ⊢ φ$ and $I ⊢ ψ$ . We can combine these with $∧$ introduction to form a proof of $I ⊢ φ∧ψ$ . If, on the other hand, $I \not ⊨ φ∧ψ$ then either $I \not⊨ φ$ or $I \not⊨ ψ$ . Without loss of generality, assume that $I \not⊨ φ$ . By $P(φ)$ , we have $I ⊢ ¬φ$ . The proof tree that $I ⊢ ¬(φ∧ψ)$ is similar to a homework exercise.

The remaining cases are similar; we examine the definition of $eval$ , and then apply the inductive hypothesis to get pieces of a proof tree, which we then assemble using a proof rule.