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

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.