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

# Lecture 40: Soundness and completeness

## Soundness and completeness

In the last two lectures, we have looked at propositional formulas from two perspectives: truth and provability. Our goal now is to (meta) prove that the two interpretations match each other. We will prove:

1. Soundness: if something is provable, it is valid. If $$⊢ φ$$ then $$⊨ φ$$.
2. Completeness: if something is valid, it is provable. If $$⊨ φ$$ then $$⊢ φ$$.

In this lecture, we will outline proofs for both of these facts for the propositional logic we have been developing.

Note that this is analogous to Kleene's theorem: there we examined language from two different perspectives (recognizability and regularity) and then proved that they gave the same answers.

## Proof of Soundness

To show that our proof system is sound, we prove something stronger: if $$φ_1, φ_2, \dots ⊢ ψ$$ then $$φ_1, φ_2, \dots ⊨ ψ$$.

Assume $$φ_1, φ_2, \dots ⊢ ψ$$, so that there exists a proof tree $$T$$ terminating with this line. Note that proof trees are inductively defined structures, so we can actually do a meta-inductive proof on the structure of the object proof.

Let $$T$$ be a proof tree, and let $$P(T)$$ say "if $$T$$ is a complete proof tree showing that $$φ_1, φ_2, \dots ⊢ ψ$$, then $$φ_1, φ_2, \dots ⊨ ψ$$. To prove $$∀T, P(T)$$, we will consider trees that end with each of the possible rules. If the proof tree has subtrees $$T_1, T_2, \dots$$, we will inductively assume $$P(T_1), P(T_2), \dots$$.

For example, if the rule at the root of the tree is the and introduction rule:

$\infer[(∧ intro)]{\cdots ⊢ φ ∧ ψ}{ \cdots ⊢ φ & \cdots ⊢ ψ }$

then there are valid proof subtrees ending in $$\cdots ⊢ φ$$ and $$\cdots ⊢ ψ$$, so we will inductively assume that $$\cdots ⊨ φ$$ and $$\cdots ⊨ ψ$$.

We examine a few of the rules; the remaining cases are left as review exercises. We refer to the list of rules here

1. $$P(T)$$ where $$T$$ ends with law of excluded middle to show $$\cdots ⊢ φ ∨ ¬φ$$. We wish to show that in any $$I$$ satisfying the assumptions, $$I ⊨ φ ∨ ¬φ$$. Well, $$φ[I]$$ is either T or F. If it is T, then the rule for evaluating $$∨$$ says that $$φ∨¬φ$$ evaluates to T, while if it is $$F$$, then $$¬φ$$ evaluates to true, and again we see that $$φ∨¬φ$$ evaluates to T.

2. $$P(T)$$ where $$T$$ ends with reductio ad absurdum:

$\infer[(absurd)]{A ⊢ ψ}{A ⊢ φ & A ⊢ ¬φ}$

We wish to show that in any interpretation $$I$$ satisfying the assumptions $$A$$, that $$I ⊨ ψ$$. Inductively, we assume that $$I ⊨ φ$$ and $$I ⊨ ¬φ$$. But this is impossible, because $$φ$$ either evaluates to T or F in $$I$$. So the conclusion for all $$I$$ satisfying $$A$$, $$I ⊨ ψ$$ is vacuously true: there are no interpretations satisfying $$A$$.

1. $$P(T)$$ where $$T$$ ends with $$∧$$ introduction:

$\infer[(∧ intro)]{A ⊢ φ ∧ ψ}{A ⊢ φ & A ⊢ ψ}$

We wish to show that in any interpretation $$I$$ satisfying $$A$$, that $$I ⊨ φ∧ψ$$. We inductively assume $$I ⊨ φ$$ and $$I ⊨ ψ$$. The rules for evaluating $$φ∧ψ[I]$$ immediately show that $$I ⊨ φ∧ψ$$ as required.

The remaining cases are left as review exercises.

## Completeness

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.

1. $$P(Q)$$. 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.

2. $$P(¬φ)$$. 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:

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

1. $$P(φ∧ψ)$$. We inductively assume $$P(φ)$$ and $$P(ψ)$$. If $$I ⊨ φ∧ψ$$ then we see that $$I ⊨ φ$$ and $$I ⊨ ψ$$, 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 induction, $$I ⊢ ¬φ$$. The proof tree that $$I ⊢ ¬(φ∧ψ)$$ is in the homework.

The remaining cases are left as review exercises.