Lecture 2: Models

Teacher: Gilles Dowek

$\lbrace 0, 1 \rbrace$-valued Models

The algebra $\lbrace 0, 1 \rbrace$:
  • $ℬ \, ≝ \, \lbrace 0, 1 \rbrace$
  • $≤$: “natural” order on this set: $0 ≤ 1$
  • $\widetilde{⊤} ≝ 1, \, \widetilde{⊥} ≝ 0$

In a model:

  • for each sort $s$, you have a non-empty set $ℳ_s$
  • for each function $f: s_1 × ⋯ × s_n ⟶ s’$, there is a function $\hat f: ℳ_{s_1} × ⋯ × ℳ_{s_n} ⟶ ℳ_{s’}$
  • for each predicate $P$ on $s_1, …, s_n$, there is a function $\hat P: ℳ_{s_1} × ⋯ × ℳ_{s_n} ⟶ ℬ$
Interpretation $⟦\bullet⟧$:
  • $⟦f(t_1, …, t_n)⟧ = \hat f (⟦t_1⟧, …, ⟦t_n⟧)$
  • $⟦P(t_1, …, t_n)⟧ = \hat P (⟦t_1⟧, …, ⟦t_n⟧)$
  • $⟦A ∧ B⟧ = ⟦A⟧ \tilde ∧ ⟦B⟧$

⟶ completely defined by its image on the variables (i.e. for each valuation, you have such an interpretation): so there are $\vert ℳ \vert^{\vert Var \vert}$ of them

Interpretation of propositions: for example,

\[⟦∀x \, (P(x) ∧ P(y) ⇒ P(x))⟧_{y = 7} = 1\]

if and only if for all $a ∈ ℳ$:

\[⟦P(x) ∧ P(y) ⇒ P(x)⟧_{y=7, \, x=a} = 1\]

i.e.

\[\lbrace ⟦P(x) ∧ P(y) ⇒ P(x)⟧_{y=7, \, x=a} \, | \, a ∈ ℳ\rbrace = \lbrace 1 \rbrace\]

Similarly:

\[⟦∃x \, (P(x) ∧ P(y) ⇒ P(x))⟧_{y = 7} = 1\]

if and only if there exists $a ∈ ℳ$ such that:

\[⟦P(x) ∧ P(y) ⇒ P(x)⟧_{y=7, \, x=a} = 1\]

i.e.

\[\lbrace ⟦P(x) ∧ P(y) ⇒ P(x)⟧_{y=7, \, x=a} \, | \, a ∈ ℳ\rbrace \ni 1\]
Soundness:

If the proposition $A$ has a classical proof in a theory, then it is valid in all models of this theory.

Completeness (Gödel, 1930):

If the proposition $A$ is valid in all models of a theory, then it has a classical proof in this theory.

NB: for constructive proofs/intuitionistic logic: we have soundness, but not completeness (ex: the excluded middle is valid in every model but not provable).

Contrapositive of soundness: example:

  • you have

    • two proposition (= 0-arity predicates) symbols $P$ and $Q$
    • a single axiom $P$: $𝒯 \, ≝ \, \lbrace P \rbrace$

  • then any model in which $\hat P = 1 ∈ ℬ$ is a model of $𝒯$, and:

    • $Q$ is not provable: consider the model where $\hat P = 1 ∈ ℬ, \, \hat Q = 0 ∈ ℬ$

    • $¬Q$ is not provable either: consider the model where $\hat P = 1 ∈ ℬ, \, \hat Q = 1 ∈ ℬ$

$Q$ is said to be independent of a theory $𝒯$:

if neither $Q$ nor $¬ Q$ are provable in this theory.

NB: Problem: so to prove that a formula is independent of a theory, naively: you need to provide at least two models (one in which $Q$ is valid, the other one where it is not): we’d like to provide only one model.

Model in an arbitrary algebra

Algebra

Notations:

  • greatest lower bound (glb) denoted by $\bigwedge$
  • lowest upper bound (lub) denoted by $\bigvee$
Algebra:
  • a set $ℬ$
  • a reflexive and transitive (= preorder) relation $≤$ on $ℬ$
  • two elements $\tilde ⊤$ and $\tilde ⊥$ of $ℬ$
  • three functions $\tilde ∧, \tilde ∨, \, \tilde ⇒$ from $ℬ × ℬ$ to $ℬ$, where
  • a subset $\cal A$ of $𝒫^+(B)$, a function $\tilde ∀: \cal A ⟶ ℬ$
  • a subset $\cal E$ of $𝒫^+(B)$, $\tilde ∃: \cal E ⟶ ℬ$ where

Pre-boolean algebras

Pre-boolean algebra:
  • $a \tilde ∧ b ≝ \bigwedge \lbrace a, b \rbrace$
  • $a \tilde ∨ b ≝ \bigvee \lbrace a, b \rbrace$
  • $\tilde ∀ A ≝ \bigwedge A$
  • $\tilde ∃ A ≝ \bigvee A$
  • $a ≤ b \tilde ⇒ c$ if and only if $a \tilde ∧ b ≤ c$
    • this is the introduction rule of $⇒$ (categorically: $\bullet ∧ a$ left adjoint to $a ⇒ \bullet$)
  • $\tilde ⊤≤ (a ∨ \tilde (a \tilde ⇒ b))$

NB: if $≤$ were antisymmetric (hence an order), we would have a boolean algebra.

Examples of pre-Boolean algebra: $\lbrace 0, 1 \rbrace, \, 𝒫(\lbrace 3, 4 \rbrace), \lbrace 0 \rbrace$, …

Example of preorder used: $A ≤ B \text{ iff } ⊢ A ⇒ B$

Validity: $A$ is valid iff $⟦A⟧ ≥ \tilde ⊤$

Coming back to our previous example: now, let’s set $ℬ ≝ 𝒫(\lbrace 3, 4 \rbrace)$.

  • $\hat P = \lbrace 3, 4 \rbrace$
  • $\hat Q = \lbrace 4 \rbrace$, so $\hat {¬ Q} = \lbrace 3 \rbrace$

so neither $Q$ nor $¬Q$ is valid: that model aggregates the two previous models

Now: completeness is easier to prove now, but soundness is harder (as we have more models now).

Sketch of soundness for $∧$: by induction on the formula structure, and then: $⟦A⟧ ≥ \tilde ⊤$ and $⟦B⟧ ≥ \tilde ⊤$ iff $⟦A ∧ B⟧ ≥ \tilde ⊤$.

From pre-Boolean to pre-Heyting algebras

The condition $\tilde ⊤≤ (a ∨ \tilde (a \tilde ⇒ b))$ is only there to make the excluded middle valid in Boolean algebras: if you drop it, you get a pre-Heyting algebra (intuitionistic).

Key example of a pre-Heyting algebra that is not a pre-Boolean

Standard topology over $ℝ$:

  • instead of $𝒫(ℝ)$, take the open sets
  • pre-order: $⊆$

Negation is taking the interior of the complement.

So the excluded middle is not valid in this pre-Heyting algebra: the \(\hat P ∨ \tilde ¬ \hat P = (−∞, 0) ∪ (0, +∞) = ℝ \backslash \lbrace 0 \rbrace ≠ ℝ\)

Soundness and Completeness

Soundness: If $Γ ⊢ A$ has a constructive proof, then it is valid in all pre-Heyting models (induction over proof structure).

Completeness (weaker theorem than before): very unusual proof. Usually, we saturate our theory, and then build a model satisfying it.

But here, we don’t need that (as for propositions not appearing in the theory: neither themselves nor their negation may be valid): Lindenbaum model (where the algebra is the set of propositions).

Actually: usually you take the set of propositions quotiented by $A \sim B$ iff $A \Leftrightarrow B$ is provable (to turn the pre-order into an order). But here we don’t care, as we work with pre-orders.

Deduction modulo

Congruence relation $≡$ valid in $ℳ$:

if for all $A, B$ s.t. $A ≡ B$: for all $φ$, $⟦A⟧_φ = ⟦B⟧_φ$

$𝒯,≡$ is valid in $ℳ$ if all axioms of $𝒯$ and $≡$ are valid in $ℳ$

  • if $A⇔B$ provable in $𝒯,≡$: then $⟦A⟧_φ ≤ ⟦B⟧_φ$ and $⟦A⟧_φ ≥ ⟦B⟧_φ$ ⟶ they are provably equal
  • if $A ≡ B$ provable in $𝒯,≡$: then $⟦A⟧_φ = ⟦B⟧_φ$ ⟶ they are computationally equal
Consistency of a theory $𝒯$:

iff $𝒯$ has a model

$𝒯 , ≡$ is consistent:

iff it has a model whose pre-Heyting algebra is non trivial

Super-consistent theory:

iff it has a model valued in every pre-Heyting algebra

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