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 nonempty 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 (= 0arity 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
Preboolean algebras
 Preboolean 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 preBoolean 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 preBoolean to preHeyting 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 preHeyting algebra (intuitionistic).
Key example of a preHeyting algebra that is not a preBoolean
Standard topology over $ℝ$:
 instead of $𝒫(ℝ)$, take the open sets
 preorder: $⊆$
Negation is taking the interior of the complement.
So the excluded middle is not valid in this preHeyting 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 preHeyting 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 preorder into an order). But here we don’t care, as we work with preorders.
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 preHeyting algebra is non trivial
 Superconsistent theory:

iff it has a model valued in every preHeyting algebra
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