Phase semantics

Introduction

The semantics given by phase spaces is a kind of "formula and provability semantics", and is thus quite different in spirit from the more usual denotational semantics of linear logic. (Those are rather some "formulas and proofs semantics".)

 --- probably a whole lot more of blabla to put here... ---

Preliminaries: relation and closure operators

Part of the structure obtained from phase semantics works in a very general framework and relies solely on the notion of relation between two sets.

Relations and operators on subsets

The starting point of phase semantics is the notion of duality. The structure needed to talk about duality is very simple: one just needs a relation R between two sets X and Y. Using standard mathematical practice, we can write either $(a,b) \in R$ or $a\mathrel{R} b$ to say that $a\in X$ and $b\in Y$ are related.

Definition

If $R\subseteq X\times Y$ is a relation, we write $R^\sim\subseteq Y\times X$ for the converse relation: $(b,a)\in R^\sim$ iff $(a,b)\in R$.

Such a relation yields three interesting operators sending subsets of X to subsets of Y:

Definition

Let $R\subseteq X\times Y$ be a relation, define the operators $\langle R\rangle$, [R] and _^R taking subsets of X to subsets of Y as follows:

1. $b\in\langle R\rangle(x)$ iff $\exists a\in x,\ (a,b)\in R$
2. $b\in[R](x)$ iff $\forall a\in X,\ (a,b)\in R \implies a\in x$
3. $b\in x^R$ iff $\forall a\in x, (a,b)\in R$
   

The operator $\langle R\rangle$ is usually called the direct image of the relation, [R] is sometimes called the universal image of the relation.

It is trivial to check that $\langle R\rangle$ and [R] are covariant (increasing for the $\subseteq$ relation) while _^R is contravariant (decreasing for the $\subseteq$ relation). More interesting:

Lemma (Galois Connections)

1. $\langle R\rangle$ is right-adjoint to [R^˜]: for any $x\subseteq X$ and $y\subseteq Y$, we have $[R^\sim]y \subseteq x$ iff $y\subseteq \langle R\rangle(x)$
2. we have $y\subseteq x^R$ iff $x\subseteq y^{R^\sim}$
   

This implies directly that $\langle R\rangle$ commutes with arbitrary unions and [R] commutes with arbitrary intersections. (And in fact, any operator commuting with arbitrary unions (resp. intersections) is of the form $\langle R\rangle$ (resp. [R]).

Remark: the operator _^R sends unions to intersections because $\_^R : \mathcal{P}(X) \to \mathcal{P}(Y)^\mathrm{op}$ is right adjoint to $\_^{R^\sim} : \mathcal{P}(Y)^{\mathrm{op}} \to \mathcal{P}(X)$...

Closure operators

Definition

A closure operator on $\mathcal{P}(X)$ is a monotonic operator P on the subsets of X which satisfies:

1. for all $x\subseteq X$, we have $x\subseteq P(x)$
2. for all $x\subseteq X$, we have $P(P(x))\subseteq P(x)$
   

Closure operators are quite common in mathematics and computer science. They correspond exactly to the notion of monad on a preorder...

It follows directly from the definition that for any closure operator P, the image P(x) is a fixed point of P. Moreover:

Lemma

P(x) is the smallest fixed point of P containing x.

One other important property is the following:

Lemma

Write $\mathcal{F}(P) = \{x\ |\ P(x)\subseteq x\}$ for the collection of fixed points of a closure operator P. We have that $\left(\mathcal{F}(P),\bigcap\right)$ is a complete inf-lattice.

Remark: A closure operator is in fact determined by its set of fixed points: we have $P(x) = \bigcup \{ y\ |\ y\in\mathcal{F}(P),\,y\subseteq x\}$

Since any complete inf-lattice is automatically a complete sup-lattice, $\mathcal{F}(P)$ is also a complete sup-lattice. However, the sup operation isn't given by plain union:

Lemma

If P is a closure operator on $\mathcal{P}(X)$, and if $(x_i)_{i\in I}$ is a (possibly infinite) family of subsets of X, we write $\bigvee_{i\in I} x_i = P\left(\bigcup_{i\in I} x_i\right)$.

We have $\left(\mathcal{F}(P),\bigcap,\bigvee\right)$ is a complete lattice.

Proof.  easy.

A rather direct consequence of the Galois connections of the previous section is:

Lemma

The operator and $\langle R\rangle \circ [R^\sim]$ and the operator $x\mapsto {x^R}^{R^\sim}$ are closures.

A last trivial lemma:

Lemma

We have $x^R = {{x^R}^{R^\sim}}^{R}$.

As a consequence, a subset $x\subseteq X$ is in $\mathcal{F}({\_^R}^{R^\sim})$ iff it is of the form $y^{R^\sim}$.

Remark: everything gets a little simpler when R is a symmetric relation on X.

Phase Semantics

Phase spaces

Definition (monoid)

A monoid is simply a set X equipped with a binary operation $\_\cdot\_$ s.t.:

1. the operation is associative
2. there is a neutral element $1\in X$

The monoid is commutative when the binary operation is commutative.

Definition (Phase space)

A phase space is given by:

1. a commutative monoid $(X,1,\cdot)$,
2. together with a subset $\Bot\subseteq X$.

The elements of X are called phases.

We write $\bot$ for the relation $\{(a,b)\ |\ a\cdot b \in \Bot\}$. This relation is symmetric.

A fact in a phase space is simply a fixed point for the closure operator $x\mapsto x\biorth$.

Thanks to the preliminary work, we have:

Corollary

The set of facts of a phase space is a complete lattice where:

1. $\bigwedge_{i\in I} x_i$ is simply $\bigcap_{i\in I} x_i$,
2. $\bigvee_{i\in I} x_i$ is $\left(\bigcup_{i\in I} x_i\right)\biorth$.
   

Additive connectives

The previous corollary makes the following definition correct:

Definition (additive connectives)

If $(X,1,\cdot,\Bot)$ is a phase space, we define the following facts and operations on facts:

1. $\top = X = \emptyset\orth$
2. $\zero = \emptyset\biorth = X\orth$
3. $x\with y = x\cap y$
4. $x\plus y = (x\cup y)\biorth$
   

Once again, the next lemma follows from previous observations:

Lemma (additive de Morgan laws)

We have

1. $\zero\orth = \top$
2. $\top\orth = \zero$
3. $(x\with y)\orth = x\orth \plus y\orth$
4. $(x\plus y)\orth = x\orth \with y\orth$
   

Multiplicative connectives

In order to define the multiplicative connectives, we actually need to use the monoid structure of our phase space. One interpretation that is reminiscent in phase semantics is that our spaces are collections of tests / programs / proofs / strategies that can interact with each other. The result of the interaction between a and b is simply $a\cdot b$.

The set $\Bot$ can be thought of as the set of "good" things, and we thus have $a\in x\orth$ iff "a interacts correctly with all the elements of x".

Definition

If x and y are two subsets of a phase space, we write $x\cdot y$ for the set $\{a\cdot b\ |\ a\in x, b\in y\}$.

Thus $x\cdot y$ contains all the possible interactions between one element of x and one element of y.

The tensor connective of linear logic is now defined as:

Definition (multiplicative connectives)

If x and y are facts in a phase space, we define

• $\one = \{1\}\biorth$;
• $\bot = \one\orth$;
• the tensor $x\tens y$ to be the fact $(x\cdot y)\biorth$;
• the par connective is the de Morgan dual of the tensor: $x\parr y = (x\orth \tens y\orth)\orth$;
• the linear arrow is just $x\limp y = x\orth\parr y = (x\tens y\orth)\orth$.
  

Note that by unfolding the definition of $\limp$, we have the following, "intuitive" definition of $x\limp y$:

Lemma

If x and y are facts, we have $a\in x\limp y$ iff $\forall b\in x,\,a\cdot b\in y$

Proof.  easy exercise.

Readers familiar with realisability will appreciate...

Remark: Some people say that this idea of orthogonality was implicitly present in Tait's proof of strong normalisation. More recently, Jean-Louis Krivine and Alexandre Miquel have used the idea explicitly to do realisability...

Properties

All the expected properties hold:

Lemma

• The operations $\tens$, $\parr$, $\plus$ and $\with$ are commutative and associative,
• They have respectively $\one$, $\bot$, $\zero$ and $\top$ for neutral element,
• $\zero$ is absorbant for $\tens$,
• $\top$ is absorbant for $\parr$,
• $\tens$ distributes over $\plus$,
• $\parr$ distributes over $\with$.
  

Exponentials

Definition (Exponentials)

Write I for the set of idempotents of a phase space: $I=\{a\ |\ a\cdot a=a\}$. We put:

1. $\oc x = (x\cap I\cap \one)\biorth$,
2. $\wn x = (x\orth\cap I\cap\one)\orth$.
   

This definition captures precisely the intuition behind the exponentials:

• we need to have contraction, hence we restrict to indempotents in x,
• and weakening, hence we restrict to $\one$.

Since I isn't necessarily a fact, we then take the biorthogonal to get a fact...

Soundness

Definition

Let $(X, 1, \cdot)$ be a commutative monoid.

Given a formula A of linear logic and an assignation ρ that associate a fact to any variable, we can inductively define the interpretation $\sem{A}_\rho$ of A in X as one would expect. Interpretation is lifted to sequents as .

Theorem

Let Γ be a provable sequent in linear logic. Then $1_X \in \sem{\Gamma}$.

Proof. By induction on $\vdash\Gamma$.

Completeness

Phase semantics is complete w.r.t. linear logic. In order to prove this, we need to build a particular commutative monoid.

Definition

We define the syntactic monoid as follows:

• Its elements are sequents Γ quotiented by the equivalence relation $\cong$ generated by the rules:
  1. $\Gamma \cong \Delta$ if Γ is a permutation of Δ
  2. $\wn{A}, \wn{A}, \Gamma \cong \wn{A}, \Gamma$

• Product is concatenation: $\Gamma \cdot \Delta := \Gamma, \Delta$

• Neutral element is the empty sequent: $1 := \emptyset$.
  

The equivalence relation intuitively means that we do not care about the multiplicity of $\wn$-formulae.

Lemma

The syntactic monoid is indeed a commutative monoid.

Definition

The syntactic assignation is the assignation that sends any variable α to the fact $\{\alpha\}\orth$.

We instantiate the pole as $\Bot$$:= \{\Gamma \mid \vdash\Gamma\}$.

Theorem

If $\Gamma\in\sem{\Gamma}\orth$, then $\vdash\Gamma$.

Proof. By induction on Γ.

Cut elimination

Actually, the completeness result is stronger, as the proof does not use the cut-rule in the reconstruction of $\vdash\Gamma$. By refining the pole as the set of cut-free provable formulae, we get:

Theorem

If $\Gamma\in\sem{\Gamma}\orth$, then Γ is cut-free provable.

From soundness, one can retrieve the cut-elimination theorem.

Corollary

Linear logic enjoys the cut-elimination property.