## Thursday, August 09, 2018

### Intersection Types, Sub-formula Property, and the Functional Character of the Lambda Calculus

Intersection Types, Sub-formula Property, and the Functional Character of the Lambda Calculus

Last December I proved that my graph model of the lambda calculus, once suitable restricted, is deterministic. That is, I defined a notion of consistency between values, written $$v_1 \sim v_2$$, and showed that any two outputs of the same program are consistent.
Theorem (Determinism)
If $$v \in {\mathcal{E}{[\![ e ]\!]}}\rho$$, $$v' \in {\mathcal{E}{[\![ e ]\!]}}\rho'$$, and $$\rho \sim \rho'$$, then $$v \sim v'$$.
Recall that values are integers or finite relations; consistency for integers is equality and consistency for relations means mapping consistent inputs to consistent outputs. I then restricted values to be well formed, meaning that they must be consistent with themselves (and similarly for their parts).

Having proved the Determinism Theorem, I thought it would be straightforward to prove the following related theorem about the join of two values.
Theorem (Join)
If $$v \in {\mathcal{E}{[\![ e ]\!]}}\rho$$, $$v' \in {\mathcal{E}{[\![ e ]\!]}}\rho'$$, $$\rho$$ is well formed, $$\rho'$$ is well formed, and $$\rho \sim \rho'$$,
then $$v \sqcup v' \in {\mathcal{E}{[\![ e ]\!]}}(\rho\sqcup\rho')$$.
I am particularly interested in this theorem because $$\beta$$-equality can be obtained as a corollary. ${\mathcal{E}{[\![ ({\lambda x.\,}e){\;}e' ]\!]}}\rho = {\mathcal{E}{[\![ [x{:=}e']e ]\!]}}\rho$ This would enable the modeling of the call-by-name $$\lambda$$-calculus and it would also enable the use of $$\beta$$-equality in a call-by-value setting when $$e'$$ is terminating (instead of restricting $$e'$$ to be a syntactic value).

Recall that we have defined a partial order $$\sqsubseteq$$ on values, and that, in most partial orders, there is a close connection between notions of consistency and least upper bounds (joins). One typically has that $$v \sim v'$$ iff $$v \sqcup v'$$ exists. So my thinking was that it should be easy to adapt my proof of the Determinism Theorem to prove the Join Theorem, and I set out hoping to finish in a couple weeks. Hah! Here we are 8 months later and the proof is complete; it was a long journey that ended up depending on a result that was published just this summer, concerning intersection types, the sub-formula property, and cut elimination by Olivier Laurent. In this blog post I’ll try to recount the journey and describe the proof, hopefully remembering the challenges and motivations. Here is a tar ball of the mechanization in Isabelle and in pdf form.

Many of the challenges revolved around the definitions of $$\sqsubseteq$$, consistency, and $$\sqcup$$. Given that I already had definitions for $$\sqsubseteq$$ and consistency, the obvious thing to try was to define $$\sqcup$$ such that it would be the least upper bound of $$\sqsubseteq$$. So I arrived at this partial function: \begin{aligned} n \sqcup n &= n \\ f_1 \sqcup f_2 &= f_1 \cup f_2\end{aligned} Now suppose we prove the Join Theorem by induction on $$e$$ and consider the case for application: $$e = (e_1 {\;}e_2)$$. From $$v \in {\mathcal{E}{[\![ e_1 {\;}e_2 ]\!]}}$$ and $$v' \in {\mathcal{E}{[\![ e_1 {\;}e_2 ]\!]}}$$ we have

• $$f \in {\mathcal{E}{[\![ e_1 ]\!]}}\rho$$, $$v_2 \in {\mathcal{E}{[\![ e_2 ]\!]}}\rho$$, $$v_3 \mapsto v_4 \in f$$, $$v_3 \sqsubseteq v_2$$, and $$v \sqsubseteq v_4$$ for some $$f, v_2, v_3, v_4$$.

• $$f' \in {\mathcal{E}{[\![ e_2 ]\!]}}\rho'$$, $$v'_2 \in {\mathcal{E}{[\![ e_2 ]\!]}}\rho'$$, $$v'_3 \mapsto v'_4 \in f$$, $$v'_3 \sqsubseteq v'_2$$, and $$v' \sqsubseteq v'_4$$ for some $$f', v'_2, v'_3, v'_4$$.

By the induction hypothesis we have $$f \sqcup f' \in {\mathcal{E}{[\![ e_1 ]\!]}}$$ and $$v_2 \sqcup v'_2 \in {\mathcal{E}{[\![ e_2 ]\!]}}$$. We need to show that $v''_3 \mapsto v''_4 \in f \sqcup f' \qquad v''_3 \sqsubseteq v_2 \sqcup v'_2 \qquad v \sqcup v' \sqsubseteq v''_4$ But here we have a problem. Given our definition of $$\sqcup$$ in terms of set union, there won’t necessarily be a single entry in $$f \sqcup f'$$ that combines the information from both $$v_3 \mapsto v_4$$ and $$v'_3 \mapsto v'_4$$. After all, $$f \sqcup f'$$ contains all the entries of $$f$$ and all the entries of $$f'$$, but the set union operation does not mix together information from entries in $$f$$ and $$f'$$ to form new entries.

# Intersection Types to the Rescue

At this point I started thinking that my definitions of $$\sqsubseteq$$, consistency, and $$\sqcup$$ were too simple, and that I needed to incorporate ideas from the literature on filter models and intersection types. As I’ve written about previously, my graph model corresponds to a particular intersection type system, and perhaps a different intersection type system would do the job. Recall that the correspondence goes as follows: values correspond to types, $$\sqsubseteq$$ corresponds to subtyping $$<:$$ (in reverse), and $$\sqcup$$ corresponds to intersection $$\sqcap$$. The various intersection type systems primarily differ in their definitions of subtyping. Given the above proof attempt, I figured that I would need the usual co/contra-variant rule for function types and also the following rule for distributing intersections over function types. $(A\to B) \sqcap (A \to C) <: A \to (B \sqcap C)$ This distributivity rule enables the “mixing” of information from two different entries.

So I defined types as follows: $A,B,C,D ::= n \mid A \to B \mid A \sqcap B$ and defined subtyping according to the BCD intersection type system (Lambda Calculus with Types, Barendregt et al. 2013). $\begin{gathered} A <: A \qquad \frac{A <: B \quad B <: C}{A <: C} \\[2ex] A \sqcap B <: A \qquad A \sqcap B <: B \qquad \frac{C <: A \quad C <: B}{C <: A \sqcap B} \\[2ex] \frac{C <: A \quad B <: D}{A \to B <: C \to D} \qquad (A\to B) \sqcap (A \to C) <: A \to (B \sqcap C)\end{gathered}$ I then adapted the definition of consistency to work over types. (Because this definition uses negation, it is easier to define consistency as a recursive function in Isabelle instead of as an inductively defined relation.) \begin{aligned} n \sim n' &= (n = n') \\ n \sim (C \to D) &= \mathit{false} \\ n \sim (C \sqcap D) &= n \sim C \text{ and } n \sim D \\ (A \to B) \sim n' &= \mathit{false} \\ (A \to B) \sim (C \to D) &= (A \sim C \text{ and } B \sim D) \text{ or } A \not\sim C \\ (A \to B) \sim (C \sqcap D) &= (A \to B) \sim C \text{ and } (A \to B) \sim D \\ (A \sqcap B) \sim n' &= A \sim n' \text{ and } B \sim n' \\ (A \sqcap B) \sim (C \sqcap D) &= A \sim C \text{ and } A \sim D \text{ and } B \sim C \text{ and } B \sim D\end{aligned}

Turning back to the Join Theorem, I restated it in terms of the intersection type system and rebranded it the Meet Theorem. Instead of using the letter $$\rho$$ for environments, we shall switch to $$\Gamma$$ because they now contain types instead of values.
Theorem (Meet)
If $$\Gamma \vdash e : A$$, $$\Gamma' \vdash e : B$$, and $$\Gamma \sim \Gamma'$$, then $$\Gamma\sqcap\Gamma' \vdash e : A \sqcap B$$.
By restating the theorem in terms of intersection types, we have essentially arrived at the rule for intersection introduction. In other words, if we can prove this theorem we will have shown that the intersection introduction rule is admissible in our system.

While the switch to intersection types and subtyping enabled this top-level proof to go through, I got stuck on one of the lemmas that it requires, which is an adaptation of Proposition 3 of the prior blog post.
Lemma (Consistency and Subtyping)

1. If $$A \sim B$$, $$A <: C$$, and $$B <: D$$, then $$C \sim D$$.

2. If $$A \not\sim B$$, $$C <: A$$, $$D <: B$$, then $$C \not\sim D$$.

In particular, I got stuck in the cases where the subtyping $$A <: C$$ or $$B <: D$$ was derived using the transitivity rule.

# Subtyping and the Sub-formula Property

For a long time I’ve disliked definitions of subtyping in which transitivity is given as a rule instead of proved as a theorem. There are several reasons for this: a subtyping algorithm can’t directly implement a transitivity rule (or any rule that is not syntax directed), reasoning by induction or cases (inversion) is more difficult, and it is redundant. Furthermore, the presence of the transitivity rule means that subtyping does not satisfy the sub-formula property. This term sub-formula property comes from logic, and means that a derivation (proof) of a formula only mentions propositions that are a part of the formulate to be proved. The transitivity rule breaks this property because the type $$B$$ comes out of nowhere, it is not part of $$A$$ or $$C$$, the types in the conclusion of the rule.

So I removed the transitivity rule and tried to prove transitivity. For most type systems, proving the transitivity of subtyping is straightforward. But I soon realized that the addition of the distributivity rule makes it significantly more difficult. After trying and failing to prove transitivity for some time, I resorted to reading the literature. Unfortunately, it turns out that none of the published intersection type systems satisfied the sub-formula property and vast majority of them included the transitivity rule. However, there was one paper that offered some hope. In a 2012 article in Fundamenta Informaticae titled Intersection Types with Subtyping by Means of Cut Elimination, Olivier Laurent defined subtyping without transitivity and instead proved it, but his system still did not satisfy the sub-formula property because of an additional rule for function types. Nevertheless, Olivier indicated that he was interested in finding a version of the system that did, writing

“it would be much nicer and much more natural to go through a sub-formula property”

A lot of progress can happen in six years, so I sent an email to Olivier. He replied,

“Indeed! I now have two different sequent-calculus systems which are equivalent to BCD subtyping and satisfy the sub-formula property. I am currently writting a paper on this but it is not ready yet.”

and he attached the paper draft and the Coq mechanization. What great timing! Furthermore, Olivier would be presenting the paper, titled Intersection Subtyping with Constructors, at the Workshop on Intersection Types and Related System in Oxford on July 8, part of the Federated Logic Conference (FLOC). I was planning to attend FLOC anyways, for the DOMAINS workshop to celebrate Dana Scott’s 85th birthday.

Olivier’s systems makes two important changes compared to prior work: he combines the distributivity rule and the usual arrow rule into a single elegant rule, and to enable this, he generalizes the form of subtyping from $$A <: B$$ to $$A_1,\ldots,A_n \vdash B$$, which should be interpreted as meaning $$A_1 \sqcap \cdots \sqcap A_n <: B$$. Having a sequence of formulas (types) on the left is characteristic of proof systems in logic, including both natural deduction systems and sequence calculi. (Sequent calculi, in addition, typically have a sequence on the right that means the disjunction of the formulas.) Here is one of Olivier’s systems, adapted to my setting, which I’ll describe below. Let $$\Gamma$$ range over sequences of types. $\begin{gathered} \frac{\Gamma_1, \Gamma_2 \vdash A} {\Gamma_1 , n, \Gamma_2 \vdash A} \qquad \frac{\Gamma_1, \Gamma_2 \vdash A} {\Gamma_1 , B \to C, \Gamma_2 \vdash A} \\[2ex] \frac{\Gamma \vdash A \quad \Gamma \vdash B}{\Gamma \vdash A \sqcap B} \qquad \frac{\Gamma_1,B,C,\Gamma_2 \vdash A}{\Gamma_1,B\sqcap C,\Gamma_2 \vdash A} \\[2ex] \frac{}{n \vdash n} \qquad \frac{A \vdash C_1, \ldots, C_n \quad D_1, \ldots, D_n \vdash B} {C_1\to D_1,\ldots, C_n\to D_n \vdash A \to B}\end{gathered}$ The first two rules are weakening rules for singleton integers and function types. There is no weakening rule for intersections. The third and fourth rules are introduction and elimination rules for intersection. The fifth rule is reflexivity for integers, and the last is the combined rule for function types.

The combined rule for function types says that the intersection of a sequence of function types $${\sqcap}_{i=1\ldots n} (C_i\to D_i)$$ is a subtype of $$A \to B$$ if $A <: {\sqcap}_{i\in\{1\ldots n\}} C_i \qquad \text{and}\qquad {\sqcap}_{i\in\{1\ldots n\}} D_i <: B$ Interestingly, the inversion principle for this rule is the $$\beta$$-sound property described in Chapter 14 of Lambda Calculus with Types by Barendregt et al., and is the key to proving $$\beta$$-equality. In Olivier’s system, $$\beta$$-soundness falls out immediately, instead of by a somewhat involved proof.

The regular subtyping rule for function types is simply an instance of the combined rule in which the sequence on the left contains just one function type.

The next step for me was to enter Olivier’s definitions into Isabelle and prove transitivity via cut elimination. That is, I needed to prove the following generalized statement via a sequence of lemmas laid out by Olivier in his draft.
Theorem (Cut Elimination)
If $$\Gamma_2 \vdash B$$ and $$\Gamma_1,B,\Gamma_3 \vdash C$$, then $$\Gamma_1,\Gamma_2,\Gamma_3 \vdash C$$.
The transitivity rule is the instance of cut elimination where $$\Gamma_2 = A$$ and both $$\Gamma_1$$ and $$\Gamma_3$$ are empty.

Unfortunately, I couldn’t resist making changes to Olivier’s subtyping system as I entered it into Isabelle, which cost me considerable time. Some of Olivier’s lemmas show that the collection of types on the left, that is, the $$A's$$ in $$A_1,\ldots, A_n \vdash B$$, behave like a set instead of a sequence. I figured that if the left-hand-side was represented as a set, then I would be able to bypass several lemmas and obtain a shorter proof. I got stuck in proving Lemma $$\cap L_e$$ which states that $$\Gamma_1,A\sqcap B,\Gamma_2 \vdash C$$ implies $$\Gamma_1,A, B,\Gamma_2 \vdash C$$. Olivier’s subtyping rules are carefully designed to minimize the amount of overlap between the rules, and switching to a set representation increases the amount of overlap, making the proof of this lemma more difficult (perhaps impossible?).

So after struggling with the set representation for some time, I went back to sequences and was able to complete the proof of cut elimination, with a little help from Olivier at FLOC. I proved the required lemmas in the following order.
Lemma (Weakening)
If $$\Gamma_1,\Gamma_2 \vdash A$$, then $$\Gamma_1,B,\Gamma_2 \vdash A$$.
(Proved by induction on $$A$$.)
Lemma (Axiom)
$$A \vdash A$$
(Proved by induction on $$A$$.)
Lemma (Permutation)
If $$\Gamma_1 \vdash A$$ and $$\Gamma_2$$ is a permutation of $$\Gamma_1$$, then $$\Gamma_2 \vdash A$$.
(Proved by induction on the derivation of $$\Gamma_1 \vdash A$$, using many lemmas about permutations.)
Lemma ($$\cap L_e$$)
If $$\Gamma_1,A\sqcap B,\Gamma_2 \vdash C$$, then $$\Gamma_1,A, B,\Gamma_2 \vdash C$$.
(Proved by induction on the derivation of $$\Gamma_1,A\sqcap B,\Gamma_2 \vdash C$$.)
Lemma (Collapse Duplicates)
If $$\Gamma_1,A,A,\Gamma_2 \vdash C$$, then $$\Gamma_1,A,\Gamma_2 \vdash C$$.
(This is proved by well-founded induction on the lexicographical ordering of the pair $$(n,k)$$ where $$n$$ is the size of $$A$$ and $$k$$ is the depth of the derivation of $$\Gamma_1,A,A,\Gamma_2 \vdash C$$. Proof assistants such as Isabelle and Coq do not directly provide the depth of a derivation, but the depth can be manually encoded as an extra argument of the relation, as in $$\Gamma_1,A,A,\Gamma_2 \vdash_k C$$.)
The Cut Elimination Theorem is then proved by well-founded induction on the triple $$(n,k_1,k_2)$$ where $$n$$ is the size of B, $$k_1$$ is the depth of the derivation of $$\Gamma_2 \vdash B$$, and $$k_2$$ is the depth of the derivation of $$\Gamma_1,B,\Gamma_3 \vdash C$$.

We define subtyping as follows. $A <: B \quad = \quad A \vdash B$

The BCD subtyping rules and other derived rules follow from the above lemmas.
Proposition (Properties of Subtyping)

1. $$A <: A$$.

2. If $$A <: B$$ and $$B <: C$$, then $$A <: C$$.

3. If $$C <: A$$ and $$B <: D$$, then $$A \to B <: C \to D$$.

4. If $$A_1 <: B$$, then $$A_1 \sqcap A_2 <: B$$.

5. If $$A_2 <: B$$, then $$A_1 \sqcap A_2 <: B$$.

6. If $$B <: A_1$$ and $$B <: A_2$$, then $$B <: A_1 \sqcap A_2$$.

7. If $$A <: C$$ and $$B <: D$$, then $$A \sqcap B <: C \sqcap D$$.

8. $$(A\to B) \sqcap (A \to C) <: A \to (B \sqcap C)$$.

9. $$(A \to C) \sqcap (B \to D) <: (A\sqcap B) \to (C \sqcap D)$$

# Consistency and Subtyping, Resolved

Recall that my switch to intersection types was motivated by my failure to prove the Consistency and Subtyping Lemma. We now return to the proof of that Lemma. We start with a handful of lemmas that are needed for that proof.
Lemma (Consistency is Symmetric and Reflexive)

1. If $$A \sim B$$, then $$B \sim A$$.

2. If $${\mathsf{wf}(A)}$$, then $$A \sim A$$.

It will often be convenient to decompose a type into its set of atoms, defined as follows. \begin{aligned} {\mathit{atoms}(n)} &= \{ n \} \\ {\mathit{atoms}(A\to B)} &= \{ A \to B \} \\ {\mathit{atoms}(A \sqcap B)} &= {\mathit{atoms}(A)} \cup {\mathit{atoms}(B)}\end{aligned}

The consistency of two types is determined by the consistency of its atoms.
Lemma (Atomic Consistency)

1. If $$A \sim B$$, $$C \in {\mathit{atoms}(A)}$$, and $$D \in {\mathit{atoms}(B)}$$, then $$C \sim D$$.

2. If (for any $$C \in {\mathit{atoms}(A)}$$ and $$D \in {\mathit{atoms}(B)}$$, $$C \sim D$$), then $$A \sim B$$.

3. If $$A \not\sim B$$, then $$C \not\sim D$$ for some $$C \in {\mathit{atoms}(A)}$$ and $$D \in {\mathit{atoms}(B)}$$.

4. If $$C \not\sim D$$, $$C \in {\mathit{atoms}(A)}$$, and $$D \in {\mathit{atoms}(B)}$$, then $$A \not\sim B$$.

There are also several properties of subtyping and the atoms of a type.
Lemma (Atomic Subtyping)

1. If $$A <: B$$ and $$C \in {\mathit{atoms}(B)}$$, then $$A <: C$$.

2. If $$A <: n$$, then $$n \in {\mathit{atoms}(A)}$$.

3. $$n <: A$$ if and only if $${\mathit{atoms}(A)} \subseteq \{ n \}$$.

4. If $$C <: A \to B$$, then $$D\to E \in {\mathit{atoms}(C)}$$ for some $$D,E$$.

5. If $$\Gamma \vdash A$$ and every atom in $$\Gamma$$ is a function type, then every atom of $$A$$ is a function type.

And we have the following important inversion lemma for function types. We use the following abbreviations: \begin{aligned} \mathrm{dom}(\Gamma) &= \{ A \mid \exists B.\; A \to B \in \Gamma \}\\ \mathrm{cod}(\Gamma) &= \{ B \mid \exists A.\; A \to B \in \Gamma \}\end{aligned}

Lemma (Subtyping Inversion for Function Types)
If $$C <: A \to B$$, then there is a sequence of function types $$\Gamma$$ such that

1. each element of $$\Gamma$$ is an atom of $$C$$,

2. For every $$D\to E \in \Gamma$$, we have $$A <: D$$, and

3. $${\sqcap}\mathrm{cod}(\Gamma) <: B$$.

Note that item 2 above implies that $$A <: {\sqcap}\mathrm{dom}(\Gamma)$$.

Lemma (Consistency and Subtyping)

1. If $$A \sim B$$, $$A <: C$$, and $$B <: D$$, then $$C \sim D$$.

2. If $$A \not\sim B$$, $$C <: A$$, $$D <: B$$, then $$C \not\sim D$$.

(1) The proof is by strong induction on the sum of the depths of $$A$$, $$B$$, $$C$$, and $$D$$. We define the depth of a type as follows. \begin{aligned} \mathit{depth}(n) &= 0 \\ \mathit{depth}(A \to B) &= 1 + \mathrm{max}(\mathit{depth}(A),\mathit{depth}(B)) \\ \mathit{depth}(A \sqcap B) &= \mathrm{max}(\mathit{depth}(A),\mathit{depth}(B)) \end{aligned} To show that $$C \sim D$$ it suffices to show that all of their atoms are consistent. Suppose $$C' \in {\mathit{atoms}(C)}$$ and $$D'\in{\mathit{atoms}(D)}$$. So we need to show that $$C' \sim D'$$. We proceed by cases on $$C'$$.

• Case $$C'=n_1$$:
We have $$A <: C'$$ and therefore $$n_1 \in {\mathit{atoms}(A)}$$. Then because $$A \sim B$$, we have $${\mathit{atoms}(B)} \subseteq \{n_1\}$$. We have $$B <: D'$$, so we also have $${\mathit{atoms}(D)} \subseteq \{n_1\}$$. Therefore $$C' \sim D'$$.

• Case $$C'=C_1\to C_2$$:
We have $$A <: C_1 \to C_2$$, so by inversion we have some sequence of function types $$\Gamma_1$$ such that every element of $$\Gamma_1$$ is an atom of $$A$$, $$C_1 <: {\sqcap}\mathrm{dom}(\Gamma_1)$$, and $${\sqcap}\mathrm{cod}(\Gamma_1) <: C_2$$.

We also know that $$D'$$ is a function type, say $$D'=D_1 \to D_2$$. (This is because we have $$A <: C'$$, so we know that $$A_1\to A_2 \in {\mathit{atoms}(A)}$$ for some $$A_1,A_2$$. Then because $$A \sim B$$, we know that all the atoms in $$B$$ are function types. Then because $$B <: D$$ and $$D' \in {\mathit{atoms}(D)}$$, we have that $$D'$$ is a function type.) So by inversion on $$B <: D_1 \to D_2$$, we have some sequence of function types $$\Gamma_2$$ such that every element of $$\Gamma_2$$ is an atom of $$B$$, $$D_1 <: {\sqcap}\mathrm{dom}(\Gamma_2)$$, and $${\sqcap}\mathrm{cod}(\Gamma_2) <: D_2$$.

It’s the case that either $$C_1 \sim D_1$$ or $$C_1 \not\sim D_1$$.

• Sub-case $$C_1 \sim D_1$$.
It suffices to show that $$C_2 \sim D_2$$. By the induction hypothesis, we have $${\sqcap}\mathrm{dom}(\Gamma_1) \sim {\sqcap}\mathrm{dom}(\Gamma_2)$$.

As an intermediate step, we shall prove that $${\sqcap}\mathrm{cod}(\Gamma_1) \sim {\sqcap}\mathrm{cod}(\Gamma_2)$$, which we shall do by showing that all their atoms are consistent. Suppose $$A' \in {\mathit{atoms}({\sqcap}\mathrm{cod}(\Gamma_1))}$$ and $$B' \in {\mathit{atoms}({\sqcap}\mathrm{cod}(\Gamma_2))}$$. There is some $$A_1\to A_2 \in \Gamma_1$$ where $$A' \in {\mathit{atoms}(A_2)}$$. Similarly, there is $$B_1 \to B_2 \in \Gamma_2$$ where $$B' \in {\mathit{atoms}(B_2)}$$. Also, we have $$A_1 \to A_2 \in {\mathit{atoms}(A)}$$ and $$B_1 \to B_2 \in {\mathit{atoms}(B)}$$. Then because $$A \sim B$$, we have $$A_1 \to A_2 \sim B_1 \to B_2$$. Furthermore, we have $$A_1 \sim B_1$$ because $${\sqcap}\mathrm{dom}(\Gamma_1) \sim {\sqcap}\mathrm{dom}(\Gamma_2)$$, so it must be the case that $$A_2 \sim B_2$$. Then because $$A' \in {\mathit{atoms}(A_2)}$$ and $$B' \in {\mathit{atoms}(B_2)}$$, we have $$A' \sim B'$$. Thus concludes this intermediate step.

By another use of the induction hypothesis, we have $$C_2 \sim D_2$$, and this case is finished.

• Sub-case $$C_1 \not\sim D_1$$.
Then we immediately have $$C_1 \to C_2 \sim D_1 \to D_2$$.

• Case $$C'=C_1\sqcap C_2$$:
We already know that $$C'$$ is an atom, so we have a contradiction and this case is vacously true.

The next two lemmas follow from the Consistency and Subtyping Lemma and help prepare to prove the case for application in the Join Theorem.
Lemma (Application Consistency)
If $$A_1 \sim A_2$$, $$B_1 \sim B_2$$, $$A_1 <: B_1 \to C_1$$, $$A_2 <: B_2 \to C_2$$, and all these types are well formed, then $$C_1 \sim C_2$$.
(This lemma is proved directly, without induction.)
Lemma (Application Intersection)
If $$A_1 <: B_1 \to C_1$$, $$A_2 <: B_2 \to C_2$$, $$A_1 \sim A_2$$, $$B_1 \sim B_2$$, and $$C_1 \sim C_2$$, then $$(A_1\sqcap A_2) <: (B_1 \sqcap B_2) \to (C_1 \sqcap C_2)$$.
(This lemma is proved directly, without induction.)

# Updating the Denotational Semantics

Armed with the Consistency and Subtyping Lemma, I turned back to the proof of the Join Theorem, but first I needed to update my denotational semantics to use intersection types instead of values. For this we’ll need the definition of well formed types that we alluded to earlier.

$\begin{gathered} \frac{}{{\mathsf{wf}(n)}} \qquad \frac{{\mathsf{wf}(A)} \quad {\mathsf{wf}(B)}}{{\mathsf{wf}(A \to B)}} \qquad \frac{A \sim B \quad {\mathsf{wf}(A)} \quad {\mathsf{wf}(B)}}{{\mathsf{wf}(A \sqcap B)}}\end{gathered}$

Here are some examples and non-examples of well-formed types. $\begin{gathered} {\mathsf{wf}(4)} \qquad {\mathsf{wf}(3 \sqcap 3)} \qquad \neg {\mathsf{wf}(3 \sqcap 4)} \\ {\mathsf{wf}((0\to 1) \sqcap (2 \to 3))} \qquad \neg {\mathsf{wf}((0 \to 1) \sqcap (0 \to 2))}\end{gathered}$ It is sometimes helpful to think of well-formed types in terms of the equivalence classes determined by subtype equivalence: $A \approx B \quad = \quad A <: B \text{ and } B <: A$ For example, we have $$3 \approx (3 \sqcap 3)$$, so they are in the same equivalence class and $$3$$ would be the representative.

We also introduce the following notation for all the well-formed types that are super-types of a given type. ${\mathord{\uparrow} A} \quad = \quad \{ B\mid A <: B \text{ and } {\mathsf{wf}(B)} \}$

We shall represent variables with de Bruijn indices, so an environment $$\Gamma$$ is a sequence of types. The denotational semantics of the CBV $$\lambda$$-calculus is defined as follows. \begin{aligned} {\mathcal{E}{[\![ n ]\!]}}\Gamma &= {\mathord{\uparrow} n} \\ {\mathcal{E}{[\![ x ]\!]}}\Gamma &= {\mathrm{if}\;}x < |\Gamma| {\;\mathrm{then}\;}{\mathord{\uparrow} \Gamma[k]} {\;\mathrm{else}\;}\emptyset \\ {\mathcal{E}{[\![ \lambda e ]\!]}}\Gamma &= \{ A \mid {\mathsf{wf}(A)} \text{ and } {\mathcal{F}{[\![ A ]\!]}}e\Gamma \} \\ {\mathcal{E}{[\![ e_1{\;}e_2 ]\!]}}\Gamma &= \left\{ C\, \middle| \begin{array}{l} \exists A,B.\; A \in {\mathcal{E}{[\![ e_1 ]\!]}}\Gamma, B \in {\mathcal{E}{[\![ e_2 ]\!]}}\Gamma,\\ A <: B \to C, \text{ and } {\mathsf{wf}(C)} \end{array} \right\} \\ {\mathcal{E}{[\![ f(e_1,e_2) ]\!]}}\Gamma &= \left\{ C\, \middle| \begin{array}{l} \exists A,B,n_1,n_2.\; A \in {\mathcal{E}{[\![ e_1 ]\!]}}\Gamma, B \in {\mathcal{E}{[\![ e_2 ]\!]}}\Gamma,\\ A <: n_1, B <: n_2, {[\![ f ]\!]}(n_1,n_2) <: C, {\mathsf{wf}(C)} \end{array} \right\} \\ {\mathcal{E}{[\![ {\mathrm{if}\;}e_1 {\;\mathrm{then}\;}e_2 {\;\mathrm{else}\;}e_3 ]\!]}}\Gamma &= \left\{ B\, \middle| \begin{array}{l} \exists A, n.\; A \in {\mathcal{E}{[\![ e_1 ]\!]}}\Gamma, A <: n,\\ n = 0 \Rightarrow B \in {\mathcal{E}{[\![ e_3 ]\!]}}\Gamma,\\ n \neq 0 \Rightarrow B \in {\mathcal{E}{[\![ e_2 ]\!]}}\Gamma \end{array} \right\} \\[2ex] {\mathcal{F}{[\![ n ]\!]}}e\Gamma &= \mathit{false} \\ {\mathcal{F}{[\![ A \sqcap B ]\!]}}e \Gamma &= {\mathcal{F}{[\![ A ]\!]}}e\Gamma \text{ and } {\mathcal{F}{[\![ B ]\!]}}e\Gamma\\ {\mathcal{F}{[\![ A \to B ]\!]}}e \Gamma &= B \in {\mathcal{E}{[\![ e ]\!]}} (A, \Gamma)\end{aligned}

It is easy to show that swapping in a “super” environment does not change the semantics.

Lemma (Weakening)

1. If $${\mathcal{F}{[\![ A ]\!]}}e \Gamma_1$$, $$\Gamma_1 <: \Gamma_2$$ and $$(\forall B, \Gamma_1, \Gamma_2.\; B \in {\mathcal{E}{[\![ e ]\!]}}\Gamma_1, \Gamma_2 <: \Gamma_1 \Rightarrow B \in {\mathcal{E}{[\![ e ]\!]}}\Gamma_2)$$, then $${\mathcal{F}{[\![ A ]\!]}}e \Gamma_2$$.

2. If $$A \in {\mathcal{E}{[\![ e ]\!]}}\Gamma_1$$ and $$\Gamma_2 <: \Gamma_1$$, then $$A \in {\mathcal{E}{[\![ e ]\!]}}\Gamma_2$$.

(Part 1 is proved by induction on $$A$$. Part 2 is proved by induction on $$e$$ and uses part 1.)

# The Home Stretch

Now for the main event, the proof of the Meet Theorem!
Theorem (Meet)
If $$A_1 \in {\mathcal{E}{[\![ e ]\!]}}\Gamma_1$$, $$A_2 \in {\mathcal{E}{[\![ e ]\!]}}\Gamma_2$$, both $$\Gamma_1$$ and $$\Gamma_2$$ are well formed, and $$\Gamma_1 \sim \Gamma_2$$,
then $$A_1 \sqcap A_2 \in {\mathcal{E}{[\![ e ]\!]}}(\Gamma_1\sqcap\Gamma_2)$$ and $${\mathsf{wf}(A_1 \sqcap A_2)}$$.
Proof We proceed by induction on $$e$$.

• Case $$e=k$$ ($$k$$ is a de Bruijn index for a variable):
We have $$\Gamma_1[k] <: A_1$$ and $$\Gamma_2[k] <: A_2$$, so $$\Gamma_1[k] \sqcap \Gamma_2[k] <: A_1 \sqcap A_2$$. Also, because $$\Gamma_1 \sim \Gamma_2$$ we have $$\Gamma_1[k] \sim \Gamma_2[k]$$ and therefore $$A_1 \sim A_2$$, by the Consistency and Subtyping Lemma. So we have $${\mathsf{wf}(A_1 \sqcap A_2)}$$ and this case is finished.

• Case $$e=n$$:
We have $$n <: A_1$$ and $$n <: A_2$$, so $$n <: A_1 \sqcap A_2$$. Also, we have $$A_1 \sim A_2$$ by the Consistency and Subtyping Lemma. So we have $${\mathsf{wf}(A_1 \sqcap A_2)}$$ and this case is finished.

• Case $$e=\lambda e$$:
We need to show that $${\mathsf{wf}(A_1 \sqcap A_2)}$$ and $${\mathcal{F}{[\![ A_1 \sqcap A_2 ]\!]}}e(\Gamma_1\sqcap\Gamma_2)$$. For the later, it suffices to show that $$A_1 \sim A_2$$, which we shall do by showing that their atoms are consistent. Suppose $$A'_1 \in {\mathit{atoms}(A_1)}$$ and $$A'_2 \in {\mathit{atoms}(A_2)}$$. Because $${\mathcal{F}{[\![ A_1 ]\!]}}e\Gamma_1$$ we have $$A'_1 =A'_{11} \to A'_{12}$$ and $$A'_{12} \in {\mathcal{E}{[\![ e ]\!]}}(A'_{11},\Gamma_1)$$. Similarly, from $${\mathcal{F}{[\![ A_2 ]\!]}}e\Gamma_2$$ we have $$A'_2 =A'_{21} \to A'_{22}$$ and $$A'_{22} \in {\mathcal{E}{[\![ e ]\!]}}(A'_{21},\Gamma_2)$$. We proceed by cases on whether $$A'_{11} \sim A'_{21}$$.

• Sub-case $$A'_{11} \sim A'_{21}$$:
By the induction hypothesis, we have $${\mathsf{wf}(A'_{12} \sqcap A'_{22})}$$ from which we have $$A'_{12} \sim A'_{22}$$ and therefore $$A'_{11}\to A'_{12} \sim A'_{21} \to A'_{22}$$.

• Sub-case $$A'_{11} \not\sim A'_{21}$$:
It immediately follows that $$A'_{11}\to A'_{12} \sim A'_{21} \to A'_{22}$$.

It remains to show $${\mathcal{F}{[\![ A_1 \sqcap A_2 ]\!]}}e(\Gamma_1\sqcap\Gamma_2)$$. This follows from two uses of the Weakening Lemma to obtain $${\mathcal{F}{[\![ A_1 ]\!]}}e(\Gamma_1\sqcap\Gamma_2)$$ and $${\mathcal{F}{[\![ A_2 ]\!]}}e(\Gamma_1\sqcap\Gamma_2)$$.

• Case $$e = (e_1 {\;}e_2)$$:
We have $B_1 \in {\mathcal{E}{[\![ e_1 ]\!]}}\Gamma_1 \quad C_1 \in {\mathcal{E}{[\![ e_2 ]\!]}}\Gamma_1 \quad B_1 <: C_1 \to A_1 \quad {\mathsf{wf}(A_1)}$ and $B_2 \in {\mathcal{E}{[\![ e_1 ]\!]}}\Gamma_2 \quad C_2 \in {\mathcal{E}{[\![ e_2 ]\!]}}\Gamma_2 \quad B_2 <: C_2 \to A_2 \quad {\mathsf{wf}(A_2)}$ By the induction hypothesis, we have $B_1 \sqcap B_2 \in {\mathcal{E}{[\![ e_1 ]\!]}}(\Gamma_1 \sqcap \Gamma_2) \quad {\mathsf{wf}(B_1 \sqcap B_2)}$ and $C_1 \sqcap C_2 \in {\mathcal{E}{[\![ e_2 ]\!]}}(\Gamma_1 \sqcap \Gamma_2) \quad {\mathsf{wf}(C_1 \sqcap C_2)}$ We obtain $$A_1 \sim A_2$$ by the Application Consistency Lemma, and then by the Application Intersection Lemma we have $B_1 \sqcap B_2 <: (C_1 \sqcap C_2) \to (A_1 \sqcap A_2)$ So we have $$A_1 \sqcap A_2 \in {\mathcal{E}{[\![ e ]\!]}}(\Gamma_1 \sqcap \Gamma_2)$$.

Also, from $$A_1 \sim A_2$$, $${\mathsf{wf}(A_1)}$$, and $${\mathsf{wf}(A_2)}$$, we conclude that $${\mathsf{wf}(A_1 \sqcap A_2)}$$.

• Case $$e= f(e_1,e_2)$$:
(This case is not very interesting. See the Isabelle proof for the details.)

• Case $$e= {\mathrm{if}\;}e_1 {\;\mathrm{then}\;}e_2 {\;\mathrm{else}\;}e_3$$:
(This case is not very interesting. See the Isabelle proof for the details.)

I thought that the following Subsumption Theorem would be needed to prove the Meet Theorem, but it turned out not to be necessary, which is especially nice because the proof of the Subsumption Theorem turned out to depend on the Meet Theorem!
Theorem (Subsumption)
If $$A \in {\mathcal{E}{[\![ e ]\!]}}\Gamma$$, $$A <: B$$, and both $$B$$ and $$\Gamma$$ are well-formed, then $$B \in {\mathcal{E}{[\![ e ]\!]}}\Gamma$$.
The proof is by induction on $$e$$ and all but the case $$e=\lambda e'$$ are straightforward. For that case, we use the following lemmas.
Lemma (Distributivity for $$\mathcal{F}$$)
If $${\mathcal{F}{[\![ (A \to B)\sqcap (C \to D) ]\!]}} e \Gamma$$, $$A \sim C$$, and everything is well formed, then $${\mathcal{F}{[\![ (A\sqcap C) \to (B\sqcap D) ]\!]}} e \Gamma$$.
(The proof is direct, using the Meet Theorem and the Weakening Lemma.)
Lemma ($$\mathcal{F}$$ and Intersections)
Suppose $$\Gamma_1$$ is a non-empty sequence of well-formed and consistent function types. If $${\mathcal{F}{[\![ {\sqcap}\Gamma_1 ]\!]}} e \Gamma_2$$, then $${\mathcal{F}{[\![ {\sqcap}\mathrm{dom}(\Gamma_1) \to {\sqcap}\mathrm{cod}(\Gamma_1) ]\!]}} e \Gamma_2$$.
(The proof is by induction on $$\Gamma_1$$ and uses the previous lemma.)

# Conclusion

This result can be viewed a couple ways. As discussed at the beginning of this post, establishing the Meet Theorem means that the this call-by-value denotational semantics respects $$\beta$$-equality for any terminating argument expression. This is useful in proving the correctness of a function inlining optimizer. Also, it would be straightforward to define a call-by-name (or need) version of the semantics that respects $$\beta$$-equality unconditionally.

Secondly, from the viewpoint of intersection type systems, this result shows that, once we require types to be well formed (i.e. self consistent), we no longer need the intersection introduction rule because it is a consequence of having the subtyping rule for distributing intersections through function types.