How to prove a compiler fully abstract

by Daniel Patterson on April 19, 2018

A compiler that preserves and reflects equivalences is called a fully abstract compiler. This is a powerful property for a compiler that is different (but complimentary) to the more common notion of compiler correctness. So what does it mean, and how do we prove it?

All the code for this post, along with instructions to get it running, is in the repository https://github.com/dbp/howtoprovefullabstraction. If you have any trouble getting it going, please open an issue on that repository and I’ll help figure it out with you.

Both equivalence preservation and equivalence reflection (what make a compiler fully abstract) relate to how the compiler treats program equivalences, which in this case I’m considering observational equivalence. Two programs p1 and p2 are observationally equivalent if you cannot tell any difference between the result of running them, including any side effects.

For example, if the only observable behavior about programs in your language that you can make is see what output they print, this means that the two programs that print the same output, even if they are implemented in completely different ways are equivalent. Observational equivalence is extremely useful, especially for compilers, which when optimizing may change how a particular program is implemented but should not change the observable behavior. But it is also useful for programmers, who commonly refactor code, which means they change how the code is implemented (to make it easier to maintain, or extend, or better support some future addition), without changing any functionality. Refactoring is an equivalence-preserving transformation. We write observational equivalence on programs formally as:

p1 ≈ p1

Contextual equivalence

But we often also want to compile not just whole programs, but particular modules, expressions, or in the general sense, components, and in that case, we want an analogous notion of equivalence. Two components are contextually equivalent if in all program contexts they produce the same observable behavior. In other words, if you have two modules, but any way you combine those modules with the rest of a program (so the rest is syntactically identical, but the modules differ), the results are observationally equivalent, then those two modules are contextually equivalent. We will write this, overloading the for both observational and contextual equivalence, as:

e1 ≈ e1

As an example, if we consider a simple functional language and consider our components to be individual expressions, it should be clear that these two expressions are contextually equivalent:

λx. x * 2 ≈ λx. x + x

While they are implemented differently, no matter how they are used, the result will always be the same (as the only thing we can do with these functions is call them on an argument, and when we do, each will double its argument, even though in a different way). It’s important to note that contextual equivalence always depends on what is observable within the language. For example, in Javascript, you can reflect over the syntax of functions, and so the above two functions, written as:

function(x){ return x * 2; }function(x){ return x + x; }

Would not be contextually equivalent, because there exists a program context that can distinguish them. What is that context? Well, if we imagine plugging in the functions above into the “hole” written as [·] below, the result will be different for the two functions! This is because the toString() method on functions in Javascript returns the source code of the function.

([·]).toString()

From the perspective of optimizations, this is troublesome, as you can’t be sure that a transformation between the above programs was safe (assuming one was much faster than the other), as there could be code that relied upon the particular way that the source code had been written. There are more complicated things you can do (like optimizing speculatively and falling back to unoptimized versions when reflection was needed). In general though, languages with that kind of reflection are both harder to write fast compilers for and harder to write secure compilers for, and while it’s not the topic of this post, it’s always important to know what you mean by contextual equivalence, which usually means: what can program contexts determine about components.

Part 1. Equivalence reflection

With that in mind, what does equivalence reflection and equivalence preservation for a compiler mean? Let’s start with equivalence reflection, as that’s the property that all your correct compilers already have. Equivalence reflection means that if two components, when compiled, are equivalent, then the source components must have been equivalent. We can write this more formally as (where we write s ↠ t to mean a component s is compiled to t):

∀ s1 s2 t1 t2. s1 ↠ t1 ∧ s2 ↠ t2 ∧ t1 ≈ t2 ⇒ s1 ≈ s2

What are the consequences of this definition? And why do correct compilers have this property? Well, the contrapositive is actually easier to understand: it says that if the source components weren’t equivalent then the target components would have to be different, or more formally:

∀ s1 s2 t1 t2. s1 ↠ t1 ∧ s2 ↠ t2 ∧ s1 ≉ s2 ⇒ t1 ≉ t2

If this didn’t hold, then the compiler could take different source components and compile them to the same target component! Which means you could have different source programs you wrote, which have observationally different behavior, and your compiler would produce the same target program! Any correct compiler has to preserve observational behavior, and it couldn’t do that in this case, as the target program only has one behavior, so it can’t have both the behavior of s1 and s2 (for pedants, not considering non-deterministic targets).

So equivalence reflection should be thought of as related to compiler correctness. Note, however, that equivalence reflection is not the same as compiler correctness: as long as your compiler produced different target programs for different source programs, all would be fine – your compiler could hash the source program and produce target programs that just printed the hash to the screen, and it would be equivalence reflecting, since it would produce different target programs not only for source programs that were observationally different, but even syntactically different! That would be a pretty bad compiler, and certainly not correct, but it would be equivalence reflecting.

Part 2. Equivalence preservation

Equivalence preservation, on the other hand, is the hallmark of fully abstract compilers, and it is a property that even most correct compilers do not have, though it would certainly be great if they did. It says that if two source components are equivalent, then the compiled versions must still be equivalent. Or, more formally:

∀ s1 s2 t1 t2. s1 ↠ t1 ∧ s2 ↠ t2 ∧ s1 ≈ s2 ⇒ t1 ≈ t2

(See, I just reversed the implication. Neat trick! But now it means something totally different). One place where this has been studied extensively is by security researchers, because what it tells you is that observers in the target can’t make observations that aren’t possible to distinguish in the source. Let’s make that a lot more concrete, where we will also see why it’s not frequently true, even of proven correct compilers.

Say your language has some information hiding feature, like a private field, and you have two source components that are identical except they have different values stored in the private field. If the compiler does not preserve the fact that it is private (because, for example, it translates the higher level object structure into a C struct or just a pile of memory accessed by assembly), then other target code could read the private values, and these two components will no longer be equivalent.

This also has implications for programmer refactoring and compiler optimizations: I (or my compiler) might think that it is safe to replace one version of the program with another, because I know that in my language these are equivalent, but what I don’t know is that the compiler reveals distinguishing characteristics, and perhaps some target-level library that I’m linking with relies upon details (that were supposed to be hidden) of how the old code worked. If that’s the case, I can have a working program, and make a change that does not change the meaning of the component in my language, but the whole program can no longer work.

Proving a compiler fully abstract, therefore, is all about proving equivalence preservation. So how do we do it?

How to prove equivalence preservation

Looking at what we have to prove, we see that given contextually equivalent source components s1 and s2, we need to show that t1 and t2 are contextually equivalent. We can expand this to explicitly quantify over the contexts that combine with the components to make whole programs:

∀ s1 s2 t1 t2. s1 ↠ t1 ∧ s2 ↠ t2 ∧ (∀Cs. Cs[s1] ≈ Cs[s2]) ⇒ (∀Ct. Ct[t1] ≈ Ct[t2])

Noting that as mentioned above, I am overloading to now mean whole-program observational equivalence (so, running the program produces the same observations).

First I’ll outline how the proof will go in general, and then we’ll consider an actual example compiler and do the proof for the concrete example.

We can see that in order to prove this, we need to consider an arbitrary target context Ct and show that Ct[t1] and Ct[t2] are observationally equivalent. We do this by showing that Ct[t1] is observationally equivalent to Cs'[s1] – that is, we produce a source context Cs' that we claim is equivalent to Ct. We do this by way of a “back-translation”, which will be a sort of compiler in reverse. Assuming that we can produce such a Cs' and that Cs'[s1] and Ct[t1] (and correspondingly Cs'[s2] and Ct[t2]) are indeed observationally equivalent (noting that this relies upon a cross-language notion of observations), we can prove that Ct[t1] and Ct[t2] are observationally equivalent by instantiating our hypothesis ∀Cs. Cs[s1] ≈ Cs[s2] with Cs'. This tells us that Cs'[s1] ≈ Cs'[s2], and by transitivity, Ct[t1] ≈ Ct[t2].

It can be helpful to see it in a diagram, where the top line is given by the hypothesis (once instantiated with the source context we come up with by way of backtranslation) and coming up with the back-translation and showing that Ct and Cs' are equivalent is the hard part of the proof.

Cs'[s1]  ≈  Cs'[s2]
  ≈           ≈
Ct[t1]   ?  Ct[t2]

Concrete example of languages, compiler, & proof of full abstraction

Let’s make this concrete with an example. This will be presented some in english and some in the proof assistant Coq. This post isn’t an introduction to Coq; for that, see e.g., Bertot and Casteron’s Coq’Art, Chlipala’s CPDT, or Pierce et al’s Software Foundations.

Our source language is arithmetic expressions over integers with addition and subtraction:

e ::= n
    | e + e
    | e - e

This is written down in Coq as:

Inductive Expr : Set := 
  | Num : Z -> Expr
  | Plus : Expr -> Expr -> Expr 
  | Minus : Expr -> Expr -> Expr.

Evaluation is standard (if you wanted to parse this, you would need to deal with left/right associativity, and probably add parenthesis to disambiguate, but we consider the point where you already have a tree structure, so it is unambiguous). We can write the evaluation function as:

Fixpoint eval_Expr (e : Expr) : Z := 
  match e with
  | Num n => n                               
  | Plus e1 e2 => eval_Expr e1 + eval_Expr e2
  | Minus e1 e2 => eval_Expr e1 - eval_Expr e2
  end.

Our target language is a stack machine which uses a stack of integers to evaluate the sequence of instructions. In addition to having instructions to add and subtract, our stack machine has an extra instruction: OpCount. This instruction returns how many operations remain on the stack machine, and it puts that integer on the top of the stack. This is the simplest abstraction I could think of that will provide an interesting case study for problems of full abstraction, and is a stand-in for both reflection (as it allows the program to inspect other parts of the program), and also somewhat of a proxy for execution time (remaining). Our stack machine requires that the stack be empty at the end of execution.

Inductive Op : Set :=
| Push : Z -> Op
| Add : Op
| Sub : Op
| OpCount : Op.

Let’s see the compiler and the evaluation function (note that we reverse the order when we pop values off the stack from when we put them on in the compiler).

Fixpoint compile_Expr (e : Expr) : list Op :=
  match e with
  | Num n => [Push n]
  | Plus e1 e2 => compile_Expr e1 ++ compile_Expr e2 ++ [Add]
  | Minus e1 e2 => compile_Expr e1 ++ compile_Expr e2 ++ [Sub]
  end.

Fixpoint eval_Op (s : list Z) (ops : list Op) : option Z :=
  match (ops, s) with
  | ([], [n]) => Some n
  | (Push z :: rest, _) => eval_Op (z :: s) rest 
  | (Add :: rest, n2 :: n1 :: ns) => eval_Op (n1 + n2 :: ns)%Z rest
  | (Sub :: rest, n2 :: n1 :: ns) => eval_Op (n1 - n2 :: ns)%Z rest
  | (OpCount :: rest, _) => eval_Op (Z.of_nat (length rest) :: s) rest
  | _ => None
  end.

We can prove a basic (whole program) compiler correctness result for this (for more detail on this type of result, see this post), where first we prove a general eval_step lemma and then use that to prove correctness (note: the hint and hint_rewrite tactics are from an experimental literatecoq library that adds support for proof-local hinting, which some might think is a hack but I think makes the proofs much more readable/maintainable).

Lemma eval_step : forall a : Expr, forall s : list Z, forall xs : list Op,
      eval_Op s (compile_Expr a ++ xs) = eval_Op (eval_Expr a :: s) xs.
Proof.
  hint_rewrite List.app_assoc_reverse.
  induction a; intros; iauto; simpl;
  hint_rewrite IHa2, IHa1;
  iauto'.
Qed.

Theorem compiler_correctness : forall a : Expr,
    eval_Op [] (compile_Expr a) = Some (eval_Expr a).
Proof.
  hint_rewrite eval_step.
  hint_simpl.
  induction a; iauto'.
Qed.

Now, before we can state properties about equivalences, we need to define what we mean by equivalence for our source and target languages. Both produce no side effects, so the only observation is the end result. Thus, observational equivalence is pretty straightforward; it follows from evaluation:

Definition equiv_Expr (e1 e2 : Expr) : Prop := eval_Expr e1 = eval_Expr e2.
Definition equiv_Op (p1 p2 : list Op) : Prop := eval_Op [] p1 = eval_Op [] p2.

But, we want to talk not just about whole programs, but about partial programs that can get linked with other parts to create whole programs. In order to do that, we create a new type of “evaluation context” for our Expr, that has a hole (typically written on paper as [·]). This is a program that is missing an expression, which must be filled into the hole. Given how simple our language is, any expression can be filled in to the hole and that will produce a valid program. We only want to have one hole per partial program, so in the cases for + and -, one branch must be a normal Expr (so it contains no hole), and the other can contain one hole. Our link_Expr function takes a context and an expression and fills in the hole.

Inductive ExprCtxt : Set := 
| Hole : ExprCtxt
| Plus1 : ExprCtxt -> Expr -> ExprCtxt
| Plus2 : Expr -> ExprCtxt -> ExprCtxt 
| Minus1 : ExprCtxt -> Expr -> ExprCtxt
| Minus2 : Expr -> ExprCtxt -> ExprCtxt.

Fixpoint link_Expr (c : ExprCtxt) (e : Expr) : Expr :=
  match c with
  | Hole => e
  | Plus1 c' e' => Plus (link_Expr c' e) e'
  | Plus2 e' c' => Plus e' (link_Expr c' e)
  | Minus1 c' e' => Minus (link_Expr c' e) e'
  | Minus2 e' c' => Minus e' (link_Expr c' e)
  end.

For our stack machine, partial programs are much easier, since a program is just a list of Op, which means that any program can be extended by adding new Ops on either end (or inserting in the middle).

With ExprCtxt, we can now define “contextual equivalence” for our source language:

Definition ctxtequiv_Expr (e1 e2 : Expr) : Prop :=
  forall c : ExprCtxt, eval_Expr (link_Expr c e1) = eval_Expr (link_Expr c e2).

We can do the same with our target, simplifying slightly and saying that we will allow adding arbitrary Ops before and after, but not in the middle, of an existing sequence of Ops.

Definition ctxtequiv_Op (p1 p2 : list Op) : Prop :=
  forall c1 c2 : list Op, eval_Op [] (c1 ++ p1 ++ c2) = eval_Op [] (c1 ++ p2 ++ c2).

To prove our compiler fully abstract, remember we need to prove that it preserves and reflects equivalences. Since we already proved that it is correct, proving that it reflects equivalences should be relatively straightforward, so lets start there. The lemma we want is:

Lemma equivalence_reflection :
  forall e1 e2 : Expr,
  forall p1 p2 : list Op,
  forall comp1 : compile_Expr e1 = p1,
  forall comp2 : compile_Expr e2 = p2,
  forall eqtarget : ctxtequiv_Op p1 p2,
    ctxtequiv_Expr e1 e2.
Proof.
  unfold ctxtequiv_Expr, ctxtequiv_Op in *.
  intros.
  induction c; simpl; try solve [hint_rewrite IHc; iauto];
    (* NOTE(dbp 2018-04-16): Only the base case, for Hole, remains *)
    [idtac].
  (* NOTE(dbp 2018-04-16): In the hole case, specialize the target ctxt equiv hypothesis to empty *)
  specialize (eqtarget [] []); simpl in eqtarget; repeat rewrite app_nil_r in eqtarget.

  (* NOTE(dbp 2018-04-16): At this point, we know e1 -> p1, e2 -> p2, & p1 ≈ p2,
  and want e1 ≈ e2, which follows from compiler correctness *)
  rewrite <- comp1 in eqtarget. rewrite <- comp2 in eqtarget.
  repeat rewrite compiler_correctness in eqtarget.
  inversion eqtarget.
  reflexivity.
Qed.

This lemma is a little more involved, but not by much; we proceed by induction on the structure of the evaluation contexts, and in all but the case for Hole, the induction hypothesis gives us exactly what we need. In the base case, we need to appeal to the compiler_correctness lemma we proved earlier, but otherwise it follows easily.

So what about equivalence preservation? We can state the lemma quite easily:

Lemma equivalence_preservation :
  forall e1 e2 : Expr,
  forall p1 p2 : list Op,
  forall comp1 : compile_Expr e1 = p1,
  forall comp2 : compile_Expr e2 = p2,
  forall eqsource : ctxtequiv_Expr e1 e2,
    ctxtequiv_Op p1 p2.
Proof.
Abort.

But proving it is another matter. In fact, it’s not provable, because it’s not true. We can come up with a counter-example, using that OpCount instruction we (surreptitiously) added to our target language. These two expressions are contextually equivalent in our source language (should be obvious, but putting a proof):

Example src_equiv : ctxtequiv_Expr (Plus (Num 1) (Num 1)) (Num 2).
Proof.
  unfold ctxtequiv_Expr.
  induction c; simpl; try rewrite IHc; iauto.
Qed.

But they are not contextually equivalent in the target; in particular, if we put the OpCount instruction before and then the Add instruction afterwards, the result will be the value plus the number of instructions it took to compute it:

Example target_not_equiv :
  eval_Op [] (OpCount :: compile_Expr (Plus (Num 1) (Num 1)) ++ [Add]) <>
  eval_Op [] (OpCount :: compile_Expr (Num 2) ++ [Add]).
Proof.
  simpl.
  congruence.
Qed.

The former evaluating to 6, while the latter evaluates to 4. This means that there is no way we are going to be able to prove equivalence preservation (as we have a counter-example!).

So what do we do? Well, this scenario is not uncommon, and it’s the reason why many, even correct, compilers are not fully abstract. It’s also related to why many of these compilers may still have security problems! The solution is to somehow protect the compiled code from having the equivalences disrupted. If this were a real machine, we might want to have some flag on instructions that meant that they should not be counted, and OpCount would just not return anything if it saw any of those (or would count them as 0). Alternately, we might give our target language a type system that is able to rule out linking with code that uses the OpCount instruction, or perhaps restricts how it can be used.

Because this is a blog-post sized example, and I wanted to keep the proofs as short as possible, and the unstructured and untyped nature of our target (which, indeed, is much less structured than our source language; the fact that the source is so well-structured is why our whole-program correctness result was so easy!) will mean the proofs get relatively complex (or require us to add various auxiliary definitions), so the solution I’m going to take is somewhat extreme. Rather than, say, restricting how OpCount is used, or even ruling out linking with OpCount, we’re going to highly restrict what we can link with. This is very artificial, and done entirely so that the proofs can fit into a few lines. In this case, rather than a list, we are going to allow one Op before and one Op after our compiled program, neither of which can be OpCount, and further, we still want the resulting program to be well-formed (i.e., no errors, only one number on stack at end), so either there should be nothing before and after, or there is a Push n before and either Add or Sub after. (You should be able to verify that no other combination of Op before or after will fulfill our requirement).

We can define these possible linking contexts and a helper to combine them with programs as the following:

Inductive OpCtxt : Set :=
| PushAdd : Z -> OpCtxt
| PushSub : Z -> OpCtxt
| Empty : OpCtxt.

Definition link_Op  (c : OpCtxt) (p : list Op) : list Op :=
  match c with
  | PushAdd n => Push n :: p ++ [Add]
  | PushSub n => Push n :: p ++ [Sub]
  | Empty => p
  end.

Using that, we can redefine contextual equivalence for our target language, only permitting these contexts:

Definition ctxtequiv_Op (p1 p2 : list Op) : Prop :=
  forall c : OpCtxt, eval_Op [] (link_Op c p1) = eval_Op [] (link_Op c p2).

The only change to our proof of equivalence reflection is on one line, to change our specialization of the target contexts, now to the Empty context:

specialize (eqtarget Empty) (* Empty rather than [] [] *)

With that change, we now believe that our compiler, when linked against these restricted contexts, is indeed fully abstract. So let’s prove it. If you recall from earlier in this post, proving equivalence preservation means proving that the top line implies the bottom, in the following diagram:

Cs'[s1]  ≈  Cs'[s2]
  ≈           ≈
Ct[t1]   ?  Ct[t2]

In order to do that, we rely upon a backtranslation to get from Ct to Cs', where Ct is a target context, in this tiny example our restricted OpCtxt. We can write that backtranslation as:

Definition backtranslate (c : OpCtxt) : ExprCtxt :=
  match c with
  | PushAdd n => Plus2 (Num n) Hole
  | PushSub n => Minus2 (Num n) Hole
  | Empty => Hole
  end.

The second part of the proof is showing that the vertical equivalences in the diagram hold — that is, that if s1 is compiled to t1 and Ct is backtranslated to Cs' then Ct[t1] is equivalent to Cs'[s1]. We can state and prove that as the following lemma, which follows from straightforward case analysis on the structure of our target context and backtranslation (using our eval_step lemmas):

Lemma back_translation_equiv :
  forall c : OpCtxt,
  forall p : list Op,
  forall e : Expr,
  forall c' : ExprCtxt, 
    compile_Expr e = p ->
    backtranslate c = c' ->
    eval_Op [] (link_Op c p) = Some (eval_Expr (link_Expr c' e)).
Proof.
  hint_rewrite eval_step, eval_step'.
  intros.
  match goal with
  | [ c : OpCtxt |- _] => destruct c
  end; 
    match goal with
    | [ H : backtranslate _ = _ |- _] => invert H
    end; simpl; iauto. 
Qed.

Once we have that lemma, we can prove equivalence preservation directly. We do this by doing case analysis on the target context we are given, backtranslating it and then using the lemma we just proved to get the equivalence that we need.

Lemma equivalence_preservation :
  forall e1 e2 : Expr,
  forall p1 p2 : list Op,
  forall comp1 : compile_Expr e1 = p1,
  forall comp2 : compile_Expr e2 = p2,
  forall eqsource : ctxtequiv_Expr e1 e2,
    ctxtequiv_Op p1 p2.
Proof.
  unfold ctxtequiv_Expr, ctxtequiv_Op in *.
  intros.

  remember (backtranslate c) as c'.
  destruct c; iauto;

  erewrite back_translation_equiv with (e := e1) (c' := c'); iauto;
  erewrite back_translation_equiv with (e := e2) (c' := c'); iauto;
  specialize (eqsource c'); simpl in *; congruence.
Qed.

This was obviously a very tiny language and a very restrictive linker that only allowed very restrictive contexts, but the general shape of the proof is the same as that used in more realistic languages published in research conferences today!

So next time you see a result about a correct (or even hoped to be correct) compiler, ask if it is fully abstract! And if it’s not, are the violations of equivalences something that could be exploited? Or something that would invalidate optimizations?

Some recent conference publications include Devriese et al, Fully-abstract compilation by approximate back-translation published in POPL’16 and New at al, Fully Abstract Compilation via Universal Embedding, published in ICFP’16.

As stated at the top of the post, all the code in this post is available at https://github.com/dbp/howtoprovefullabstraction. If you have any trouble getting it going, please open an issue on that repository and I’ll help figure it out with you.