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Danila Fedorin 826dde759f Finish a draft of the UCC evaluator article

2021-11-28 16:50:28 -08:00

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Defining the Syntax

Expressions and Intrinsics

This is mostly the easy part. A UCC expression is one of three things:

An "intrinsic", written \(i\), which is akin to a built-in function or command.
A "quote", written \([e]\), which takes a UCC expression \(e\) and moves it onto the stack (UCC is stack-based).
A composition of several expressions, written \(e_1\ e_2\ \ldots\ e_n\), which effectively evaluates them in order.

This is straightforward to define in Coq, but I'm going to make a little simplifying change. Instead of making "composition of \(n\) expressions" a core language feature, I'll only allow "composition of \(e_1\) and \(e_2\)", written \(e_1\ e_2\). This change does not in any way reduce the power of the language; we can still {{< sidenote "right" "assoc-note" "write e_1\ e_2\ \ldots\ e_n as (e_1\ e_2)\ \ldots\ e_n." >}} The same expression can, of course, be written as e_1\ \ldots\ (e_{n-1}\ e_n). So, which way should we really use when translating the many-expression composition from the Dawn article into the two-expression composition I am using here? Well, the answer is, it doesn't matter; expression composition is associative, so both ways effectively mean the same thing.

This is quite similar to what we do in algebra: the regular old addition operator, + is formally only defined for pairs of numbers, like a+b. However, no one really bats an eye when we write 1+2+3, because we can just insert parentheses any way we like, and get the same result: (1+2)+3 is the same as 1+(2+3). {{< /sidenote >}} With that in mind, we can translate each of the three types of expressions in UCC into cases of an inductive data type in Coq.

Why do we need e_int? We do because a token like \(\text{swap}\) can be viewed as belonging to the set of intrinsics \(i\), or the set of expressions, \(e\). While writing down the rules in mathematical notation, what exactly the token means is inferred from context - clearly \(\text{swap}\ \text{drop}\) is an expression built from two other expressions. In statically-typed functional languages like Coq or Haskell, however, the same expression can't belong to two different, arbitrary types. Thus, to turn an intrinsic into an expression, we need to wrap it up in a constructor, which we called e_int here. Other than that, e_quote accepts as argument another expression, e (the thing being quoted), and e_comp accepts two expressions, e1 and e2 (the two sub-expressions being composed).

The definition for intrinsics themselves is even simpler:

We simply define a constructor for each of the six intrinsics. Since none of the intrinsic names are reserved in Coq, we can just call our constructors exactly the same as their names in the written formalization.

Values and Value Stacks

Values are up next. My initial thought was to define a value much like I defined an intrinsic expression: by wrapping an expression in a constructor for a new data type. Something like:

Inductive value :=
    | v_quot (e : expr).

Then, v_quot (e_int swap) would be the Coq translation of the expression \([\text{swap}]\). However, I didn't decide on this approach for two reasons:

There are now two ways to write a quoted expression: either v_quote e to represent a quoted expression that is a value, or e_quote e to represent a quoted expression that is just an expression. In the extreme case, the value \(e\) would be represented by v_quote (e_quote e) - two different constructors for the same concept, in the same expression!
When formalizing the lambda calculus, Programming Language Foundations uses an inductively-defined property to indicate values. In the simply typed lambda calculus, much like in UCC, values are a subset of expressions.

I took instead the approach from Programming Language Foundations: a value is merely an expression for which some predicate, IsValue, holds. We will define this such that IsValue (e_quote e) is provable, but also such that here is no way to prove IsValue (e_int swap), since that expression is not a value. But what does "provable" mean, here?

By the Curry-Howard correspondence, a predicate is just a function that takes something and returns a type. Thus, if \(\text{Even}\) is a predicate, then \(\text{Even}\ 3\) is actually a type. Since \(\text{Even}\) takes numbers in, it is a predicate on numbers. Our \(\text{IsValue}\) predicate will be a predicate on expressions, instead. In Coq, we can write this as:

You might be thinking,

Huh, Prop? But you just said that predicates return types!

This is a good observation; In Coq, Prop is a special sort of type that corresponds to logical propositions. It's special for a few reasons, but those reasons are beyond the scope of this post; for our purposes, it's sufficient to think of IsValue e as a type.

Alright, so what good is this new IsValue e type? Well, we will define IsValue such that this type is only inhabited if e is a value according to the UCC specification. A type is inhabited if and only if we can find a value of that type. For instance, the type of natural numbers, nat, is inhabited, because any number, like 0, has this type. Uninhabited types are harder to come by, but take as an example the type 3 = 4, the type of proofs that three is equal to four. Three is not equal to four, so we can never find a proof of equality, and thus, 3 = 4 is uninhabited. As I said, IsValue e will only be inhabited if e is a value per the formal specification of UCC; specifically, this means that e is a quoted expression, like e_quote e'.

To this end, we define IsValue as follows:

Now, IsValue is a new data type with only only constructor, ValQuote. For any expression e, this constructor creates a value of type IsValue (e_quote e). Two things are true here:

Since Val_quote accepts any expression e to be put inside e_quote, we can use Val_quote to create an IsValue instance for any quoted expression.
Because Val_quote is the only constructor, and because it always returns IsValue (e_quote e), there's no way to get IsValue (e_int i), or anything else.

Thus, IsValue e is inhabited if and only if e is a UCC value, as we intended.

Just one more thing. A value is just an expression, but Coq only knows about this as long as there's an IsValue instance around to vouch for it. To be able to reason about values, then, we will need both the expression and its IsValue proof. Thus, we define the type value to mean a pair of two things: an expression v and a proof that it's a value, IsValue v:

A value stack is just a list of values:

Semantics

Remember our IsValue predicate? Well, it's not just any predicate, it's a unary predicate. Unary means that it's a predicate that only takes one argument, an expression in our case. However, this is far from the only type of predicate. Here are some examples:

Equality, =, is a binary predicate in Coq. It takes two arguments, say x and y, and builds a type x = y that is only inhabited if x and y are equal.
The mathematical "less than" relation is also a binary predicate, and it's called le in Coq. It takes two numbers n and m and returns a type le n m that is only inhabited if n is less than or equal to m.
The evaluation relation in UCC is a ternary predicate. It takes two stacks, vs and vs', and an expression, e, and creates a type that's inhabited if and only if evaluating e starting at a stack vs results in the stack vs'.

Binary predicates are just functions of two inputs that return types. For instance, here's what Coq has to say about the type of eq:

eq : ?A -> ?A -> Prop

By a similar logic, ternary predicates, much like UCC's evaluation relation, are functions of three inputs. We can thus write the type of our evaluation relation as follows:

We define the constructors just like we did in our IsValue predicate. For each evaluation rule in UCC, such as:

{{< latex >}} \langle V, v, v'\rangle\ \text{swap}\ \rightarrow\ \langle V, v', v \rangle {{< /latex >}}

We introduce a constructor. For the swap rule mentioned above, the constructor looks like this:

Although the stacks are written in reverse order (which is just a consequence of Coq's list notation), I hope that the correspondence is fairly clear. If it's not, try reading this rule out loud:

The rule Sem_swap says that for every two values v and v', and for any stack vs, evaluating swap in the original stack v' :: v :: vs, aka \(\langle V, v, v'\rangle\), results in a final stack v :: v' :: vs, aka \(\langle V, v', v\rangle\).

With that in mind, here's a definition of a predicate Sem_int, the evaluation predicate for intrinsics:

Hey, what's all this with v_quote and projT1? It's just a little bit of bookkeeping. Given a value -- a pair of an expression e and a proof IsValue e -- the function projT1 just returns the expression e. That is, it's basically a way of converting a value back into an expression. The function v_quote takes us in the other direction: given an expression \(e\), it constructs a quoted expression \([e]\), and combines it with a proof that the newly constructed quote is a value.

The above two function in combination help us define the quote intrinsic, which wraps a value on the stack in an additional layer of quotes. When we create a new quote, we need to push it onto the value stack, so it needs to be a value; we thus use v_quote. However, v_quote needs an expression to wrap in a quote, so we use projT1 to extract the expression from the value on top of the stack.

In addition to intrinsics, we also define the evaluation relation for actual expressions.

Here, we may as well go through the three constructors to explain what they mean:

Sem_e_int says that if the expression being evaluated is an intrinsic, and if the intrinsic has an effect on the stack as described by Sem_int above, then the effect of the expression itself is the same.
Sem_e_quote says that if the expression is a quote, then a corresponding quoted value is placed on top of the stack.
Sem_e_comp says that if one expression e1 changes the stack from vs1 to vs2, and if another expression e2 takes this new stack vs2 and changes it into vs3, then running the two expressions one after another (i.e. composing them) means starting at stack vs1 and ending in stack vs3.

\(\text{true}\), \(\text{false}\), \(\text{or}\) and Proofs

Now it's time for some fun! The UCC language specification starts by defining two values: true and false. Why don't we do the same thing?

UCC Spec	Coq encoding
\(\text{false}\)=\([\text{drop}]\)	{{< codelines "Coq" "dawn/Dawn.v" 41 42 >}}
\(\text{true}\)=\([\text{swap} \ \text{drop}]\)	{{< codelines "Coq" "dawn/Dawn.v" 44 45 >}}

Let's try prove that these two work as intended.

This is the first real proof in this article. Rather than getting into the technical details, I invite you to take a look at the "shape" of the proof:

After the initial use of intros, which brings the variables v, v, and vs into scope, we start by applying Sem_e_comp. Intuitively, this makes sense - at the top level, our expression, \(\text{false}\ \text{apply}\), is a composition of two other expressions, \(\text{false}\) and \(\text{apply}\). Because of this, we need to use the rule from our semantics that corresponds to composition.
The composition rule requires that we describe the individual effects on the stack of the two constituent expressions (recall that the first expression takes us from the initial stack v1 to some intermediate stack v2, and the second expression takes us from that stack v2 to the final stack v3). Thus, we have two "bullet points":
- The first expression, \(\text{false}\), is just a quoted expression. Thus, the rule Sem_e_quote applies, and the contents of the quote are puhsed onto the stack.
- The second expression, \(\text{apply}\), is an intrinsic, so we need to use the rule Sem_e_int, which handles the intrinsic case. This, in turn, requires that we show the effect of the intrinsic itself; the apply intrinsic evaluates the quoted expression on the stack. The quoted expression contains the body of false, or \(\text{drop}\). This is once again an intrinsic, so we use Sem_e_int; the intrinsic in question is \(\text{drop}\), so the Sem_drop rule takes care of that.

Following these steps, we arrive at the fact that evaluating false on the stack simply drops the top element, as specified. The proof for \(\text{true}\) is very similar in spirit:

We can also formalize the \(\text{or}\) operator:

UCC Spec	Coq encoding
\(\text{or}\)=\(\text{clone}\ \text{apply}\)	{{< codelines "Coq" "dawn/Dawn.v" 65 65 >}}

We can write two top-level proofs about how this works: the first says that \(\text{or}\), when the first argument is \(\text{false}\), just returns the second argument (this is in agreement with the truth table, since \(\text{false}\) is the identity element of \(\text{or}\)). The proof proceeds much like before:

To shorten the proof a little bit, I used the Proof with construct from Coq, which runs an additional tactic (like apply) whenever ... is used. Because of this, in this proof writing apply Sem_apply... is the same as apply Sem_apply. apply Sem_e_int. Since the Sem_e_int rule is used a lot, this makes for a very convenient shorthand.

Similarly, we prove that \(\text{or}\) applied to \(\text{true}\) always returns \(\text{true}\).

Finally, the specific facts (like \(\text{false}\ \text{or}\ \text{false}\) evaluating to \(\text{false}\)) can be expressed using our two new proofs, or_false_v and or_true.

Derived Expressions

Quotes

The UCC specification defines \(\text{quote}_n\) to make it more convenient to quote multiple terms. For example, \(\text{quote}_2\) composes and quotes the first two values on the stack. This is defined in terms of other UCC expressions as follows:

{{< latex >}} \text{quote}n = \text{quote}{n-1}\ \text{swap}\ \text{quote}\ \text{swap}\ \text{compose} {{< /latex >}}

We can write this in Coq as follows:

This definition diverges slightly from the one given in the UCC specification; particularly, UCC's spec mentions that \(\text{quote}_n\) is only defined for \(n \geq 1\).However, this means that in our code, we'd have to somehow handle the error that would arise if the term \(\text{quote}_0\) is used. Instead, I defined quote_n n to simply mean \(\text{quote}_{n+1}\); thus, in Coq, no matter what n we use, we will have a valid expression, since quote_n 0 will simply correspond to \(\text{quote}_1 = \text{quote}\).

We can now attempt to prove that this definition is correct by ensuring that the examples given in the specification are valid. We may thus write,

We used a new tactic here, repeat, but overall, the structure of the proof is pretty straightforward: the definition of quote_n consists of many intrinsics, and we apply the corresponding rules one-by-one until we arrive at the final stack. Writing this proof was kind of boring, since I just had to see which intrinsic is being used in each step, and then write a line of apply code to handle that intrinsic. This gets worse for \(\text{quote}_3\):

It's so long! Instead, I decided to try out Coq's Ltac2 mechanism to teach Coq how to write proofs like this itself. Here's what I came up with:

You don't have to understand the details, but in brief, this checks what kind of proof we're asking Coq to do (for instance, if we're trying to prove that a \(\text{swap}\) instruction has a particular effect), and tries to apply a corresponding semantic rule. Thus, it will try Sem_swap if the expression is \(\text{swap}\), Sem_clone if the expression is \(\text{clone}\), and so on. Then, the two proofs become:

Rotations

There's a little trick to formalizing rotations. Values have an important property: when a value is run against a stack, all it does is place itself on a stack. We can state this as follows:

{{< latex >}} \langle V \rangle\ v = \langle V\ v \rangle {{< /latex >}}

Or, in Coq,

This is the trick to how \(\text{rotate}_n\) works: it creates a quote of \(n\) reordered and composed values on the stack, and then evaluates that quote. Since evaluating each value just places it on the stack, these values end up back on the stack, in the same order that they were in the quote. When writing the proof, solve_basic () gets us almost all the way to the end (evaluating a list of values against a stack). Then, we simply apply the composition rule over and over, following it up with eval_value to prove that the each value is just being placed back on the stack.

`e_comp` is Associative

When composing three expressions, which way of inserting parentheses is correct? Is it \((e_1\ e_2)\ e_3\)? Or is it \(e_1\ (e_2\ e_3)\)? Well, both! Expression composition is associative, which means that the order of the parentheses doesn't matter. We state this in the following theorem, which says that the two ways of writing the composition, if they evaluate to anything, evaluate to the same thing.

Conclusion

That's all I've got in me for today. However, we got pretty far! The UCC specification says:

One of my long term goals for UCC is to democratize formal software verification in order to make it much more feasible and realistic to write perfect software.

I think that UCC is definitely getting there: formally defining the semantics outlined on the page was quite straightforward. We can now have complete confidence in the behavior of \(\text{true}\), \(\text{false}\), \(\text{or}\), \(\text{quote}_n\) and \(\text{rotate}_n\). The proof of associativity is also enough to possibly argue for simplifying the core calculus' syntax even more. All of this we got from an official source, with only a little bit of tweaking to get from the written description of the language to code! I'm looking forward to reading the next post about the multistack concatenative calculus.

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