The evaluators we have seen so far (e.g., the ones for aexps, bexps, and commands) have been formulated in a "big-step" style — they specify how a given expression can be evaluated to its final value (or a command plus a store to a final store) "all in one big step." This style is simple and natural for many purposes (indeed, Gilles Kahn, who popularized its use, called it natural semantics), but there are some things it does not do well. In particular, it does not give us a natural way of talking about concurrent programming languages, where the "semantics" of a program — i.e., the essence of how it behaves — is not just which input states get mapped to which output states, but also includes the intermediate states that it passes through along the way, since these states can also be observed by concurrently executing code. Another shortcoming of the big-step style is more technical, but critical for some applications. Consider the variant of Imp with lists that we introduced in ImpList.v. We chose to define the meaning of programs like 0 + nil by specifying that a list should be interpreted as 0 when it occurs in a context expecting a number, but this was a bit of a hack. It would be better simply to say that the behavior of such a program is undefined — it doesn't evaluate to any result. We could easily do this: we'd just have to use the formulations of aeval and beval as inductive propositions (rather than Fixpoints), so that we can make them partial functions instead of total ones. However, this way of defining Imp has a serious deficiency. In this language, a command might fail to map a given starting state to any ending state for two quite different reasons: either because the execution gets into an infinite loop or because, at some point, the program tries to do an operation that makes no sense, such as taking the successor of a boolean variable, and none of the evaluation rules can be applied. These two outcomes — nontermination vs. getting stuck in an erroneous configuration — are quite different. In particular, we want to allow the first (permitting the possibility of infinite loops is the price we pay for the convenience of programming with general looping constructs like while) but prevent the second (which is just wrong), for example by adding some form of typechecking to the language. Indeed, this will be a major topic for the rest of the course. As a first step, we need a different way of presenting the semantics that allows us to distinguish nontermination from erroneous "stuck states." So, for lots of reasons, we'd like to have a finer-grained way of defining and reasoning about program behaviors. This is the topic of the present chapter. We replace the "big-step" eval relation with a "small-step" relation that specifies, for a given program, how the "atomic steps" of computation are performed.

A Toy Language

To save space in the discussion, let's go back to an incredibly simple language containing just constants and addition. At the end of the chapter, we'll see how to apply the same techniques to the full Imp language.

Here is a standard evaluator for this language, written in the same (big-step) style as we've been using up to this point.

Now, here is the same evaluator, written in exactly the same style, but formulated as an inductively defined relation. Again, we use the notation t ⇓ n for "t evaluates to n."

	(E_Const)
n ⇓ n

t1 ⇓ n1
t2 ⇓ n2	(E_Plus)
t1 + t2 ⇓ plus n1 n2

We write plus n1 n2 in the second rule, rather than n1 + n2, to emphasize the fact that the + on the left of the ⇓ is an abstract-syntax-tree node while the addition on the right of the ⇓ is the mathematical sum of the numbers n1 and n2. The formal Coq version of the definition is more pedantic about this distinction:

Here is a slight variation, still in big-step style, where the final result of evaluating a term is also a term, rather than a bare number.

(Note that this doesn't change anything in the informal rules, where we are eliding the distinction between the constant term n and the bare number n.)

Now, here is a small-step version.

	(ST_PlusConstConst)
n1 + n2 ⇒ plus n1 n2

t1 ⇓ t1'	(ST_Plus1)
t1 + t2 ⇓ t1' + t2

t2 ⇓ t2'	(ST_Plus2)
n1 + t2 ⇓ n1 + t2'

Note that we're using variable names here to lighten the notation: by convention, n1 and n2 refer only to constants (constructed with tm_const), while t1 and t2 refer to arbitrary terms. In the formal rules, we use explicit constructors to make the same distinction.

Things to notice:

• We are defining just a single reduction step, in which one tm_plus node is replaced by its value.
• Each step finds the leftmost tm_plus node that is ready to go (both of its operands are constants) and rewrites it in place. The first rule tells how to rewrite this tm_plus node itself; the other two rules tell how to find it.
• A term that is just a constant cannot take a step.

A couple of examples of reasoning with the step relation... If t1 can take a step to t1', then tm_plus t1 t2 steps to plus t1' t2:

Exercise: 2 stars (test_step_2)

Right-hand sides of sums can take a step only when the left-hand side is finished: if t2 can take a step to t2', then tm_plus (tm_const n) t2 steps to tm_plus (tm_const n) t2':

☐ One interesting property of the ⇒ relation is that, like the evaluation relation for our language of Imp programs, it is deterministic. Theorem: For each t, there is at most one t' such that t steps to t' (t ⇒ t' is provable). Formally, this is the same as saying that ⇒ is a partial function. Proof sketch: We show that if x steps to both y1 and y2 then y1 and y2 are equal, by induction on a derivation of step x y1. There are several cases to consider, depending on the last rule used in this derivation and in the given derivation of step x y2.

• If both are ST_PlusConstConst, the result is immediate.
• The cases when both derivations end with ST_Plus1 or ST_Plus2 follow by the induction hypothesis.
• It cannot happen that one is ST_PlusConstConst and the other is ST_Plus1 or ST_Plus2, since this would imply that x has the form tm_plus t1 t2 where both t1 and t2 are constants (by ST_PlusConstConst) and one of t1 or t2 has the form tm_plus ....
• Similarly, it cannot happen that one is ST_Plus1 and the other is ST_Plus2, since this would imply that x has the form tm_plus t1 t2 where t1 has both the form tm_plus t1 t2 and the form tm_const n. ☐

Values

Let's take a moment to slightly generalize the way we state the definition of single-step reduction. It is useful to think of the ⇒ relation as defining an abstract machine:

• At any moment, the state of the machine is a term.
• A step of the machine is an atomic unit of computation — here, a single "add" operation.
• The halting states of the machine are ones where there is no more computation to be done.

We can then execute a term t as follows:

• Take t as the starting state of the machine.
• Repeatedly use the ⇒ relation to find a sequence of machine states, starting with t, where each state steps to the next.
• When no more reduction is possible, "read out" the final state of the machine as the result of execution.

Intuitively, it is clear that the final states of the machine are always terms of the form tm_const n for some n. We call such terms values.

Having introduced the idea of values, we can use it in the definition of the ⇒ relation to write ST_Plus2 rule in a slightly more intuitive way:

	(ST_PlusConstConst)
n1 + n2 ⇒ plus n1 n2

t1 ⇓ t1'	(ST_Plus1)
t1 + t2 ⇓ t1' + t2

t2 ⇓ t2'	(ST_Plus2)
v1 + t2 ⇓ v1 + t2'

Again, the variable names here carry important information: by convention, v1 ranges only over values, while t1 and t2 range over arbitrary terms. In the Coq version of the rules, we use an explicit value hypothesis for the same purpose.

Exercise: 3 stars, recommended (redo_determinacy)

As a sanity check on this change, let's re-verify determinacy Proof sketch: We must show that if x steps to both y1 and y2 then y1 and y2 are equal. Consider the final rules used in the derivations of step x y1 and step x y2.

• If both are ST_PlusConstConst, the result is immediate.
• It cannot happen that one is ST_PlusConstConst and the other is ST_Plus1 or ST_Plus2, since this would imply that x has the form tm_plus t1 t2 where both t1 and t2 are constants (by ST_PlusConstConst) AND one of t1 or t2 has the form tm_plus ....
• Similarly, it cannot happen that one is ST_Plus1 and the other is ST_Plus2, since this would imply that x has the form tm_plus t1 t2 where t1 both has the form tm_plus t1 t2 and is a value (hence has the form tm_const n).
• The cases when both derivations end with ST_Plus1 or ST_Plus2 follow by the induction hypothesis. ☐

Most of this proof is the same as the one above. But to get maximum benefit from the exercise you should try to write it from scratch and just use the earlier one if you get stuck.

☐