More Automation

Time to take the gloves off...

The auto and eauto Tactics

The auto tactic solves goals that are solvable by any combination of

• intros
• apply (with a local hypothesis, by default)
• reflexivity

The eauto tactic works just like auto, except that it uses eapply instead of apply. Using auto is always "safe" in the sense that it will never fail and will never change the proof state: either it completely solves the current goal, or it does nothing. Here is a contrived example:

By default, auto and eauto consider just the hypotheses in the current context. But we can also give a list of lemmas and constructors to consider in addition to these, by writing auto using ....

There are some constructors and lemmas that are applied very often in proofs. We can tell auto to always consider these, instead of mentioning them explicitly each time. This is accomplished by writing at the top level, where l is a top-level lemma theorem or a constructor of an inductively defined proposition (i.e., anything whose type is an implication). As a shorthand, we can write to tell Coq to add all of the constructors from the inductive definition of c to the database used by auto. It is also sometimes necessary to add where d is a defined symbol, so that auto knows to expand uses of d and enable further possibilities for applying lemmas that it knows about. Here are some Hints we will find useful. There is an example of the effect of the and and or hints below.

Warning: Just as with Coq's other automation facilities, it is easy to overuse auto and eauto and wind up with proofs that are impossible to understand later! Also, overuse of eauto can make proof scripts very slow. Get in the habit of using auto most of the time and eauto more sparingly.

The Proof with Tactic

If you start a proof with Proof with (tactic) then writing ... instead of . after a tactic in the body of the proof will try to solve all generated subgoals with tactic (and fail if this doesn't work). One common use of this facility is "Proof with auto" (or eauto). We'll see many examples of this later in the file.

The solve by inversion Tactic

Here's another nice automation feature: it often arises that the context contains a contradictory assumption and we want to use inversion on it to solve the goal. We'd like to be able to say to Coq, "find a contradictory assumption and invert it" without giving its name explicitly. Unfortunately, Coq does not provide a built-in tactic that does this. However, it is not too hard to define one ourselves. (Thanks to Aaron Bohannon for this nice hack.)

Again, the details of how the Tactic Notation definition works are not important. All we need to know is that doing solve by inversion will find a hypothesis that can be inverted to solve the goal, if there is one. The tactics solve by inversion 2 and solve by inversion 3 are slightly fancier versions which will perform two or three inversions in a row, if necessary, to solve the goal.

The try solve Tactic

If t is a tactic, then try solve [t] is a tactic that

• if t solves the goal, behaves just like t, or
• if t cannot completely solve the goal, does nothing.

More generally, try solve [t1 | t2 | ...] will try to solve the goal by using t1, t2, etc. If none of them succeeds in completely solving the goal, then try solve [t1 | t2 | ...] does nothing. The fact that it does nothing when it doesn't completely succeed in solving the goal means that there is no harm in using try solve T in situations where T might actually make no sense. In particular, if we put it after a ; it will solve as many subgoals as it can and leave the rest for us to solve by other methods. It will not leave any of the subgoals in a mangled state.

The normalize Tactic

When experimenting when the languages defined in this file, we will sometimes want to see what a particular concrete term steps to, so we want to solve goals of the form t -->* t' where t is a completely concrete term. These proofs are simple but repetitive. Consider for example reducing an arithmetic expression using the small-step relation astep defined in the previous chapter:

We repeatedly applied rsc_step until we got to a normal form. The proofs that the intermediate steps are possible are simple enough that auto, with appropriate hints, can solve them.

The following custom Ltac captures this pattern. In addition, before each rsc_step we print out the current goal, so that the user can follow how the term is being evaluated.

Finally, this is also useful as a simple way to calculate what the normal form of a term is, by proving a goal with an existential variable in it.

Typed Arithmetic Expressions

To motivate the discussion of types, let's begin with an extremely simple language of arithmetic and boolean expressions -- a language just barely complex enough to have the possibility of programs "going wrong" because of runtime type errors.

Syntax

Small-step reduction

Notice that the step relation doesn't care at all about whether expressions make sense or not. For example, it is easy to see that is nonsensical, but the step relation will happily reduce it one step to

Normal Forms and Values

The first interesting thing about the step relation is that the progress theorem as we stated it before fails! That is, there are terms that are normal forms (they can't take a step) but not values -- i.e., we have not included them in our definition of possible "results of evaluation." Such terms are said to be stuck.

Exercise: 2 stars, optional (some_tm_is_stuck)

☐ Fortunately, things are not completely messed up: values and normal forms are not the same in this language, but at least the former set is included in the latter (i.e., we did not accidentally define things so that some value could still take a step).

Exercise: 3 stars, optional (value_is_nf)

Hint: You will reach a point in this proof where you need to use an induction to reason about a term that is known to be a numeric value. This induction can be performed either over the term itself or over the evidence that it is a numeric value. The proof goes through in either case, but you will find that one way is quite a bit shorter than the other. For the sake of the exercise, try to complete the proof both ways.

☐

Exercise: 3 stars, optional

Using value_is_nf, we can show that the step relation is also deterministic...

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Typing

Critical observation: although there are stuck terms in this language, they are all "nonsensical", mixing booleans and numbers in a way that we don't even want to have a meaning. We can easily exclude such ill-typed terms by defining a "typing relation" that relates terms to the types (either numeric or boolean) of their final results.

The typing relation In informal notation, the typing relation is often written |- t : T, pronounced "t has type T."

	(T_True)
|- true : Bool

	(T_False)
|- false : Bool

|- t1 : Bool    |- t2 : T    |- t3 : T	(T_If)
|- if t1 then t2 else t3 : T

	(T_Zero)
|- 0 : Nat

|- t1 : Nat	(T_Succ)
|- succ t1 : Not

|- t1 : Nat	(T_Pred)
|- pred t1 : Not

|- t1 : Nat	(T_IsZero)
|- iszero t1 : Bool

Examples

It's important to realize that the typing relation is a conservative (or static) approximation: it does not calculate the type of the normal form of a term.

Exercise: 1 star

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Progress

The typing relation enjoys two critical properties. The first is that well-typed normal forms are values.

Exercise: 2 stars (finish_progress)

☐ This is a bit more interesting than the progress theorem that we saw in Smallstep.v, where all normal forms were values. Here, a term can be stuck, but only if it is ill typed.

Exercise: 1 star (step_review)

Quick review. Answer true or false. (As usual, no need to hand these in.)

• In this language, every well-typed normal form is a value.
• Every value is a normal form.
• The single-step evaluation relation is a partial function.
• The single-step evaluation relation is a TOTAL function.

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Type Preservation

The second critical property of typing is that, when a well-typed term takes a step, the result is also a well-typed term. This theorem is also sometimes called the subject reduction property, because it tells us what happens when the "subject" of the typing relation is reduced.

Exercise: 2 stars (finish_preservation)

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Exercise: 3 stars, optional (preservation_alternate_proof)

Now prove the same property again by induction on the evaluation derivation instead of on the typing derivation. Begin by carefully reading and thinking about the first few lines of the above proof to make sure you understand what each one is doing. The set-up for this proof is similar, but not exactly the same.

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Type Soundness

Putting progress and preservation together, we can see that a well-typed term can never reach a stuck state.

Indeed, in the present -- extremely simple -- language, every well-typed term can be reduced to a normal form: this is the normalization property. In richer languages, this property often fails, though there are some interesting languages (such as Coq's Fixpoint language, and the simply typed lambda-calculus, which we'll be looking at next) where all well-typed terms can be reduced to normal forms.

Additional exercises

Exercise: 2 stars (subject_expansion)

Having seen the subject reduction property, it is reasonable to wonder whether the opposity property -- subject EXPANSION -- also holds. That is, is it always the case that, if t --> t' and has_type t' T, then has_type t T? If so, prove it. If not, give a counter-example. (* FILL IN HERE *)
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Exercise: 2 stars (variation1)

Suppose we add the following two new rules to the evaluation relation: Which of the following properties remain true in the presence of these rules? For each one, write either "remains true" or else "becomes false." If a property becomes false, give a counterexample.

• Determinacy of step
• Normalization of step for well-typed terms
• Progress
• Preservation

☐

Exercise: 2 stars (variation2)

Suppose, instead, that we add this new rule to the typing relation: Which of the following properties remain true in the presence of this rule? (Answer in the same style as above.)

• Determinacy of step
• Normalization of step for well-typed terms
• Progress
• Preservation

☐

Exercise: 2 stars (variation3)

Suppose, instead, that we add this new rule to the typing relation: Which of the following properties remain true in the presence of this rule? (Answer in the same style as above.)

• Determinacy of step
• Normalization of step for well-typed terms
• Progress
• Preservation

☐

Exercise: 2 stars (variation4)

Suppose, instead, that we add this new rule to the step relation: Which of the above properties become false in the presence of this rule? For each one that does, give a counter-example. ☐

Exercise: 2 stars (variation5)

Suppose instead that we add this rule: Which of the above properties become false in the presence of this rule? For each one that does, give a counter-example. ☐

Exercise: 2 stars (variation6)

Suppose instead that we add this rule: Which of the above properties become false in the presence of this rule? For each one that does, give a counter-example. ☐

Exercise: 2 stars (variation7)

Suppose instead that we add this rule: Which of the above properties become false in the presence of this rule? For each one that does, give a counter-example. ☐

Exercise: 2 stars (variation8)

Suppose instead that we add this rule: Which of the above properties become false in the presence of this rule? For each one that does, give a counter-example. ☐

Exercise: 3 stars, optional (more_variations)

Make up some exercises of your own along the same lines as the ones above. Try to find ways of selectively breaking properties -- i.e., ways of changing the definitions that break just one of the properties and leave the others alone. ☐

Exercise: 1 star (remove_predzero)

The evaluation rule E_PredZero is a bit counter-intuitive: we might feel that it makes more sense for the predecessor of zero to be undefined, rather than being defined to be zero. Can we achieve this simply by removing the rule from the definition of step? (* FILL IN HERE *)
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Exercise: 3 stars, optional (prog_pres_bigstep)

Suppose our evaluation relation is defined in the big-step style. What are the appropriate analogs of the progress and preservation properties? (* FILL IN HERE *)
☐

The Simply Typed Lambda-Calculus

The simply typed lambda-calculus (STLC) can be described as the simplest possible functional programming language -- it contains anonymous functions, application, variables, and nothing else. Since it is essentially a small fragment of Coq's built-in functional language, we could informally use the same notation for programs. For example, is the identity function at the type A, and is a function that takes an argument of type A, throws it away, and returns the constant-x function (of type B->A). However, this "fun ... => ..." notation is a little heavy, especially for writing on the blackboard. So when we discuss terms informally, we'll use another common way of writing function abstractions in this language: \x:A.x, where \ stands for the greek letter "lambda". To keep things simple, we'll start with the pure simply typed lambda-calculus, where base types like A, B, etc. are completely uninterpreted -- i.e., there are no constants belonging to base types and no primitive operations over base types. At the end of the chapter, we'll see how to extend the pure system to something closer to a real programming language by adding base types like numbers. The main new technical issues that we'll have to face in this section are dealing with variable binding and substitution.

• We've already seen how to formalize a language with variables (Imp). There, however, the variables were all global. In STLC, we need to handle bound variables because of function parameters.
• Moreover, instead of just looking up global variables in the store, we'll need to reduce function applications by substituting arguments for parameters in function bodies.

Syntax and Operational Semantics

Types

The function types of the STLC are just like the function types of Coq. We begin with a collection of base types, which we write A, B, etc., in informal examples (formally, they are designated by the constructor ty_base together with an id). Then, for every pair of types T1 and T2, we construct the arrow type T1->T2 (formally, ty_arrow T1 T2).

Some base types for use in examples...

Terms

The terms of the STLC are just variables, function application, and function abstraction. We use identifiers as variable names.

One important thing to note here is that an abstraction \x:T.t (formally, tm_abs x T t) is always annotated with the type (T) of its parameter. This is in contrast to Coq (and other functional languages like ML, Haskell, etc.), where we can write either or and trust Coq to fill in an appropriate annotation in the latter case.

Some variables for examples...

Some identity functions on various types... id_A = \a:A. a

id_AarrowA = \a:A->A. a

id_AarrowA_arrow_AarrowA = \a:(A->A)->(A->A). a

A function that takes an A and returns a constant function of type B->A: k = \a:A. \b:B. a

(We make these Notations rather than Definitions to make things easier for auto.)

Values

When we come to defining the values of the STLC, we have to think a little. First, an application is clearly not a value: It represents a function being invoked on some argument, and the term cannot be finished computing till this invocation has actually happened. For abstractions, on the other hand, we have a choice:

• We can say that \a:A.t is a value only if t is a value -- i.e., only if the function's body has been reduced (as much as it can be without knowing what argument it is going to be applied to).
• Or we can say that \a:A.t is always a value, no matter whether t is one or not -- in other words, we can say that reduction stops at abstractions.

Coq makes the first choice -- for example, yields fun a:bool => 7. But most real functional programming languages make the second choice -- reduction of a function's body only begins when the function is actually applied to an argument. We also make the latter choice here. Having decided this, we don't need to worry about whether variables are values, since we'll always be reducing programs "from the outside in," and that means the step relation will always be working with closed terms (ones with no free variables).

Substitution

Now we come to the most interesting part of the definition: the operation of substituting one term for the free occurrences of some variable in another term. This operation will be used in a moment to define the operational semantics of function application, where we will need to substitute the argument term for the function parameter in the function's body. Informally, substitution is often written [s/x]t, pronounced "substitute s for x in t". For example: Note that, in the last two examples, we do not substitute for the a in the body of an abstraction whose bound variable is named a. Here is the formal definition:

Technical note: Substitution becomes trickier to define if we consider the case where s, the term being substituted in, may itself contain free variables. Since we are only interested in defining the step relation on closed terms here, we can avoid this extra complexity.

Reduction

The small-step reduction relation for STLC follows the same pattern as the ones we have seen before. Intuitively, to reduce a function application, we first reduce its left-hand side until it becomes a literal function; then we reduce its right-hand side (the argument) until it is also a value; and finally we substitute the argument for the bound variable in the body of the function. This last rule, written informally as is traditionally called "beta-reduction". Informally:

value v2	(ST_AppAbs)
(\a:T.t12) v2 --> [v2/a]t12

t1 --> t1'	(ST_App1)
t1 t2 --> t1' t2

value v1     t2 --> t2'	(ST_App2)
v1 t2 --> v1 t2'

Formally:

Examples

We can use the normalize tactic from above to simplify the proof.

Exercise: 2 stars (step_example3)

Try to do this one both with and without normalize.

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Typing

Contexts

Question: What is the type of the term "x y"? Answer: It depends on the types of x and y! I.e., in order to assign a type to a term, we need to know what assumptions we should make about the types of its free variables. This leads us to a three-place "typing judgment", informally written Gamma |- t : T, where Gamma is a "typing context" a mapping from variables to their types.

Typing Relation

Informally:

Gamma x = T	(T_Var)
Gamma |- x : T

Gamma, x:T11 |- t12 : T12	(T_Abs)
Gamma |- \x:T11.t12 : T11->T12

Gamma |- t1 : T11->T12
Gamma |- t2 : T11	(T_App)
Gamma |- t1 t2 : T12

The notation Gamma, x:T means "extend the partial function Gamma to also map x to T."

Examples

Note that since we added the has_type constructors to the hints database, auto can actually solve this one immediately.

Written informally, the next one is:

Exercise: 2 stars, optional

Prove the same result without using auto, eauto, or eapply.

☐

Exercise: 2 stars (typing_example_3)

Formally prove the following typing derivation holds:

☐ We can also show that terms are not typable. For example, let's formally check that there is no typing derivation assigning a type to the term \a:A. \b:B, a b -- i.e.,

Exercise: 2 stars (typing_nonexample_2)

Now you prove this one doesn't hold:

☐

Exercise: 3 stars (typing_nonexample_3)

Another nonexample:

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Exercise: 1 star (typing_statements)

Which of the following propositions are provable?

• b:B |- \a:A.a : A->A
• exists T, empty |- (\b:B->B. \a:B. b a) : T
• exists T, empty |- (\b:B->B. \a:B. a b) : T
• exists S, a:S |- (\b:B->B. b) a : S
• exists S, exists T, a:S |- (a a a) : T

☐

Exercise: 1 star (more_typing_statements)

Which of the following propositions are provable? For the ones that are, give witnesses for the existentially bound variables.

• exists T, empty |- (\b:B->B->B. \a:B, b a) : T
• exists T, empty |- (\a:A->B, \b:B-->C, \c:A, b (a c)):T
• exists S, exists U, exists T, a:S, b:U |- (\c:A. a b c) : T
• exists S, exists T, a:S |- \b:A. a (a b) : T
• exists S, exists U, exists T, a:S |- a (\c:U. c a) : T

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Properties

Free Variables

Substitution

Exercise: 2 stars

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Preservation

Exercise: 2 stars (subject_expansion_stlc)

An exercise earlier in this file asked about the subject expansion property for the simple language of arithmetic and boolean expressions. Does this property hold for STLC? That is, is it always the case that, if t --> t' and has_type t' T, then has_type t T? If so, prove it. If not, give a counter-example. (* FILL IN HERE *)
☐

Progress

Exercise: 3 stars, optional

Show that progress can also be proved by induction on terms instead of types.

☐

Uniqueness of types

Exercise: 3 stars

Another pleasant property of the STLC is that types are unique: a given term (in a given context) has at most one type. Formalize this statement and prove it.

☐

Additional exercises

Exercise: 1 star (progress_preservation_statement)

Without peeking, write down the progress and preservation theorems for the simply typed lambda-calculus. ☐

Exercise: 2 stars (stlc_variation1)

Suppose we add the following new rule to the evaluation relation of the STLC: Which of the following properties of the STLC remain true in the presence of this rule? For each one, write either "remains true" or else "becomes false." If a property becomes false, give a counterexample.

• Determinacy of step
• Progress
• Preservation

☐

Exercise: 2 stars (stlc_variation2)

Suppose we remove the rule ST_App1 from the step relation. Which of the three properties in the previous exercise become false in the absence of this rule? For each that becomes false, give a counterexample. ☐

Exercise: STLC with Arithmetic

To see how the STLC might function as the core of a real programming language, let's extend it with a concrete base type of numbers and some constants and primitive operators.

Syntax and operational semantics

To types, we add a base type of natural numbers (and remove the other base types, for brevity)

To terms, we add natural number constants, along with successor, predecessor, and zero-testing...

Exercise: 4 stars (STLCArith)

Finish formalizing the definition and properties of the STLC extended with arithmetic. Specifically:

• Copy the whole development of STLC that we went through above (from the definition of values through the Progress theorem), and paste it into the file at this point.
• Extend the definitions of the subst operation and the step relation to include appropriate clauses for the arithmetic operators.
• Extend the proofs of all the properties of the original STLC to deal with the new syntactic forms. Make sure Coq accepts the whole file.

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Version of 4/26/2010