StlcProp: Properties of STLC

In this chapter, we develop the fundamental theory of the Simply Typed Lambda Calculus -- in particular, the type safety theorem.

Canonical Forms

As we saw for the very simple language in the Types chapter, the first step in establishing basic properties of reduction and types is to identify the possible canonical forms (i.e., well-typed values) belonging to each type. For Bool, these are again the boolean values true and false; for arrow types, they are lambda-abstractions. Formally, we will need these lemmas only for terms that are not only well typed but closed -- i.e., well typed in the empty context.

Progress

The progress theorem tells us that closed, well-typed terms are not stuck: either a well-typed term is a value, or it can take a reduction step. The proof is a relatively straightforward extension of the progress proof we saw in the Types chapter. We give the proof in English first, then the formal version.

Proof: By induction on the derivation of |-- t \in T.

• The last rule of the derivation cannot be T_Var, since a variable is never well typed in an empty context.
• The T_True, T_False, and T_Abs cases are trivial, since in each of these cases we can see by inspecting the rule that t is a value.
• If the last rule of the derivation is T_App, then t has the form t1 t2 for some t1 and t2, where |-- t1 \in T2 → T and |-- t2 \in T2 for some type T2. The induction hypothesis for the first subderivation says that either t1 is a value or else it can take a reduction step.
  • If t1 is a value, then consider t2, which by the induction hypothesis for the second subderivation must also either be a value or take a step.
    • Suppose t2 is a value. Since t1 is a value with an arrow type, it must be a lambda abstraction; hence t1 t2 can take a step by ST_AppAbs.
    • Otherwise, t2 can take a step, and hence so can t1 t2 by ST_App2.
  • If t1 can take a step, then so can t1 t2 by ST_App1.
• If the last rule of the derivation is T_If, then t = if t1 then t2 else t3, where t1 has type Bool. The first IH says that t1 either is a value or takes a step.
  • If t1 is a value, then since it has type Bool it must be either true or false. If it is true, then t steps to t2; otherwise it steps to t3.
  • Otherwise, t1 takes a step, and therefore so does t (by ST_If).

Exercise: 3 stars, advanced (progress_from_term_ind)

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

☐

Preservation

The other half of the type soundness property is the preservation of types during reduction. For this part, we'll need to develop some technical machinery for reasoning about variables and substitution. Working from top to bottom (from the high-level property we are actually interested in to the lowest-level technical lemmas that are needed by various cases of the more interesting proofs), the story goes like this:

• The preservation theorem is proved by induction on a typing derivation, pretty much as we did in the Types chapter. The one case that is significantly different is the one for the ST_AppAbs rule, whose definition uses the substitution operation. To see that this step preserves typing, we need to know that the substitution itself does. So we prove a...
• substitution lemma, stating that substituting a (closed, well-typed) term s for a variable x in a term t preserves the type of t. The proof goes by induction on the form of t and requires looking at all the different cases in the definition of substitition. This time, for the variables case, we discover that we need to deduce from the fact that a term s has type S in the empty context the fact that s has type S in every context. For this we prove a...
• weakening lemma, showing that typing is preserved under "extensions" to the context Gamma.

To make Coq happy, of course, we need to formalize the story in the opposite order, starting with weakening...

The Weakening Lemma

First, we show that typing is preserved under "extensions" to the context Gamma. (Recall the definition of "includedin" from Maps.v.)

The following simple corollary is what we actually need below.

The Substitution Lemma

Now we come to the conceptual heart of the proof that reduction preserves types -- namely, the observation that substitution preserves types. Formally, the so-called substitution lemma says this: Suppose we have a term t with a free variable x, and suppose we've assigned a type T to t under the assumption that x has some type U. Also, suppose that we have some other term v and that we've shown that v has type U. Then, since v satisfies the assumption we made about x when typing t, we can substitute v for each of the occurrences of x in t and obtain a new term that still has type T. Lemma: If x|->U; Gamma |-- t \in T and |-- v \in U, then Gamma |-- [x:=v]t \in T.

The substitution lemma can be viewed as a kind of "commutation property." Intuitively, it says that substitution and typing can be done in either order: we can either assign types to the terms t and v separately (under suitable contexts) and then combine them using substitution, or we can substitute first and then assign a type to [x:=v] t ; the result is the same either way. Proof: We show, by induction on t, that for all T and Gamma, if x|->U; Gamma |-- t \in T and |-- v \in U, then Gamma |-- [x:=v]t \in T.

• If t is a variable there are two cases to consider, depending on whether t is x or some other variable.
  • If t = x, then from the fact that x|->U; Gamma |-- x \in T we conclude that U = T. We must show that [x:=v]x = v has type T under Gamma, given the assumption that v has type U = T under the empty context. This follows from the weakening lemma.
  • If t is some variable y that is not equal to x, then we need only note that y has the same type under x|->U; Gamma as under Gamma.
• If t is an abstraction \y:S, t0, then T = S→T1 and the IH tells us, for all Gamma' and T0, that if x|->U; Gamma' |-- t0 \in T0, then Gamma' |-- [x:=v]t0 \in T0. Moreover, by inspecting the typing rules we see it must be the case that y|->S; x|->U; Gamma |-- t0 \in T1.
  The substitution in the conclusion behaves differently depending on whether x and y are the same variable.
  First, suppose x = y. Then, by the definition of substitution, [x:=v]t = t, so we just need to show Gamma |-- t \in T. Using T_Abs, we need to show that y|->S; Gamma |-- t0 \in T1. But we know y|->S; x|->U; Gamma |-- t0 \in T1, and the claim follows since x = y.
  Second, suppose x ≠ y. Again, using T_Abs, we need to show that y|->S; Gamma |-- [x:=v]t0 \in T1. Since x ≠ y, we have y|->S; x|->U; Gamma = x|->U; y|->S; Gamma. So we have x|->U; y|->S; Gamma |-- t0 \in T1. Then, the the IH applies (taking Gamma' = y|->S; Gamma), giving us y|->S; Gamma |-- [x:=v]t0 \in T1, as required.
• If t is an application t1 t2, the result follows straightforwardly from the definition of substitution and the induction hypotheses.
• The remaining cases are similar to the application case.

One technical subtlety in the statement of the above lemma is that we assume v has type U in the empty context -- in other words, we assume v is closed. (Since we are using a simple definition of substition that is not capture-avoiding, it doesn't make sense to substitute non-closed terms into other terms. Fortunately, closed terms are all we need!)

Exercise: 3 stars, advanced (substitution_preserves_typing_from_typing_ind)

Show that substitution_preserves_typing can also be proved by induction on typing derivations instead of induction on terms.

☐

Main Theorem

We now have the ingredients we need to prove preservation: if a closed, well-typed term t has type T and takes a step to t', then t' is also a closed term with type T. In other words, the small-step reduction relation preserves types.

Proof: By induction on the derivation of |-- t \in T.

• We can immediately rule out T_Var, T_Abs, T_True, and T_False as final rules in the derivation, since in each of these cases t cannot take a step.
• If the last rule in the derivation is T_App, then t = t1 t2, and there are subderivations showing that |-- t1 \in T2→T and |-- t2 \in T2 plus two induction hypotheses: (1) t1 --> t1' implies |-- t1' \in T2→T and (2) t2 --> t2' implies |-- t2' \in T2. There are now three subcases to consider, one for each rule that could be used to show that t1 t2 takes a step to t'.
  • If t1 t2 takes a step by ST_App1, with t1 stepping to t1', then, by the first IH, t1' has the same type as t1 (|-- t1' \in T2→T), and hence by T_App t1' t2 has type T.
  • The ST_App2 case is similar, using the second IH.
  • If t1 t2 takes a step by ST_AppAbs, then t1 = \x:T0,t0 and t1 t2 steps to [x0:=t2]t0; the desired result now follows from the substitution lemma.
• If the last rule in the derivation is T_If, then t = if t1 then t2 else t3, with |-- t1 \in Bool, |-- t2 \in T1, and |-- t3 \in T1, and with three induction hypotheses: (1) t1 --> t1' implies |-- t1' \in Bool, (2) t2 --> t2' implies |-- t2' \in T1, and (3) t3 --> t3' implies |-- t3' \in T1.
  There are again three subcases to consider, depending on how t steps.
  • If t steps to t2 or t3 by ST_IfTrue or ST_IfFalse, the result is immediate, since t2 and t3 have the same type as t.
  • Otherwise, t steps by ST_If, and the desired conclusion follows directly from the first induction hypothesis.

Exercise: 2 stars, standard, especially useful (subject_expansion_stlc)

An exercise in the Types chapter asked about the subject expansion property for the simple language of arithmetic and boolean expressions. This property did not hold for that language, and it also fails for STLC. That is, it is not always the case that, if t --> t' and empty |-- t' \in T, then empty |-- t \in T. Show this by giving a counter-example that does not involve conditionals.

☐

Type Soundness

Exercise: 2 stars, standard, optional (type_soundness)

Put progress and preservation together and show that a well-typed term can never reach a stuck state.

☐

Uniqueness of Types

Exercise: 3 stars, standard (unique_types)

Another nice property of the STLC is that types are unique: a given term (in a given context) has at most one type.

☐

Context Invariance (Optional)

Another standard technical lemma associated with typed languages is context invariance. It states that typing is preserved under "inessential changes" to the context Gamma -- in particular, changes that do not affect any of the free variables of the term. In this section, we establish this property for our system, introducing some other standard terminology on the way. First, we need to define the free variables in a term -- i.e., variables that are used in the term in positions that are not in the scope of an enclosing function abstraction binding a variable of the same name. More technically, a variable x appears free in a term t if t contains some occurrence of x that is not under an abstraction labeled x. For example:

• y appears free, but x does not, in \x:T→U, x y
• both x and y appear free in (\x:T→U, x y) x
• no variables appear free in \x:T→U, \y:T, x y

Formally:

The free variables of a term are just the variables that appear free in it. This gives us another way to define closed terms -- arguably a better one, since it applies even to ill-typed terms. Indeed, this is the standard definition of the term "closed."

Conversely, an open term is one that may contain free variables. (I.e., every term is an open term; the closed terms are a subset of the open ones. "Open" precisely means "possibly containing free variables.")

Exercise: 1 star, standard, optional (afi)

(Officially optional, but strongly recommended!) In the space below, write out the rules of the appears_free_in relation in informal inference-rule notation. (Use whatever notational conventions you like -- the point of the exercise is just for you to think a bit about the meaning of each rule.) Although this is a rather low-level, technical definition, understanding it is crucial to understanding substitution and its properties, which are really the crux of the lambda-calculus.

☐ Next, we show that if a variable x appears free in a term t, and if we know t is well typed in context Gamma, then it must be the case that Gamma assigns a type to x.

Proof: We show, by induction on the proof that x appears free in t, that, for all contexts Gamma, if t is well typed under Gamma, then Gamma assigns some type to x.

• If the last rule used is afi_var, then t = x, and from the assumption that t is well typed under Gamma we have immediately that Gamma assigns a type to x.
• If the last rule used is afi_app1, then t = t1 t2 and x appears free in t1. Since t is well typed under Gamma, we can see from the typing rules that t1 must also be, and the IH then tells us that Gamma assigns x a type.
• Almost all the other cases are similar: x appears free in a subterm of t, and since t is well typed under Gamma, we know the subterm of t in which x appears is well typed under Gamma as well, and the IH gives us exactly the conclusion we want.
• The only remaining case is afi_abs. In this case t = \y:T1,t1 and x appears free in t1, and we also know that x is different from y. The difference from the previous cases is that, whereas t is well typed under Gamma, its body t1 is well typed under y|->T1; Gamma, so the IH allows us to conclude that x is assigned some type by the extended context y|->T1; Gamma. To conclude that Gamma assigns a type to x, we appeal to lemma update_neq, noting that x and y are different variables.

Exercise: 2 stars, standard (free_in_context)

Complete the following proof.

☐ From the free_in_context lemma, it immediately follows that any term t that is well typed in the empty context is closed (it has no free variables).

Exercise: 2 stars, standard, optional (typable_empty__closed)

☐ Finally, we establish context_invariance. It is useful in cases when we have a proof of some typing relation Gamma |-- t \in T, and we need to replace Gamma by a different context Gamma'. When is it safe to do this? Intuitively, it must at least be the case that Gamma' assigns the same types as Gamma to all the variables that appear free in t. In fact, this is the only condition that is needed.

Proof: By induction on the derivation of Gamma |-- t \in T.

• If the last rule in the derivation was T_Var, then t = x and Gamma x = T. By assumption, Gamma' x = T as well, and hence Gamma' |-- t \in T by T_Var.
• If the last rule was T_Abs, then t = \y:T2, t1, with T = T2 → T1 and y|->T2; Gamma |-- t1 \in T1. The induction hypothesis states that for any context Gamma'', if y|->T2; Gamma and Gamma'' assign the same types to all the free variables in t1, then t1 has type T1 under Gamma''. Let Gamma' be a context which agrees with Gamma on the free variables in t; we must show Gamma' |-- \y:T2, t1 \in T2 → T1.
  By T_Abs, it suffices to show that y|->T2; Gamma' |-- t1 \in T1. By the IH (setting Gamma'' = y|->T2;Gamma'), it suffices to show that y|->T2;Gamma and y|->T2;Gamma' agree on all the variables that appear free in t1.
  Any variable occurring free in t1 must be either y or some other variable. y|->T2; Gamma and y|->T2; Gamma' clearly agree on y. Otherwise, note that any variable other than y that occurs free in t1 also occurs free in t = \y:T2, t1, and by assumption Gamma and Gamma' agree on all such variables; hence so do y|->T2; Gamma and y|->T2; Gamma'.
• If the last rule was T_App, then t = t1 t2, with Gamma |-- t1 \in T2 → T and Gamma |-- t2 \in T2. One induction hypothesis states that for all contexts Gamma', if Gamma' agrees with Gamma on the free variables in t1, then t1 has type T2 → T under Gamma'; there is a similar IH for t2. We must show that t1 t2 also has type T under Gamma', given the assumption that Gamma' agrees with Gamma on all the free variables in t1 t2. By T_App, it suffices to show that t1 and t2 each have the same type under Gamma' as under Gamma. But all free variables in t1 are also free in t1 t2, and similarly for t2; hence the desired result follows from the induction hypotheses.

Exercise: 3 stars, standard, optional (context_invariance)

Complete the following proof.

☐ The context invariance lemma can actually be used in place of the weakening lemma to prove the crucial substitution lemma stated earlier.

Additional Exercises

Exercise: 1 star, standard, optional (progress_preservation_statement)

(Officially optional, but strongly recommended!) Without peeking at their statements above, write down the progress and preservation theorems for the simply typed lambda-calculus (as Coq theorems). You can write Admitted for the proofs.

☐

Exercise: 2 stars, standard (stlc_variation1)

Suppose we add a new term zap with the following reduction rule

---

(ST_Zap) t --> zap and the following typing rule:

---

(T_Zap) Gamma |-- zap \in T Which of the following properties of the STLC remain true in the presence of these rules? For each property, write either "remains true" or "becomes false." If a property becomes false, give a counterexample.

• Determinism of step

• Progress

• Preservation

☐

Exercise: 2 stars, standard (stlc_variation2)

Suppose instead that we add a new term foo with the following reduction rules:

---

(ST_Foo1) (\x:A, x) --> foo

---

(ST_Foo2) foo --> true 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.

• Determinism of step

• Progress

• Preservation

☐

Exercise: 2 stars, standard (stlc_variation3)

Suppose instead that we remove the rule ST_App1 from the step relation. 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.

• Determinism of step

• Progress

• Preservation

☐

Exercise: 2 stars, standard, optional (stlc_variation4)

Suppose instead that we add the following new rule to the reduction relation:

---

(ST_FunnyIfTrue) (if true then t1 else t2) --> true 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.

• Determinism of step

• Progress

• Preservation

☐

Exercise: 2 stars, standard, optional (stlc_variation5)

Suppose instead that we add the following new rule to the typing relation: Gamma |-- t1 \in Bool->Bool->Bool Gamma |-- t2 \in Bool

---

(T_FunnyApp) Gamma |-- t1 t2 \in Bool 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.

• Determinism of step

• Progress

• Preservation

☐

Exercise: 2 stars, standard, optional (stlc_variation6)

Suppose instead that we add the following new rule to the typing relation: Gamma |-- t1 \in Bool Gamma |-- t2 \in Bool

---

(T_FunnyApp') Gamma |-- t1 t2 \in Bool 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.

• Determinism of step

• Progress

• Preservation

☐

Exercise: 2 stars, standard, optional (stlc_variation7)

Suppose we add the following new rule to the typing relation of the STLC:

---

(T_FunnyAbs) |-- \x:Bool,t \in Bool 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.

• Determinism of step

• Progress

• Preservation

☐

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.

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

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

In this extended exercise, your job is to finish formalizing the definition and properties of the STLC extended with arithmetic. Specifically: Fill in the core definitions for STLCArith, by starting with the rules and terms which are the same as STLC. Then prove the key lemmas and theorems we provide. You will need to define and prove helper lemmas, as before. It will be necessary to also fill in "Reserved Notation", "Notation", and "Hint Constructors". Hint: If you get an error "STLC.tm" found instead of term "tm" then Coq is picking up the old notation for ie: subst instead of the new notation for STLCArith, so you need to overwrite the old with the notation before you can use it. Make sure Coq accepts the whole file before submitting.

Exercise: 5 stars, standard (STLCArith.subst)

(You'll need to remove the period at the end of this definition and add where "'' x ':=' s '' t" := (subst x s t) (in custom stlc_tm). when you fill it in.)

☐

The Technical Theorems

The next lemmas are proved exactly as before.

Exercise: 4 stars, standard (STLCArith.weakening)

☐

Exercise: 4 stars, standard (STLCArith.preservation)

☐

Exercise: 4 stars, standard (STLCArith.progress)

☐