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

Progress

As before, 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 an evaluation step. The proof is a relatively straightforward extension of the progress proof we saw in the Types chapter.

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 know immediately that t is a value.
• If the last rule of the derivation was T_App, then t = t1 t2, and we know that t1 and t2 are also well typed in the empty context; in particular, there exists a type T2 such that |- t1 \in T2 → T and |- t2 \in T2. By the induction hypothesis, either t1 is a value or it can take an evaluation step.
  • If t1 is a value, we now consider t2, which by the other induction hypothesis must also either be a value or take an evaluation 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 was T_If, then t = if t1 then t2 else t3, where t1 has type Bool. By the IH, 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, optional (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, we need to develop some technical machinery for reasoning about variables and substitution. Working from top to bottom (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, which is defined using 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) 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, the tricky cases are the ones for variables and for function abstractions. In both cases, we discover that we need to take a term s that has been shown to be well-typed in some context Gamma and consider the same term s in a slightly different context Gamma'. For this we prove a...
• context invariance lemma, showing 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. For this, we need a careful definition of
• the free variables of a term -- i.e., the variables occuring in the term that are not in the scope of a function abstraction that binds them.

Free Occurrences

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

A term in which no variables appear free is said to be closed.

Substitution

We first need a technical lemma connecting free variables and typing contexts. 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 was 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 was 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:T11.t12, and x appears free in t12; 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 t12 is well typed under (Gamma, y:T11), so the IH allows us to conclude that x is assigned some type by the extended context (Gamma, y:T11). To conclude that Gamma assigns a type to x, we appeal to lemma extend_neq, noting that x and y are different variables.

Next, we'll need the fact that any term t which is well typed in the empty context is closed -- that is, it has no free variables.

Exercise: 2 stars, optional (typable_empty__closed)

☐ Sometimes, when we have a proof Gamma |- t : T, we will 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:T11. t12, with T = T11 → T12 and Gamma, y:T11 |- t12 \in T12. The induction hypothesis is that for any context Gamma'', if Gamma, y:T11 and Gamma'' assign the same types to all the free variables in t12, then t12 has type T12 under Gamma''. Let Gamma' be a context which agrees with Gamma on the free variables in t; we must show Gamma' |- \y:T11. t12 \in T11 → T12.
  By T_Abs, it suffices to show that Gamma', y:T11 |- t12 \in T12. By the IH (setting Gamma'' = Gamma', y:T11), it suffices to show that Gamma, y:T11 and Gamma', y:T11 agree on all the variables that appear free in t12.
  Any variable occurring free in t12 must either be y, or some other variable. Gamma, y:T11 and Gamma', y:T11 clearly agree on y. Otherwise, we note that any variable other than y which occurs free in t12 also occurs free in t = \y:T11. t12, and by assumption Gamma and Gamma' agree on all such variables, and hence so do Gamma, y:T11 and Gamma', y:T11.
• 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. However, we note that all free variables in t1 are also free in t1 t2, and similarly for free variables in t2; hence the desired result follows by the two IHs.

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 been able to assign 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 should be able to substitute v for each of the occurrences of x in t and obtain a new term that still has type T. Lemma: If Gamma,x:U |- t \in T and |- v \in U, then Gamma |- [x:=v]t \in T.

One technical subtlety in the statement of the lemma is that we assign v the type U in the empty context -- in other words, we assume v is closed. This assumption considerably simplifies the T_Abs case of the proof (compared to assuming Gamma |- v \in U, which would be the other reasonable assumption at this point) because the context invariance lemma then tells us that v has type U in any context at all -- we don't have to worry about free variables in v clashing with the variable being introduced into the context by T_Abs. Proof: We prove, by induction on t, that, for all T and Gamma, if Gamma,x:U |- 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 Gamma, x:U |- 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 context invariance: if a closed term has type T in the empty context, it has that type in any context.
  • If t is some variable y that is not equal to x, then we need only note that y has the same type under Gamma, x:U as under Gamma.
• If t is an abstraction \y:T11. t12, then the IH tells us, for all Gamma' and T', that if Gamma',x:U |- t12 \in T' and |- v \in U, then Gamma' |- [x:=v]t12 \in T'.
  The substitution in the conclusion behaves differently, depending on whether x and y are the same variable name.
  First, suppose x = y. Then, by the definition of substitution, [x:=v]t = t, so we just need to show Gamma |- t \in T. But we know Gamma,x:U |- t : T, and since the variable y does not appear free in \y:T11. t12, the context invariance lemma yields Gamma |- t \in T.
  Second, suppose x ≠ y. We know Gamma,x:U,y:T11 |- t12 \in T12 by inversion of the typing relation, and Gamma,y:T11,x:U |- t12 \in T12 follows from this by the context invariance lemma, so the IH applies, giving us Gamma,y:T11 |- [x:=v]t12 \in T12. By T_Abs, Gamma |- \y:T11. [x:=v]t12 \in T11→T12, and by the definition of substitution (noting that x ≠ y), Gamma |- \y:T11. [x:=v]t12 \in T11→T12 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.

Another technical note: This proof is a rare case where an induction on terms, rather than typing derivations, yields a simpler argument. The reason for this is that the assumption extend Gamma x U |- t \in T is not completely generic, in the sense that one of the "slots" in the typing relation -- namely the context -- is not just a variable, and this means that Coq's native induction tactic does not give us the induction hypothesis that we want. It is possible to work around this, but the needed generalization is a little tricky. The term t, on the other hand, is completely generic.

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.

Main Theorem

We now have the tools we need to prove preservation: if a closed term t has type T, and takes an evaluation step to t', then t' is also a closed term with type T. In other words, the small-step evaluation 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 the final rules in the derivation, since in each of these cases t cannot take a step.
• If the last rule in the derivation was T_App, then t = t1 t2. There are three cases to consider, one for each rule that could have been 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 IH t1' has the same type as t1, and hence t1' t2 has the same type as t1 t2.
  • The ST_App2 case is similar.
  • If t1 t2 takes a step by ST_AppAbs, then t1 = \x:T11.t12 and t1 t2 steps to [x:=t2]t12; the desired result now follows from the fact that substitution preserves types.
• If the last rule in the derivation was T_If, then t = if t1 then t2 else t3, and there are again three cases depending on how t steps.
  • If t steps to t2 or t3, 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 induction hypothesis.

Exercise: 2 stars (subject_expansion_stlc)

An exercise in the Types chapter 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 empty |- t \in T? If so, prove it. If not, give a counter-example not involving conditionals. ☐

Type Soundness

Exercise: 2 stars, 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 (types_unique)

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.

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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 a new term zap with the following reduction rule:

---

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

---

(T_Zap) Gamma |- zap : T 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 (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 (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, 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, 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, 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, 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...

Exercise: 4 stars (stlc_arith)

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 (up to soundness) of the original STLC to deal with the new syntactic forms. Make sure Coq accepts the whole file.

☐