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 closed values) belonging to each type. For Bool, these are again the boolean values true and false; for arrow types, they are lambda-abstractions.

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) 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

Typing is preserved under "extensions" to the context Gamma. (Recall the definition of "inclusion" from Maps.v.)

The following simple corollary is what we actually need below.

A 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 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. You can state your counterexample informally in words, with a brief explanation.

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.

Additional Exercises

Exercise: 1 star, standard (progress_preservation_statement)

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 ∈ T

Which of the following properties of the STLC remain truee 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

(* FILL IN HERE *)

• Progress

(* FILL IN HERE *)

• Preservation

(* FILL IN HERE *)

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

(* FILL IN HERE *)

• Progress

(* FILL IN HERE *)

• Preservation

(* FILL IN HERE *)

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

(* FILL IN HERE *)

• Progress

(* FILL IN HERE *)

• Preservation

(* FILL IN HERE *)

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

(* FILL IN HERE *)

• Progress

(* FILL IN HERE *)

• Preservation

(* FILL IN HERE *)
☐

Exercise: 2 stars, standard, optional (stlc_variation5)

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

Gamma ⊢ t1 ∈ Bool->Bool->Bool
Gamma ⊢ t2 ∈ Bool	(T_FunnyApp)
Gamma ⊢ t1 t2 ∈ 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

(* FILL IN HERE *)

• Progress

(* FILL IN HERE *)

• Preservation

(* FILL IN HERE *)
☐

Exercise: 2 stars, standard, optional (stlc_variation6)

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

Gamma ⊢ t1 ∈ Bool
Gamma ⊢ t2 ∈ Bool	(T_FunnyApp')
Gamma ⊢ t1 t2 ∈ 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

(* FILL IN HERE *)

• Progress

(* FILL IN HERE *)

• Preservation

(* FILL IN HERE *)
☐

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 ∈ 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

(* FILL IN HERE *)

• Progress

(* FILL IN HERE *)

• Preservation

(* FILL IN HERE *)
☐

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: 5 stars, standard (stlc_arith)

Finish formalizing the definition and properties of the STLC extended with arithmetic. This is a longer exercise. Specifically: 1. Copy the core definitions for STLC that we went through, as well as the key lemmas and theorems, and paste them into the file at this point. Do not copy examples, exercises, etc. (In particular, make sure you don't copy any of the ☐ comments at the end of exercises, to avoid confusing the autograder.) You should copy over five definitions:

• Fixpoint subst
• Inductive value
• Inductive step
• Inductive has_type
• Inductive appears_free_in

And five theorems, with their proofs:

• Lemma weakening
• Lemma weakening_empty
• Lemma substitution_preserves_typing
• Theorem preservation
• Theorem progress

It will be helpful to also copy over "Reserved Notation", "Notation", and "Hint Constructors" for these things. 2. Edit and extend the four definitions (subst, value, step, and has_type) so they are appropriate for the new STLC extended with arithmetic. 3. Extend the proofs of all the five properties of the original STLC to deal with the new syntactic forms. Make sure Coq accepts the whole file.