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 simple calculus 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 the boolean values tru and fls; 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_Tru, T_Fls, 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_Test, then t = test 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 tru or fls. If it is tru, then t steps to t₂; otherwise it steps to t₃.
  • Otherwise, t₁ takes a step, and therefore so does t (by ST_Test).

练习：3 星, 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, the tricky cases are the ones for variables and for function abstractions. In both, 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. And finally, for this, we need a careful definition of...
• 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.

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

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

Formally:

The free variables of a term are just the variables that appear free in it. A term with no free variables is said to be closed.

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.")

练习：1 星, standard (afi)

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.

Substitution

To prove that substitution preserves typing, 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 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:T₁₁.t12 and x appears free in t₁₂, 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 t₁₂ is well typed under (y⊢>T₁₁; Gamma, so the IH allows us to conclude that x is assigned some type by the extended context (y⊢>T₁₁; Gamma. To conclude that Gamma assigns a type to x, we appeal to lemma update_neq, noting that x and y are different variables.

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).

练习：2 星, standard, optional (typable_empty__closed)

Sometimes, when we have a proof of some typing relation Gamma ⊢ t \in 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 y⊢>T₁₁; Gamma ⊢ t₁₂ \in T₁₂. The induction hypothesis is that, for any context Gamma'', if y⊢>T₁₁; Gamma 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 y⊢>T₁₁; Gamma' ⊢ t₁₂ \in T₁₂. By the IH (setting Gamma'' = y⊢>T₁₁;Gamma'), it suffices to show that y⊢>T₁₁;Gamma and y⊢>T₁₁;Gamma' agree on all the variables that appear free in t₁₂.
  Any variable occurring free in t₁₂ must be either y or some other variable. y⊢>T₁₁; Gamma and y⊢>T₁₁; Gamma' clearly agree on y. Otherwise, note that any variable other than y that occurs free in t₁₂ also occurs free in t = \y:T₁₁. t₁₂, and by assumption Gamma and Gamma' agree on all such variables; hence so do y⊢>T₁₁; Gamma and y⊢>T₁₁; 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.

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.

One technical subtlety in the statement of the lemma is that we assume v has 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. 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 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 x⊢>U; Gamma as under Gamma.
• If t is an abstraction \y:T₁₁. t₁₂, then the IH tells us, for all Gamma' and T', that if x⊢>U; Gamma' ⊢ 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.
  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 x⊢>U; Gamma ⊢ t \in T, and, since y does not appear free in \y:T₁₁. t₁₂, the context invariance lemma yields Gamma ⊢ t \in T.
  Second, suppose x ≠ y. We know y⊢>T₁₁; x⊢>U; Gamma ⊢ t₁₂ \in T₁₂ by inversion of the typing relation. Since x ≠ y, we have y⊢>T₁₁; x⊢>U; Gamma = x⊢>U; y⊢>T₁₁; Gamma. So the IH applies, giving us y⊢>T₁₁; Gamma ⊢ [x:=v]t₁₂ \in T₁₂.
• 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.

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 x⊢>U; Gamma ⊢ 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.

Main Theorem

We now have the tools 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_Tru, and T_Fls 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 T₁₁→T and ⊢ t₂ \in T₁₁ plus two induction hypotheses: (1) t₁ --> t₁' implies ⊢ t₁' \in T₁₁→T and (2) t₂ --> t₂' implies ⊢ t₂' \in T₁₁. 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 T₁₁→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:T₁₁.t12 and t₁ t₂ steps to [x:=t₂]t₁₂; the desired result now follows from the substitution lemma.
• If the last rule in the derivation is T_Test, then t = test t₁ then t₂ else t₃, with ⊢ t₁ \in Bool, ⊢ t₂ \in T, and ⊢ t₃ \in T, and with three induction hypotheses: (1) t₁ --> t₁' implies ⊢ t₁' \in Bool, (2) t₂ --> t₂' implies ⊢ t₂' \in T, and (3) t₃ --> t₃' implies ⊢ t₃' \in T.
  There are again three subcases to consider, depending on how t steps.
  • If t steps to t₂ or t₃ by ST_TestTru or ST_TestFalse, the result is immediate, since t₂ and t₃ have the same type as t.
  • Otherwise, t steps by ST_Test, and the desired conclusion follows directly from the first induction hypothesis.

练习：2 星, standard, recommended (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 has_type t' 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

练习：2 星, 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

练习：3 星, 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

练习：1 星, 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.

练习：2 星, 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 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

(* 请在此处解答 *)

• Progress

(* 请在此处解答 *)

• Preservation

(* 请在此处解答 *)

练习：2 星, 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 --> tru

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

(* 请在此处解答 *)

• Progress

(* 请在此处解答 *)

• Preservation

(* 请在此处解答 *)

练习：2 星, 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

(* 请在此处解答 *)

• Progress

(* 请在此处解答 *)

• Preservation

(* 请在此处解答 *)

练习：2 星, standard, optional (stlc_variation4)

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

	(ST_FunnyTestTru)
(test tru then t1 else t2) --> tru

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

(* 请在此处解答 *)

• Progress

(* 请在此处解答 *)

• Preservation

(* 请在此处解答 *)
☐

练习：2 星, 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

(* 请在此处解答 *)

• Progress

(* 请在此处解答 *)

• Preservation

(* 请在此处解答 *)
☐

练习：2 星, 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

(* 请在此处解答 *)

• Progress

(* 请在此处解答 *)

• Preservation

(* 请在此处解答 *)
☐

练习：2 星, 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

(* 请在此处解答 *)

• Progress

(* 请在此处解答 *)

• Preservation

(* 请在此处解答 *)
☐

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.

练习：5 星, 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 susbt
• Inductive value
• Inductive step
• Inductive has_type
• Inductive appears_free_in

And five theorems, with their proofs:

• Lemma context_invariance
• Lemma free_in_context
• 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 five definitions (subst, value, step, has_type, and appears_free_in) 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.