StlcPropProperties of STLC
Require Export Stlc.
Module STLCProp.
Import STLC.
In this chapter, we develop the fundamental theory of the Simply
Typed Lambda Calculus — in particular, the type safety
theorem.
Lemma canonical_forms_bool : ∀t,
empty ⊢ t ∈ TBool →
value t →
(t = ttrue) ∨ (t = tfalse).
Proof.
intros t HT HVal.
inversion HVal; intros; subst; try inversion HT; auto.
Qed.
Lemma canonical_forms_fun : ∀t T_{1} T_{2},
empty ⊢ t ∈ (TArrow T_{1} T_{2}) →
value t →
∃x u, t = tabs x T_{1} u.
Proof.
intros t T_{1} T_{2} HT HVal.
inversion HVal; intros; subst; try inversion HT; subst; auto.
∃x0. ∃t0. auto.
Qed.
Progress
Theorem progress : ∀t T,
empty ⊢ t ∈ T →
value t ∨ ∃t', t ⇒ t'.
Proof: by induction on the derivation of ⊢ t ∈ 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 = t_{1}
t_{2}, and we know that t_{1} and t_{2} are also well typed in the
empty context; in particular, there exists a type T_{2} such that
⊢ t_{1} ∈ T_{2} → T and ⊢ t_{2} ∈ T_{2}. By the induction
hypothesis, either t_{1} is a value or it can take an evaluation
step.
- If t_{1} is a value, we now consider t_{2}, which by the other
induction hypothesis must also either be a value or take an
evaluation step.
- Suppose t_{2} is a value. Since t_{1} is a value with an
arrow type, it must be a lambda abstraction; hence t_{1}
t_{2} can take a step by ST_AppAbs.
- Otherwise, t_{2} can take a step, and hence so can t_{1}
t_{2} by ST_App2.
- Suppose t_{2} is a value. Since t_{1} is a value with an
arrow type, it must be a lambda abstraction; hence t_{1}
t_{2} can take a step by ST_AppAbs.
- If t_{1} can take a step, then so can t_{1} t_{2} by ST_App1.
- If t_{1} is a value, we now consider t_{2}, which by the other
induction hypothesis must also either be a value or take an
evaluation step.
- If the last rule of the derivation was T_If, then t = if t_{1}
then t_{2} else t_{3}, where t_{1} has type Bool. By the IH, t_{1}
either is a value or takes a step.
- If t_{1} is a value, then since it has type Bool it must be
either true or false. If it is true, then t steps
to t_{2}; otherwise it steps to t_{3}.
- Otherwise, t_{1} takes a step, and therefore so does t (by ST_If).
- If t_{1} is a value, then since it has type Bool it must be
either true or false. If it is true, then t steps
to t_{2}; otherwise it steps to t_{3}.
Proof with eauto.
intros t T Ht.
remember (@empty ty) as Γ.
has_type_cases (induction Ht) Case; subst Γ...
Case "T_Var".
(* contradictory: variables cannot be typed in an
empty context *)
inversion H.
Case "T_App".
(* t = t_{1} t_{2}. Proceed by cases on whether t_{1} is a
value or steps... *)
right. destruct IHHt1...
SCase "t_{1} is a value".
destruct IHHt2...
SSCase "t_{2} is also a value".
assert (∃x0 t0, t_{1} = tabs x0 T_{11} t0).
eapply canonical_forms_fun; eauto.
destruct H1 as [x0 [t0 Heq]]. subst.
∃([x0:=t_{2}]t0)...
SSCase "t_{2} steps".
inversion H0 as [t_{2}' Hstp]. ∃(tapp t_{1} t_{2}')...
SCase "t_{1} steps".
inversion H as [t_{1}' Hstp]. ∃(tapp t_{1}' t_{2})...
Case "T_If".
right. destruct IHHt1...
SCase "t_{1} is a value".
destruct (canonical_forms_bool t_{1}); subst; eauto.
SCase "t_{1} also steps".
inversion H as [t_{1}' Hstp]. ∃(tif t_{1}' t_{2} t_{3})...
Qed.
intros t T Ht.
remember (@empty ty) as Γ.
has_type_cases (induction Ht) Case; subst Γ...
Case "T_Var".
(* contradictory: variables cannot be typed in an
empty context *)
inversion H.
Case "T_App".
(* t = t_{1} t_{2}. Proceed by cases on whether t_{1} is a
value or steps... *)
right. destruct IHHt1...
SCase "t_{1} is a value".
destruct IHHt2...
SSCase "t_{2} is also a value".
assert (∃x0 t0, t_{1} = tabs x0 T_{11} t0).
eapply canonical_forms_fun; eauto.
destruct H1 as [x0 [t0 Heq]]. subst.
∃([x0:=t_{2}]t0)...
SSCase "t_{2} steps".
inversion H0 as [t_{2}' Hstp]. ∃(tapp t_{1} t_{2}')...
SCase "t_{1} steps".
inversion H as [t_{1}' Hstp]. ∃(tapp t_{1}' t_{2})...
Case "T_If".
right. destruct IHHt1...
SCase "t_{1} is a value".
destruct (canonical_forms_bool t_{1}); subst; eauto.
SCase "t_{1} also steps".
inversion H as [t_{1}' Hstp]. ∃(tif t_{1}' t_{2} t_{3})...
Qed.
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.Theorem progress' : ∀t T,
empty ⊢ t ∈ T →
value t ∨ ∃t', t ⇒ t'.
Proof.
intros t.
t_cases (induction t) Case; intros T Ht; auto.
(* FILL IN HERE *) Admitted.
☐
Preservation
- 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 Γ and consider the same
term s in a slightly different context Γ'. For this
we prove a...
- context invariance lemma, showing that typing is preserved
under "inessential changes" to the context Γ — 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
- 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
Inductive appears_free_in : id → tm → Prop :=
| afi_var : ∀x,
appears_free_in x (tvar x)
| afi_app1 : ∀x t_{1} t_{2},
appears_free_in x t_{1} → appears_free_in x (tapp t_{1} t_{2})
| afi_app2 : ∀x t_{1} t_{2},
appears_free_in x t_{2} → appears_free_in x (tapp t_{1} t_{2})
| afi_abs : ∀x y T_{11} t_{12},
y ≠ x →
appears_free_in x t_{12} →
appears_free_in x (tabs y T_{11} t_{12})
| afi_if1 : ∀x t_{1} t_{2} t_{3},
appears_free_in x t_{1} →
appears_free_in x (tif t_{1} t_{2} t_{3})
| afi_if2 : ∀x t_{1} t_{2} t_{3},
appears_free_in x t_{2} →
appears_free_in x (tif t_{1} t_{2} t_{3})
| afi_if3 : ∀x t_{1} t_{2} t_{3},
appears_free_in x t_{3} →
appears_free_in x (tif t_{1} t_{2} t_{3}).
Tactic Notation "afi_cases" tactic(first) ident(c) :=
first;
[ Case_aux c "afi_var"
| Case_aux c "afi_app1" | Case_aux c "afi_app2"
| Case_aux c "afi_abs"
| Case_aux c "afi_if1" | Case_aux c "afi_if2"
| Case_aux c "afi_if3" ].
Hint Constructors appears_free_in.
A term in which no variables appear free is said to be closed.
Definition closed (t:tm) :=
∀x, ¬ appears_free_in x t.
Substitution
Lemma free_in_context : ∀x t T Γ,
appears_free_in x t →
Γ ⊢ t ∈ T →
∃T', Γ x = Some T'.
Proof: We show, by induction on the proof that x appears free
in t, that, for all contexts Γ, if t is well typed
under Γ, then Γ 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 Γ we have
immediately that Γ assigns a type to x.
- If the last rule used was afi_app1, then t = t_{1} t_{2} and x
appears free in t_{1}. Since t is well typed under Γ,
we can see from the typing rules that t_{1} must also be, and
the IH then tells us that Γ 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 Γ, we
know the subterm of t in which x appears is well typed
under Γ as well, and the IH gives us exactly the
conclusion we want.
- The only remaining case is afi_abs. In this case t = \y:T_{11}.t_{12}, and x appears free in t_{12}; we also know that x is different from y. The difference from the previous cases is that whereas t is well typed under Γ, its body t_{12} is well typed under (Γ, y:T_{11}), so the IH allows us to conclude that x is assigned some type by the extended context (Γ, y:T_{11}). To conclude that Γ assigns a type to x, we appeal to lemma extend_neq, noting that x and y are different variables.
Proof.
intros x t T Γ H H0. generalize dependent Γ.
generalize dependent T.
afi_cases (induction H) Case;
intros; try solve [inversion H0; eauto].
Case "afi_abs".
inversion H1; subst.
apply IHappears_free_in in H7.
rewrite extend_neq in H7; assumption.
Qed.
intros x t T Γ H H0. generalize dependent Γ.
generalize dependent T.
afi_cases (induction H) Case;
intros; try solve [inversion H0; eauto].
Case "afi_abs".
inversion H1; subst.
apply IHappears_free_in in H7.
rewrite extend_neq in H7; assumption.
Qed.
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)
Corollary typable_empty__closed : ∀t T,
empty ⊢ t ∈ T →
closed t.
Proof.
(* FILL IN HERE *) Admitted.
empty ⊢ t ∈ T →
closed t.
Proof.
(* FILL IN HERE *) Admitted.
☐
Sometimes, when we have a proof Γ ⊢ t : T, we will need to
replace Γ by a different context Γ'. When is it safe
to do this? Intuitively, it must at least be the case that
Γ' assigns the same types as Γ to all the variables
that appear free in t. In fact, this is the only condition that
is needed.
Lemma context_invariance : ∀Γ Γ' t T,
Γ ⊢ t ∈ T →
(∀x, appears_free_in x t → Γ x = Γ' x) →
Γ' ⊢ t ∈ T.
Proof: By induction on the derivation of Γ ⊢ t ∈ T.
- If the last rule in the derivation was T_Var, then t = x
and Γ x = T. By assumption, Γ' x = T as well, and
hence Γ' ⊢ t ∈ T by T_Var.
- If the last rule was T_Abs, then t = \y:T_{11}. t_{12}, with T
= T_{11} → T_{12} and Γ, y:T_{11} ⊢ t_{12} ∈ T_{12}. The induction
hypothesis is that for any context Γ'', if Γ,
y:T_{11} and Γ'' assign the same types to all the free
variables in t_{12}, then t_{12} has type T_{12} under Γ''.
Let Γ' be a context which agrees with Γ on the
free variables in t; we must show Γ' ⊢ \y:T_{11}. t_{12} ∈
T_{11} → T_{12}.
- If the last rule was T_App, then t = t_{1} t_{2}, with Γ ⊢ t_{1} ∈ T_{2} → T and Γ ⊢ t_{2} ∈ T_{2}. One induction hypothesis states that for all contexts Γ', if Γ' agrees with Γ on the free variables in t_{1}, then t_{1} has type T_{2} → T under Γ'; there is a similar IH for t_{2}. We must show that t_{1} t_{2} also has type T under Γ', given the assumption that Γ' agrees with Γ on all the free variables in t_{1} t_{2}. By T_App, it suffices to show that t_{1} and t_{2} each have the same type under Γ' as under Γ. However, we note that all free variables in t_{1} are also free in t_{1} t_{2}, and similarly for free variables in t_{2}; hence the desired result follows by the two IHs.
Proof with eauto.
intros.
generalize dependent Γ'.
has_type_cases (induction H) Case; intros; auto.
Case "T_Var".
apply T_Var. rewrite ← H0...
Case "T_Abs".
apply T_Abs.
apply IHhas_type. intros x1 Hafi.
(* the only tricky step... the Γ' we use to
instantiate is extend Γ x T_{11} *)
unfold extend. destruct (eq_id_dec x0 x1)...
Case "T_App".
apply T_App with T_{11}...
Qed.
intros.
generalize dependent Γ'.
has_type_cases (induction H) Case; intros; auto.
Case "T_Var".
apply T_Var. rewrite ← H0...
Case "T_Abs".
apply T_Abs.
apply IHhas_type. intros x1 Hafi.
(* the only tricky step... the Γ' we use to
instantiate is extend Γ x T_{11} *)
unfold extend. destruct (eq_id_dec x0 x1)...
Case "T_App".
apply T_App with T_{11}...
Qed.
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 Γ,x:U ⊢ t ∈ T and ⊢ v ∈ U, then Γ ⊢
[x:=v]t ∈ T.
Lemma substitution_preserves_typing : ∀Γ x U t v T,
extend Γ x U ⊢ t ∈ T →
empty ⊢ v ∈ U →
Γ ⊢ [x:=v]t ∈ 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 Γ ⊢ v ∈
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
Γ, if Γ,x:U ⊢ t ∈ T and ⊢ v ∈ U, then Γ ⊢
[x:=v]t ∈ T.
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 Γ x U ⊢ t ∈ 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.
- 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 Γ, x:U ⊢ x ∈ T we
conclude that U = T. We must show that [x:=v]x = v has
type T under Γ, 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 Γ,
x:U as under Γ.
- If t = x, then from the fact that Γ, x:U ⊢ x ∈ T we
conclude that U = T. We must show that [x:=v]x = v has
type T under Γ, 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 an abstraction \y:T_{11}. t_{12}, then the IH tells us,
for all Γ' and T', that if Γ',x:U ⊢ t_{12} ∈ T'
and ⊢ v ∈ U, then Γ' ⊢ [x:=v]t_{12} ∈ T'.
- If t is an application t_{1} t_{2}, the result follows
straightforwardly from the definition of substitution and the
induction hypotheses.
- The remaining cases are similar to the application case.
Proof with eauto.
intros Γ x U t v T Ht Ht'.
generalize dependent Γ. generalize dependent T.
t_cases (induction t) Case; intros T Γ H;
(* in each case, we'll want to get at the derivation of H *)
inversion H; subst; simpl...
Case "tvar".
rename i into y. destruct (eq_id_dec x y).
SCase "x=y".
subst.
rewrite extend_eq in H2.
inversion H2; subst. clear H2.
eapply context_invariance... intros x Hcontra.
destruct (free_in_context _ _ T empty Hcontra) as [T' HT']...
inversion HT'.
SCase "x≠y".
apply T_Var. rewrite extend_neq in H2...
Case "tabs".
rename i into y. apply T_Abs.
destruct (eq_id_dec x y).
SCase "x=y".
eapply context_invariance...
subst.
intros x Hafi. unfold extend.
destruct (eq_id_dec y x)...
SCase "x≠y".
apply IHt. eapply context_invariance...
intros z Hafi. unfold extend.
destruct (eq_id_dec y z)...
subst. rewrite neq_id...
Qed.
intros Γ x U t v T Ht Ht'.
generalize dependent Γ. generalize dependent T.
t_cases (induction t) Case; intros T Γ H;
(* in each case, we'll want to get at the derivation of H *)
inversion H; subst; simpl...
Case "tvar".
rename i into y. destruct (eq_id_dec x y).
SCase "x=y".
subst.
rewrite extend_eq in H2.
inversion H2; subst. clear H2.
eapply context_invariance... intros x Hcontra.
destruct (free_in_context _ _ T empty Hcontra) as [T' HT']...
inversion HT'.
SCase "x≠y".
apply T_Var. rewrite extend_neq in H2...
Case "tabs".
rename i into y. apply T_Abs.
destruct (eq_id_dec x y).
SCase "x=y".
eapply context_invariance...
subst.
intros x Hafi. unfold extend.
destruct (eq_id_dec y x)...
SCase "x≠y".
apply IHt. eapply context_invariance...
intros z Hafi. unfold extend.
destruct (eq_id_dec y z)...
subst. rewrite neq_id...
Qed.
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
Theorem preservation : ∀t t' T,
empty ⊢ t ∈ T →
t ⇒ t' →
empty ⊢ t' ∈ T.
Proof: by induction on the derivation of ⊢ t ∈ 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 = t_{1}
t_{2}. There are three cases to consider, one for each rule that
could have been used to show that t_{1} t_{2} takes a step to t'.
- If t_{1} t_{2} takes a step by ST_App1, with t_{1} stepping to
t_{1}', then by the IH t_{1}' has the same type as t_{1}, and
hence t_{1}' t_{2} has the same type as t_{1} t_{2}.
- The ST_App2 case is similar.
- If t_{1} t_{2} takes a step by ST_AppAbs, then t_{1} =
\x:T_{11}.t_{12} and t_{1} t_{2} steps to [x:=t_{2}]t_{12}; the
desired result now follows from the fact that substitution
preserves types.
- If t_{1} t_{2} takes a step by ST_App1, with t_{1} stepping to
t_{1}', then by the IH t_{1}' has the same type as t_{1}, and
hence t_{1}' t_{2} has the same type as t_{1} t_{2}.
- If the last rule in the derivation was T_If, then t = if t_{1}
then t_{2} else t_{3}, and there are again three cases depending on
how t steps.
- If t steps to t_{2} or t_{3}, the result is immediate, since
t_{2} and t_{3} have the same type as t.
- Otherwise, t steps by ST_If, and the desired conclusion follows directly from the induction hypothesis.
- If t steps to t_{2} or t_{3}, the result is immediate, since
t_{2} and t_{3} have the same type as t.
Proof with eauto.
remember (@empty ty) as Γ.
intros t t' T HT. generalize dependent t'.
has_type_cases (induction HT) Case;
intros t' HE; subst Γ; subst;
try solve [inversion HE; subst; auto].
Case "T_App".
inversion HE; subst...
(* Most of the cases are immediate by induction,
and eauto takes care of them *)
SCase "ST_AppAbs".
apply substitution_preserves_typing with T_{11}...
inversion HT1...
Qed.
remember (@empty ty) as Γ.
intros t t' T HT. generalize dependent t'.
has_type_cases (induction HT) Case;
intros t' HE; subst Γ; subst;
try solve [inversion HE; subst; auto].
Case "T_App".
inversion HE; subst...
(* Most of the cases are immediate by induction,
and eauto takes care of them *)
SCase "ST_AppAbs".
apply substitution_preserves_typing with T_{11}...
inversion HT1...
Qed.
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 ∈ T? If so, prove it. If not, give a counter-example not involving conditionals.☐
Type Soundness
Exercise: 2 stars, optional (type_soundness)
Definition stuck (t:tm) : Prop :=
(normal_form step) t ∧ ¬ value t.
Corollary soundness : ∀t t' T,
empty ⊢ t ∈ T →
t ⇒* t' →
~(stuck t').
Proof.
intros t t' T Hhas_type Hmulti. unfold stuck.
intros [Hnf Hnot_val]. unfold normal_form in Hnf.
induction Hmulti.
(* FILL IN HERE *) Admitted.
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.(* FILL IN HERE *)
☐
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 |
(T_Zap) | |
Γ ⊢ zap : T |
- 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 |
- 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 t_{1} else t_{2}) ⇒ true |
- Determinism of step
- Progress
- Preservation
Exercise: 2 stars, optional (stlc_variation5)
Suppose instead that we add the following new rule to the typing relation:Γ ⊢ t_{1} ∈ Bool->Bool->Bool | |
Γ ⊢ t_{2} ∈ Bool | (T_FunnyApp) |
Γ ⊢ t_{1} t_{2} ∈ Bool |
- Determinism of step
- Progress
- Preservation
Exercise: 2 stars, optional (stlc_variation6)
Suppose instead that we add the following new rule to the typing relation:Γ ⊢ t_{1} ∈ Bool | |
Γ ⊢ t_{2} ∈ Bool | (T_FunnyApp') |
Γ ⊢ t_{1} t_{2} ∈ Bool |
- 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 ∈ Bool |
- Determinism of step
- Progress
- Preservation
End STLCProp.
Exercise: STLC with Arithmetic
Module STLCArith.
To types, we add a base type of natural numbers (and remove
booleans, for brevity)
Inductive ty : Type :=
| TArrow : ty → ty → ty
| TNat : ty.
To terms, we add natural number constants, along with
successor, predecessor, multiplication, and zero-testing...
Inductive tm : Type :=
| tvar : id → tm
| tapp : tm → tm → tm
| tabs : id → ty → tm → tm
| tnat : nat → tm
| tsucc : tm → tm
| tpred : tm → tm
| tmult : tm → tm → tm
| tif0 : tm → tm → tm → tm.
Tactic Notation "t_cases" tactic(first) ident(c) :=
first;
[ Case_aux c "tvar" | Case_aux c "tapp"
| Case_aux c "tabs" | Case_aux c "tnat"
| Case_aux c "tsucc" | Case_aux c "tpred"
| Case_aux c "tmult" | Case_aux c "tif0" ].
first;
[ Case_aux c "tvar" | Case_aux c "tapp"
| Case_aux c "tabs" | Case_aux c "tnat"
| Case_aux c "tsucc" | Case_aux c "tpred"
| Case_aux c "tmult" | Case_aux c "tif0" ].
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.
(* FILL IN HERE *)
☐
End STLCArith.
(* $Date: 2014-04-23 09:37:37 -0400 (Wed, 23 Apr 2014) $ *)