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Solve most of exercise sheet 1

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amethyst
Shad Amethyst 8 months ago
parent 4c5d54810b
commit 7fd0a910ae
1 changed files with 306 additions and 115 deletions
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theories/type_systems/stlc/exercises01.v
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@ -57,80 +57,44 @@ Qed. *)
Lemma step_det e e' e'':
step e e' → step e e'' → e' = e''.
Proof.
intros H_step H_step'.
(*
`induction` here seems to introduce incorrect induction hypothesis for IH_λ, IH_v, IH_a, IH_b. Those try to prove that e' = e'', when I need them to prove other equalities later introduced by `inversion H_step`.
When swapping `inversion H_step` and `induction e`,
the number of things to prove jumps to 30, and I do not know how to quickly prove the incorrect ones.
`inversion e` doesn't introduce any induction hypothesis, so it is also out of the question.
*)
induction e as [
name |
name body |
e_λ IH_λ e_v IH_v |
n |
a IH_a b IH_b
].
- inversion H_step.
- inversion H_step.
- inversion H_step.
{
rewrite <-H in H_step'.
inversion H_step'.
- reflexivity.
- inversion H7.
- val_no_step.
}
{
inversion H_step'.
- rewrite <-H4 in H3.
inversion H3.
- admit.
- val_no_step.
}
{
inversion H_step'.
- val_no_step.
- val_no_step.
- admit.
}
- inversion H_step.
- inversion H_step.
{
rewrite <-H in H_step'.
rewrite <-H1 in H_step'.
inversion H_step'.
- rewrite <-H6.
rewrite <-H2.
reflexivity.
- val_no_step.
- val_no_step.
}
{
inversion H_step'.
- rewrite <-H4 in H3.
val_no_step.
- admit.
- val_no_step.
}
{
inversion H_step'.
- rewrite <-H5 in H2.
val_no_step.
- val_no_step.
- admit.
}
Admitted.
intro H_step.
revert e''.
(* intro H_step'. *)
induction H_step as [
name e_λ e_v
| e_fn e_fn' e_v H_val H_fn_step IH
| e_fn e_arg e_arg' H_arg_step IH
| n1 n2 n_sum H_plus
| e_l e_l' e_r H_val H_l_step IH
| e_l e_r e_r' H_r_step IH
]; subst.
all: intros e' H_step'; inversion H_step'; subst; try val_no_step.
- reflexivity.
- by rewrite (IH e1').
- by rewrite (IH e2').
- reflexivity.
- by rewrite (IH e1').
- by rewrite (IH e2').
Qed.
(** Exercise 3 (LN Exercise 2): Call-by-name Semantics *)
Inductive cbn_step : expr → expr → Prop :=
| CBNStepBeta x e e' :
| CBNStepBeta x e e' :
cbn_step (App (Lam x e) e') (subst' x e' e)
(* TODO: add more constructors *)
.
| CBNStepApp e_l e_l' e_v :
cbn_step e_l e_l' →
cbn_step (App e_l e_v) (App e_l' e_v)
| CBNStepPlusR e1 e2 e2' :
cbn_step e2 e2' →
cbn_step (Plus e1 e2) (Plus e1 e2')
| CBNStepPlusL e1 e1' v :
is_val v →
cbn_step e1 e1' →
cbn_step (Plus e1 v) (Plus e1' v)
| CBNStepPlusInt n1 n2 n_sum :
(n1 + n2)%Z = n_sum →
cbn_step (Plus (LitInt n1) (LitInt n2)) (LitInt n_sum).
(* We make the eauto tactic aware of the constructors of cbn_step *)
#[global] Hint Constructors cbn_step : core.
@ -138,15 +102,38 @@ Inductive cbn_step : expr → expr → Prop :=
Lemma different_results :
∃ (e: expr) (e1 e2: expr), rtc cbn_step e e1 ∧ rtc step e e2 ∧ is_val e1 ∧ is_val e2 ∧ e1 ≠ e2.
Proof.
(* TODO: exercise *)
Admitted.
exists (App (λ: "res", (λ: "x", "res")) (App (λ: "y", "y") 0)).
exists (λ: "x", (App (λ: "y", "y") 0))%E.
exists (λ: "x", (0))%E.
split.
{
apply rtc_once.
apply CBNStepBeta.
}
split.
{
apply (rtc_l _ _ (App (λ: "res", (λ: "x", "res")) 0)).
apply StepAppR.
apply StepBeta.
unfold is_val; intuition.
apply rtc_once.
apply StepBeta.
unfold is_val; intuition.
}
split.
unfold is_val; intuition.
split.
unfold is_val; intuition.
discriminate.
Qed.
Lemma val_no_cbn_step e e':
cbn_step e e' → is_val e → False.
Proof.
(* TODO: exercise *)
Admitted.
destruct 1.
all: intuition.
Qed.
(* Same tactic as [val_no_step] but for cbn_step.*)
@ -159,10 +146,14 @@ Ltac val_no_cbn_step :=
Lemma cbn_step_det e e' e'':
cbn_step e e' → cbn_step e e'' → e' = e''.
Proof.
(* TODO: exercise *)
Admitted.
intro H_step.
revert e''.
induction H_step; subst; intros e'' H_step'; inversion H_step'; subst; try val_no_cbn_step.
all: try reflexivity.
- by rewrite (IHH_step e_l'0).
- by rewrite (IHH_step e2'0).
- by rewrite (IHH_step e1'0).
Qed.
(** Exercise 4 (LN Exercise 3): Reflexive Transitive Closure *)
Section rtc.
@ -175,20 +166,29 @@ Section rtc.
Lemma rtc_reflexive R : Reflexive (rtc R).
Proof.
unfold Reflexive.
(* TODO: exercise *)
Admitted.
apply rtc_base.
Qed.
Lemma rtc_transitive R : Transitive (rtc R).
Proof.
unfold Transitive.
(* TODO: exercise *)
Admitted.
intros x y z lhs.
induction lhs as [| u v w H_uv H_vw IH].
- intuition.
- intro H_wz.
apply IH in H_wz as H_vz.
apply (rtc_step _ u v z).
assumption.
exact H_vz.
Qed.
Lemma rtc_subrel (R: X → X → Prop) (x y : X): R x y → rtc R x y.
Proof.
(* TODO: exercise *)
Admitted.
intro H.
apply (rtc_step _ x y y).
assumption.
exact (rtc_reflexive _ _).
Qed.
Section typeclass.
@ -224,7 +224,7 @@ Qed.
Section stdpp.
(* In fact, [rtc] is so common that it is already provided by the [stdpp] library! *)
Import stdpp.relations.
Print rtc.
(* Print rtc. *)
(* The typeclass instances are also already registered. *)
Goal rtc step (LitInt 42) (LitInt 42).
@ -235,26 +235,46 @@ End stdpp.
Lemma plus_right e1 e2 e2':
rtc step e2 e2' → rtc step (Plus e1 e2) (Plus e1 e2').
Proof.
(* TODO: exercise *)
Admitted.
intro H.
induction H.
reflexivity.
eapply rtc_step.
(* eauto knows how to instance the results of step :) *)
eauto.
eauto.
Qed.
Lemma plus_left e1 e1' n:
rtc step e1 e1' → rtc step (Plus e1 (LitInt n)) (Plus e1' (LitInt n)).
Proof.
(* TODO: exercise *)
Admitted.
intro H.
induction H.
reflexivity.
eapply rtc_step; eauto.
apply StepPlusL.
unfold is_val; intuition.
assumption.
Qed.
(* The exercise: *)
Lemma plus_to_consts e1 e2 n m:
rtc step e1 (LitInt n) → rtc step e2 (LitInt m) → rtc step (e1 + e2)%E (LitInt (n + m)%Z).
Proof.
(* TODO: exercise *)
Admitted.
intro H_n.
intro H_m.
etransitivity.
(* First, reduce the rhs *)
apply plus_right.
apply H_m.
etransitivity.
(* Then, reduce the lhs *)
apply plus_left.
apply H_n.
(* Finally, reduce the addition using StepPlusRed *)
apply rtc_subrel.
eauto.
Qed.
(* Now that you have an understanding of how rtc works, we can make eauto aware of it. *)
#[local] Hint Constructors rtc : core.
@ -268,51 +288,222 @@ Abort.
(** Exercise 5 (LN Exercise 4): Big-step vs small-step semantics *)
Lemma rtc_appr e1 e2 v :
rtc step e2 (of_val v) → rtc step (App e1 e2) (App e1 (of_val v)).
Proof.
induction 1; eauto.
Qed.
Lemma rtc_appl e1 e1' v :
rtc step e1 e1' → is_val v → rtc step (App e1 v) (App e1' v).
Proof.
induction 1; eauto.
Qed.
Lemma big_step_steps e v :
big_step e v → rtc step e (of_val v).
Proof.
(* TODO: exercise *)
Admitted.
intro H_big.
induction H_big; try reflexivity.
unfold of_val.
unfold of_val in IHH_big1.
unfold of_val in IHH_big2.
apply plus_to_consts; assumption.
(* First, reduce the argument *)
transitivity (App e1 (of_val v2)).
apply rtc_appr.
assumption.
(* Then, reduce the function to a lambda *)
transitivity (App (Lam x e) (of_val v2)).
apply rtc_appl.
apply IHH_big1.
exact (is_val_of_val _).
(* Perform the β-substitution *)
apply (rtc_step _ _ (subst' x (of_val v2) e)).
eauto.
(* Finish the proof using the induction hypothesis *)
apply IHH_big3.
Qed.
Lemma is_val_to_of_val {e} :
is_val e → ∃ (v: val), e = of_val v.
Proof.
intro H.
destruct (is_val_spec e) as [ H_val' _ ].
destruct (H_val' H) as [ e_v H_val ].
exists e_v.
symmetry.
by apply of_to_val.
Qed.
Lemma steps_big_step e (v: val):
rtc step e v → big_step e v.
Proof.
(* Note how there is a coercion (automatic conversion) hidden in the
* statement of this lemma: *)
Set Printing Coercions.
(* It is sometimes very useful to temporarily print coercions if rewrites or
* destructs do not behave as expected. *)
Unset Printing Coercions.
(* TODO: exercise *)
Admitted.
intro H.
remember (of_val v) eqn:H_val in H.
induction H; subst.
apply big_step_vals.
remember ((IHrtc (eq_refl (of_val v)))) as IH.
clear HeqIH.
clear IHrtc.
(* This is ridiculous. *)
revert H0.
revert IH.
revert v.
induction H; intros v IH H_rtc.
- destruct (is_val_to_of_val H) as [ e'_v H_val ].
rewrite H_val.
eapply bs_app.
apply bs_lam.
apply big_step_vals.
rewrite <-H_val.
assumption.
- inversion IH; subst.
eapply bs_app.
apply IHstep.
exact H3.
apply big_step_steps.
all: eauto.
- inversion IH; subst.
eapply bs_app.
exact H2.
apply IHstep.
exact H3.
apply big_step_steps.
all: eauto.
- inversion IH; subst.
eauto.
- inversion IH; subst.
assert (big_step e1 (LitIntV z1)).
{
apply IHstep.
assumption.
apply big_step_steps.
assumption.
}
eauto.
- inversion IH; subst.
assert (big_step e2 (LitIntV z2)).
{
apply IHstep.
assumption.
apply big_step_steps.
assumption.
}
eauto.
Qed.
(** Exercise 6 (LN Exercise 5): left-to-right evaluation *)
Inductive ltr_step : expr → expr → Prop := .
Inductive ltr_step : expr → expr → Prop :=
| LTRStepBeta x e e' :
is_val e' →
ltr_step (App (Lam x e) e') (subst' x e' e)
| LTRStepAppL e1 e1' e2 :
ltr_step e1 e1' →
ltr_step (App e1 e2) (App e1' e2)
| LTRStepAppR e1 e2 e2' :
is_val e1 →
ltr_step e2 e2' →
ltr_step (App e1 e2) (App e1 e2')
| LTRStepPlusRed (n1 n2 n3: Z) :
(n1 + n2)%Z = n3 →
ltr_step (Plus (LitInt n1) (LitInt n2)) (LitInt n3)
| LTRStepPlusL e1 e1' e2 :
ltr_step e1 e1' →
ltr_step (Plus e1 e2) (Plus e1' e2)
| LTRStepPlusR e1 e2 e2' :
is_val e1 →
ltr_step e2 e2' →
ltr_step (Plus e1 e2) (Plus e1 e2')
.
#[global] Hint Constructors ltr_step : core.
Lemma different_steps_ltr_step :
∃ (e: expr) (e1 e2: expr), ltr_step e e1 ∧ step e e2 ∧ e1 ≠ e2.
Proof.
(* TODO: exercise *)
Admitted.
exists ((1+1) + (1+1))%E.
exists (2 + (1+1))%E.
exists ((1+1) + 2)%E.
auto.
Qed.
Lemma ltr_plus_right n e2 e2':
rtc ltr_step e2 e2' → rtc ltr_step (Plus (LitInt n) e2) (Plus (LitInt n) e2').
Proof.
induction 1; eauto.
eapply rtc_step.
apply LTRStepPlusR.
unfold is_val; intuition.
exact H.
assumption.
Qed.
Lemma ltr_plus_left e1 e1' e2:
rtc ltr_step e1 e1' → rtc ltr_step (Plus e1 e2) (Plus e1' e2).
Proof.
induction 1; eauto.
Qed.
Lemma big_step_ltr_steps e v :
big_step e v → rtc ltr_step e (of_val v).
Lemma ltr_plus_to_consts e1 e2 n m:
rtc ltr_step e1 (LitInt n)
→ rtc ltr_step e2 (LitInt m)
→ rtc ltr_step (e1 + e2)%E (LitInt (n + m)%Z).
Proof.
(* TODO: exercise *)
Admitted.
intros H_n H_m.
(* First, reduce the lhs *)
etransitivity.
apply ltr_plus_left.
apply H_n.
etransitivity.
(* Then, reduce the rhs *)
apply ltr_plus_right.
apply H_m.
(* Finally, reduce the addition using LTRStepPlusRed *)
apply rtc_subrel.
eauto.
Qed.
Lemma ltr_rtc_appr e1 e_fn v :
rtc ltr_step e_fn (of_val v) → is_val e1 → rtc ltr_step (App e1 e_fn) (App e1 (of_val v)).
Proof.
induction 1; eauto.
Qed.
Lemma ltr_rtc_appl e1 e1' e2 :
rtc ltr_step e1 e1' → rtc ltr_step (App e1 e2) (App e1' e2).
Proof.
induction 1; eauto.
Qed.
Lemma big_step_ltr_steps e v :
big_step e v → rtc ltr_step e (of_val v).
Proof.
intro H_big.
induction H_big; try reflexivity.
unfold of_val.
unfold of_val in IHH_big1.
unfold of_val in IHH_big2.
apply ltr_plus_to_consts; assumption.
(* First, reduce the argument *)
transitivity (App (Lam x e) e2).
apply ltr_rtc_appl.
assumption.
(* Then, reduce the function to a lambda *)
transitivity (App (Lam x e) (of_val v2)).
apply ltr_rtc_appr.
assumption.
unfold is_val; intuition.
(* Perform the β-substitution *)
apply (rtc_step _ _ (subst' x (of_val v2) e)).
eauto.
(* Finish the proof using the induction hypothesis *)
apply IHH_big3.
Qed.
Lemma ltr_steps_big_step e (v: val):
rtc ltr_step e v → big_step e v.

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