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A funky grammar

You are given the following context-free grammar that produces a language L:

S -> | SS | aSaSb | aSbSa | bSaSa

It is easy to see that this grammar produces exactly the language M of all words over a, b that contains twice as many a's as b's. But is it easy to prove?

We claim that L ⊆ M is while proving M ⊆ L is not, so we give you some guidance: Assume w ∈ M. We prove w ∈ L by induction on the length of w.

For the induction step, consider w ∈ M with positive length and assume that the claim holds for all w of shorter length (IH).

Consider the function h(w):=2 * |w|b − |w|a. We have h(w)=0 iff w ∈ M. If h is zero for any prefix of w, then it is easy to establish w ∈ L with (IH). So assume the contrary for the remainder of the proof.

Now consider three cases:

  1. w = bu. Thus h(u)= − 2. As the value of h for w does not have a zero, starts with b, and can only decrease by 1, we have u = vaa for some v. This means h(v)=0. Use the last production to show that the word is in the grammar.
  2. w = ub. Analogous.
  3. w = aua. As w does not have a zero, we know that h(v) never takes the value 1 for any prefix of u. With h(u)=2, we can conclude that u = v1bv2 for v1, v2 with h(v1)=h(v2)=0. Now use the fourth production to show w ∈ L.

You are given four tasks:

  1. Show L ⊆ M
  2. Prove a theorem that helps you to establish the essential step in (a), (b).
  3. Prove a theorem that helps you to establish the essential step in (c).
  4. Show M ⊆ L. You may want to exploit the symmetry of (a) and (b).

Resources

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Definitions File

theory Defs
  imports Main
begin

section ‹Grammars and Languages›

text ‹‹#w⇘c⇙› denotes the number of times the character ‹c› occurs in word ‹w››
abbreviation count :: "'a list ⇒ 'a ⇒ nat" ("#_⇘_⇙") where
  "count xs x ≡ count_list xs x"

abbreviation
  "a ≡ CHR ''a''"

abbreviation
  "b ≡ CHR ''b''"

definition prefix :: "'a list ⇒ 'a list ⇒ bool" (infix "≺" 50) where
  "prefix xs ys ≡ ∃ zs. ys = xs @ zs ∧ zs ≠ []"


section ‹Problem statement›

text ‹Consider the following grammar:›
inductive_set L :: "string set" where
  "[] ∈ L"
| "x @ y ∈ L" if "x ∈ L" "y ∈ L"
| "a # x @ a # y @ [b] ∈ L" if "x ∈ L" "y ∈ L"
| "a # x @ b # y @ [a] ∈ L" if "x ∈ L" "y ∈ L"
| "b # x @ a # y @ [a] ∈ L" if "x ∈ L" "y ∈ L"
text ‹Q: Can we delete any of these rules?›

definition
  "M = {w. set w ⊆ {a, b} ∧ #w⇘a⇙ = 2 * #w⇘b⇙}"

text ‹
  We want to prove that the grammar exactly produces the language of all words that contain
  twice as many a's than b's:›
lemma
  "L = M"
  oops

text ‹The height function›
definition
  "h w = 2 * int (#w⇘b⇙) - int (#w⇘a⇙)"

end

Template File

theory Submission
  imports Defs
begin

text ‹First task›
theorem L_sub: "L ⊆ M"
  sorry

text ‹Second task›
theorem h_first_zero:
  assumes "h (xs @ ys) < 0" "h xs ≥ 0"
  shows "∃as bs. ys = as @ bs ∧ h (xs @ as) = 0"
  sorry

text ‹Third task›
theorem h_2_split:
  assumes "h w = 2" "∀v. v ≺ w ⟶ h v ≠ 1"
  obtains x y where "h x = 0" "h y = 0" "w = x @ b # y"
  sorry

text ‹Fourth task›
theorem L_sup:
  assumes h_first_zero:
    "⋀xs ys. h (xs @ ys) < 0 ⟹ 0 ≤ h xs ⟹ ∃as bs. ys = as @ bs ∧ h (xs @ as) = 0"
  assumes h_2_split:
    "⋀t w. h w = 2 ⟹ ∀v. v ≺ w ⟶ h v ≠ 1 ⟹ (⋀x y. h x = 0 ⟹ h y = 0 ⟹ w = x @ b # y ⟹ t)
      ⟹ t"
  assumes "h w = 0" "set w ⊆ {a, b}"
  shows "w ∈ L"
  sorry

text ‹The final theorem›
corollary
  "L = M"
  using L_sub L_sup[OF h_first_zero h_2_split] unfolding h_def M_def by auto

end

Check File

theory Check
  imports Submission
begin

text ‹First task›
theorem L_sub: "L ⊆ M"
  by (rule Submission.L_sub)

text ‹Second task›
theorem h_first_zero:
  assumes "h (xs @ ys) < 0" "h xs ≥ 0"
  shows "∃as bs. ys = as @ bs ∧ h (xs @ as) = 0"
  using assms by (rule Submission.h_first_zero)

text ‹Third task›
theorem h_2_split:
  assumes "h w = 2" "∀v. v ≺ w ⟶ h v ≠ 1"
  obtains x y where "h x = 0" "h y = 0" "w = x @ b # y"
  using assms by (rule Submission.h_2_split)

text ‹Fourth task›
theorem L_sup:
  assumes h_first_zero:
    "⋀xs ys. h (xs @ ys) < 0 ⟹ 0 ≤ h xs ⟹ ∃as bs. ys = as @ bs ∧ h (xs @ as) = 0"
  assumes h_2_split:
    "⋀t w. h w = 2 ⟹ ∀v. v ≺ w ⟶ h v ≠ 1 ⟹ (⋀x y. h x = 0 ⟹ h y = 0 ⟹ w = x @ b # y ⟹ t)
      ⟹ t"
  assumes "h w = 0" "set w ⊆ {a, b}"
  shows "w ∈ L"
  using assms by (rule Submission.L_sup)

text ‹The final theorem›
corollary
  "L = M"
  using L_sub L_sup[OF h_first_zero h_2_split] unfolding h_def M_def by auto

end
Download Files

Definitions File

Require Export Ascii List.
Require Export Arith ZArith Lia.
Require Import ssreflect.
Export ListNotations.
Open Scope char_scope.

(** In this challenge, we assume classical logic. *)
Require Export Classical.

(** A simple formalization of (possibly infinite) sets as predicates. *)
Notation set A := (A -> Prop) (only parsing).

Definition Incl {U} (A B: set U) : Prop := forall x, A x -> B x.
Definition Union {U} (A B: set U): set U := fun x => A x \/ B x.
Definition Single {U} a : set U := fun x => x = a.
Definition Pair {U} a b : set U := fun x => x = a \/ x = b.
Definition Empty {U} := fun (x: U) => False.

Notation "A  ⊆  B" := (Incl A B) (at level 70).
Notation "A  ∪  B" := (Union A B) (at level 50).
Notation "'{ x }" := (Single x) (at level 1).
Notation "∅" := Empty.

Axiom set_ext : forall U (A B: set U), (forall x, A x <-> B x) -> A = B.

(* Simple goals involving sets can be solved by the [firstorder] tactic.
   This should be enough for our needs here. *)
Goal forall U (A B: set U) a b,
  '{a} ∪ ∅ ∪ B ∪ (A ∪ '{b} ∪ B) ⊆ Pair a b ∪ A ∪ B.
Proof. firstorder. (* boom. *) Qed.

(** Definitions on words (as lists of characters). *)

Notation word := (list ascii).

(* [#{c} w] denotes the number of times the character [c] occurs in the word [w]. *)
Notation "'#' '{' c '}'" := (fun w => count_occ ascii_dec w c) (at level 1).

Eval compute in #{"a"} ["a"; "b"; "c"]. (* there is one "a" in "abc" *)

Definition a := "a".
Definition b := "b".

(* The strict prefix relation *)
Definition prefix {A} (xs ys: list A) :=
  exists zs, ys = xs ++ zs /\ zs <> [].

Notation "xs  ≺  ys" := (prefix xs ys) (at level 50).

(* The set of letters in a word *)
Definition letters (w: word): set ascii := fun c => List.In c w.

Lemma letters_nil : letters [] = ∅.
Proof. reflexivity. Qed.
Lemma letters_cons x xs : letters (x :: xs) = '{x} ∪ letters xs.
Proof. rewrite /letters /Union /Single //=. apply set_ext. firstorder. Qed.
Lemma letters_app xs ys : letters (xs ++ ys) = letters xs ∪ letters ys.
Proof. rewrite /letters. apply set_ext. intro. rewrite in_app_iff //=. Qed.

(* Can be used by ssreflect's rewrite to rewrite with any of the lemmas in the
   tuple: use "rewrite lettersE" *)
Definition lettersE := (letters_nil, letters_cons, letters_app).

(** Problem statement *)

(* Consider the following grammar, which inductively defines a set of words: *)
Inductive L : set word :=
  | L_nil : L []
  | L_app : forall x y, L x -> L y ->
      L (x ++ y)
  | L_aab : forall x y, L x -> L y ->
      L (a :: x ++ a :: y ++ [b])
  | L_aba : forall x y, L x -> L y ->
      L (a :: x ++ b :: y ++ [a])
  | L_baa : forall x y, L x -> L y ->
      L (b :: x ++ a :: y ++ [a]).

(* We want to prove that the grammar exactly produces the language of all words
   that contain twice as many a's than b's: *)

Definition M : set word :=
  fun w => letters w ⊆ Pair a b /\ #{a} w = 2 * #{b} w.

Goal L = M.
Abort.

(* The height function. w is in M iff h(w) = 0. *)
Definition h (w: word): Z :=
  2 * Z.of_nat (#{b} w) - Z.of_nat (#{a} w).

Template File

Require Import Defs.

(* It is not required, but it is a good idea to import ssreflect to benefit
   from the better rewrite tactic. (by uncommenting the two lines below)

   ssreflect's rewrite cheatsheet: a rewrite invocation is of the form [rewrite foo bar baz],
   where each of "foo", "bar", "baz" can be:
   - "foo" where foo is a lemma: rewrites with the lemma ("-foo" rewrites in the other direction)
   - "!foo" where foo is a lemma: rewrites repeatedly with foo
   - "/foo" where foo is a definition: unfolds the definition ("-/foo" folds the definition)
   - "//": try to prove the goal or side-condition if it is trivial
   - "//=": like "//" but also simplify the goal (using "simp")
   - "(_: foo = bar)": ask the user to prove "foo = bar" as a subgoal, and rewrite with it
*)
(*
Require Import ssreflect.
Local Ltac done ::= first [ solve [ trivial | eauto | lia ] ].
*)

(* Prevent simpl/cbn from unfolding the multiplication, which is never a good
   idea. *)
Arguments Nat.mul : simpl never.
Arguments Z.mul : simpl never.


(** First task *)

Theorem L_sub : L ⊆ M.
Proof.
  (* todo *)
Admitted.


(** Second task *)

Theorem h_first_zero xs ys :
  (h (xs ++ ys) < 0)%Z ->
  (0 <= h xs)%Z ->
  exists zs ws, ys = zs ++ ws /\ h (xs ++ zs) = 0%Z.
Proof.
  (* todo *)
Admitted.


(** Third task *)

(* Hint: this is useful to reason by well-founded induction on words. *)
Definition word_len_lt := ltof _ (@length ascii).
Lemma word_len_wf : well_founded word_len_lt.
Proof. apply well_founded_ltof. Qed.

Definition smallest_word_st (P: word -> Prop) (w: word) :=
  P w /\ forall v, P v -> length w <= length v.

(* If [P] holds on some word [w], then we can consider "the" smallest word that
   satisfies [P]... *)
Lemma ex_smallest_word_st : forall (P: set word) (w: word),
  P w -> exists v, smallest_word_st P v.
Proof.
  (* todo *)
Admitted.

Theorem h_2_split w :
  h w = 2%Z ->
  (forall v, v ≺ w -> h v <> 1%Z) ->
  exists x y, h x = 0%Z /\ h y = 0%Z /\ w = x ++ [b] ++ y.
Proof.
  (* todo *)
Admitted.


(** Fourth task *)

Section L_sup.

Hypothesis h_first_zero : forall xs ys,
  (h (xs ++ ys) < 0)%Z ->
  (0 <= h xs)%Z ->
  exists zs ws, ys = zs ++ ws /\ h (xs ++ zs) = 0%Z.

Hypothesis h_2_split : forall w,
  h w = 2%Z ->
  (forall v, v ≺ w -> h v <> 1%Z) ->
  exists x y, h x = 0%Z /\ h y = 0%Z /\ w = x ++ [b] ++ y.

Theorem L_sup w :
  h w = 0%Z ->
  letters w ⊆ Pair a b ->
  L w.
Proof.
  (* todo *)
Admitted.

End L_sup.

(** The final theorem *)

Theorem final : L = M.
Proof.
  (* todo *)
Admitted.
Download Files

Definitions File

theory Defs
  imports Main
begin

section ‹Grammars and Languages›

text ‹‹#w⇘c⇙› denotes the number of times the character ‹c› occurs in word ‹w››
abbreviation count :: "'a list ⇒ 'a ⇒ nat" ("#_⇘_⇙") where
  "count xs x ≡ count_list xs x"

abbreviation
  "a ≡ CHR ''a''"

abbreviation
  "b ≡ CHR ''b''"

definition prefix :: "'a list ⇒ 'a list ⇒ bool" (infix "≺" 50) where
  "prefix xs ys ≡ ∃ zs. ys = xs @ zs ∧ zs ≠ []"


section ‹Problem statement›

text ‹Consider the following grammar:›
inductive_set L :: "string set" where
  "[] ∈ L"
| "x @ y ∈ L" if "x ∈ L" "y ∈ L"
| "a # x @ a # y @ [b] ∈ L" if "x ∈ L" "y ∈ L"
| "a # x @ b # y @ [a] ∈ L" if "x ∈ L" "y ∈ L"
| "b # x @ a # y @ [a] ∈ L" if "x ∈ L" "y ∈ L"
text ‹Q: Can we delete any of these rules?›

definition
  "M = {w. set w ⊆ {a, b} ∧ #w⇘a⇙ = 2 * #w⇘b⇙}"

text ‹
  We want to prove that the grammar exactly produces the language of all words that contain
  twice as many a's than b's:›
lemma
  "L = M"
  oops

text ‹The height function›
definition
  "h w = 2 * int (#w⇘b⇙) - int (#w⇘a⇙)"

end

Template File

theory Submission
  imports Defs
begin

text ‹First task›
theorem L_sub: "L ⊆ M"
  sorry

text ‹Second task›
theorem h_first_zero:
  assumes "h (xs @ ys) < 0" "h xs ≥ 0"
  shows "∃as bs. ys = as @ bs ∧ h (xs @ as) = 0"
  sorry

text ‹Third task›
theorem h_2_split:
  assumes "h w = 2" "∀v. v ≺ w ⟶ h v ≠ 1"
  obtains x y where "h x = 0" "h y = 0" "w = x @ b # y"
  sorry

text ‹Fourth task›
theorem L_sup:
  assumes h_first_zero:
    "⋀xs ys. h (xs @ ys) < 0 ⟹ 0 ≤ h xs ⟹ ∃as bs. ys = as @ bs ∧ h (xs @ as) = 0"
  assumes h_2_split:
    "⋀t w. h w = 2 ⟹ ∀v. v ≺ w ⟶ h v ≠ 1 ⟹ (⋀x y. h x = 0 ⟹ h y = 0 ⟹ w = x @ b # y ⟹ t)
      ⟹ t"
  assumes "h w = 0" "set w ⊆ {a, b}"
  shows "w ∈ L"
  sorry

text ‹The final theorem›
corollary
  "L = M"
  using L_sub L_sup[OF h_first_zero h_2_split] unfolding h_def M_def by auto

end

Check File

theory Check
  imports Submission
begin

text ‹First task›
theorem L_sub: "L ⊆ M"
  by (rule Submission.L_sub)

text ‹Second task›
theorem h_first_zero:
  assumes "h (xs @ ys) < 0" "h xs ≥ 0"
  shows "∃as bs. ys = as @ bs ∧ h (xs @ as) = 0"
  using assms by (rule Submission.h_first_zero)

text ‹Third task›
theorem h_2_split:
  assumes "h w = 2" "∀v. v ≺ w ⟶ h v ≠ 1"
  obtains x y where "h x = 0" "h y = 0" "w = x @ b # y"
  using assms by (rule Submission.h_2_split)

text ‹Fourth task›
theorem L_sup:
  assumes h_first_zero:
    "⋀xs ys. h (xs @ ys) < 0 ⟹ 0 ≤ h xs ⟹ ∃as bs. ys = as @ bs ∧ h (xs @ as) = 0"
  assumes h_2_split:
    "⋀t w. h w = 2 ⟹ ∀v. v ≺ w ⟶ h v ≠ 1 ⟹ (⋀x y. h x = 0 ⟹ h y = 0 ⟹ w = x @ b # y ⟹ t)
      ⟹ t"
  assumes "h w = 0" "set w ⊆ {a, b}"
  shows "w ∈ L"
  using assms by (rule Submission.L_sup)

text ‹The final theorem›
corollary
  "L = M"
  using L_sub L_sup[OF h_first_zero h_2_split] unfolding h_def M_def by auto

end

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