Library prosa.util.list

From mathcomp Require Import ssreflect ssrbool eqtype ssrnat seq path fintype bigop.
Require Import prosa.util.ssrlia prosa.util.tactics.
Require Export prosa.util.supremum.

We define a few simple operations on lists that return zero for empty lists: max0, first0, and last0.
Definition max0 := foldl maxn 0.
Definition first0 := head 0.
Definition last0 := last 0.

Additional lemmas about last0.
Section Last0.

Let xs be a non-empty sequence and x be an arbitrary element, then we prove that last0 (x::xs) = last0 xs.
  Lemma last0_cons :
    ∀ x xs,
      xs ≠ [::] →
      last0 (x::xs) = last0 xs.
  Proof.
    by induction xs; last unfold last0.
  Qed.

Similarly, let xs_r be a non-empty sequence and xs_l be any sequence, we prove that last0 (xs_l ++ xs_r) = last0 xs_r.
  Lemma last0_cat :
    ∀ xs_l xs_r,
      xs_r ≠ [::] →
      last0 (xs_l ++ xs_r) = last0 xs_r.
  Proof.
    induction xs_l; intros ? NEQ; first by done.
    simpl; rewrite last0_cons.
    - by apply IHxs_l.
    - by intros C; apply: NEQ; destruct xs_l.
  Qed.

We also prove that last0 xs = xs [| size xs -1 |] .
  Lemma last0_nth :
    ∀ xs,
      last0 xs = nth 0 xs (size xs).-1.
  Proof. by intros; rewrite nth_last. Qed.

We prove that for any non-empty sequence xs there is a sequence xsh such that xsh ++ [::last0 x] = [xs].
  Lemma last0_ex_cat :
    ∀ x xs,
      xs ≠ [::] →
      last0 xs = x →
      ∃ xsh, xsh ++ [::x] = xs.
  Proof.
    intros ? ? NEQ LAST.
    induction xs; first by done.
    destruct xs.
    - ∃ [::]; by compute in LAST; subst a.
    - feed_n 2 IHxs; try by done.
      destruct IHxs as [xsh EQ].
      by ∃ (a::xsh); rewrite //= EQ.
  Qed.

We prove that if x is the last element of a sequence xs and x satisfies a predicate, then x remains the last element in the filtered sequence.
  Lemma last0_filter :
    ∀ x xs (P : nat → bool),
      xs ≠ [::] →
      last0 xs = x →
      P x →
      last0 [seq x <- xs | P x] = x.
  Proof.
    clear; intros ? ? ? NEQ LAST PX.
    destruct (last0_ex_cat x xs NEQ LAST) as [xsh EQ]; subst xs.
    rewrite filter_cat last0_cat.
    all:rewrite //= PX //=.
  Qed.

End Last0.

Additional lemmas about max0.
Section Max0.

First we prove that max0 (x::xs) is equal to max {x, max0 xs}.
  Lemma max0_cons : ∀ x xs, max0 (x :: xs) = maxn x (max0 xs).
  Proof.
    have L: ∀ xs s x, foldl maxn s (x::xs) = maxn x (foldl maxn s xs).
    { induction xs; intros.
      - by rewrite maxnC.
      - by rewrite //= in IHxs; rewrite //= maxnC IHxs [maxn s a]maxnC IHxs maxnA [maxn s x]maxnC.
    }
    by intros; unfold max; apply L.
  Qed.

Next, we prove that if all the numbers of a seq xs are equal to a constant k, then max0 xs = k.
  Lemma max0_of_uniform_set :
    ∀ k xs,
      size xs > 0 →
      (∀ x, x \in xs → x = k) →
      max0 xs = k.
  Proof.
    intros ? ? SIZE ALL.
    induction xs; first by done.
    destruct xs.
    - rewrite /max0 //= max0n; apply ALL.
      by rewrite in_cons; apply/orP; left.
    - rewrite max0_cons IHxs; [ | by done | ].
      + by rewrite [a]ALL; [ rewrite maxnn | rewrite in_cons; apply/orP; left].
      + intros; apply ALL.
        move: H; rewrite in_cons; move ⇒ /orP [/eqP EQ | IN].
        × by subst x; rewrite !in_cons; apply/orP; right; apply/orP; left.
        × by rewrite !in_cons; apply/orP; right; apply/orP; right.
  Qed.

We prove that no element in a sequence xs is greater than max0 xs.
  Lemma in_max0_le :
    ∀ xs x, x \in xs → x ≤ max0 xs.
  Proof.
    induction xs; first by intros ?.
    intros x; rewrite in_cons; move ⇒ /orP [/eqP EQ | IN]; subst.
    - by rewrite !max0_cons leq_maxl.
    - apply leq_trans with (max0 xs); first by eauto.
      by rewrite max0_cons; apply leq_maxr.
  Qed.

We prove that for a non-empty sequence xs, max0 xs ∈ xs.
  Lemma max0_in_seq :
    ∀ xs,
      xs ≠ [::] →
      max0 xs \in xs.
  Proof.
    induction xs; first by done.
    intros _; destruct xs.
    - destruct a; simpl; first by done.
      by rewrite /max0 //= max0n in_cons eq_refl.
    - rewrite max0_cons.
      move: (leq_total a (max0 (n::xs))) ⇒ /orP [LE|LE].
      + by rewrite maxnE subnKC // in_cons; apply/orP; right; apply IHxs.
      + rewrite maxnE; move: LE; rewrite -subn_eq0; move ⇒ /eqP EQ.
        by rewrite EQ addn0 in_cons; apply/orP; left.
  Qed.

We prove a technical lemma stating that one can remove duplicating element from the head of a sequence.
  Lemma max0_2cons_eq :
    ∀ x xs,
      max0 (x::x::xs) = max0 (x::xs).
  Proof. by intros; rewrite !max0_cons maxnA maxnn. Qed.

Similarly, we prove that one can remove the first element of a sequence if it is not greater than the second element of this sequence.
  Lemma max0_2cons_le :
    ∀ x1 x2 xs,
      x1 ≤ x2 →
      max0 (x1::x2::xs) = max0 (x2::xs).
  Proof. by intros; rewrite !max0_cons maxnA [maxn x1 x2]maxnE subnKC. Qed.

We prove that max0 of a sequence xs is equal to max0 of sequence xs without 0s.
  Lemma max0_rem0 :
    ∀ xs,
      max0 ([seq x <- xs | 0 < x]) = max0 xs.
  Proof.
    induction xs; first by done.
    simpl; destruct a; simpl.
    - by rewrite max0_cons max0n.
    - by rewrite !max0_cons IHxs.
  Qed.

Note that the last element is at most the max element.
  Lemma last_of_seq_le_max_of_seq:
    ∀ xs, last0 xs ≤ max0 xs.
  Proof.
    intros xs.
    have EX: ∃ len, size xs ≤ len.
    { by ∃ (size xs). }
    move: EX ⇒ [len LE].
    generalize dependent xs; induction len.
    - by intros; move: LE; rewrite leqn0 size_eq0; move ⇒ /eqP EQ; subst.
    - intros ? SIZE.
      move: SIZE; rewrite leq_eqVlt; move ⇒ /orP [/eqP EQ| LE]; last by apply IHlen.
      destruct xs as [ | x1 xs]; first by inversion EQ.
      destruct xs as [ | x2 xs]; first by rewrite /max leq_max; apply/orP; right.
      have ->: last0 [:: x1, x2 & xs] = last0 [:: x2 & xs] by done.
      rewrite max0_cons leq_max; apply/orP; right; apply IHlen.
      move: EQ ⇒ /eqP; simpl; rewrite eqSS; move ⇒ /eqP EQ.
      by subst.
  Qed.

Let's introduce the notion of the nth element of a sequence.
  Notation "xs [| n |]" := (nth 0 xs n) (at level 30).

If any element of a sequence xs is less-than-or-equal-to the corresponding element of a sequence ys, then max0 of xs is less-than-or-equal-to max of ys.
  Lemma max_of_dominating_seq :
    ∀ xs ys,
      (∀ n, xs[|n|] ≤ ys[|n|]) →
      max0 xs ≤ max0 ys.
  Proof.
    intros xs ys.
    have EX: ∃ len, size xs ≤ len ∧ size ys ≤ len.
    { ∃ (maxn (size xs) (size ys)).
      by split; rewrite leq_max; apply/orP; [left | right].
    }
    move: EX ⇒ [len [LE1 LE2]].
    generalize dependent xs; generalize dependent ys.
    induction len; intros.
    { by move: LE1 LE2; rewrite !leqn0 !size_eq0; move ⇒ /eqP E1 /eqP E2; subst. }
    { destruct xs, ys; try done.
      { have L: ∀ xs, (∀ n, xs [| n |] = 0) → max0 xs = 0.
        { clear; intros.
          induction xs; first by done.
          rewrite max0_cons.
          apply/eqP; rewrite eqn_leq; apply/andP; split; last by done.
          rewrite geq_max; apply/andP; split.
          - by specialize (H 0); simpl in H; rewrite H.
          - rewrite leqn0; apply/eqP; apply: IHxs.
            by intros; specialize (H n.+1); simpl in H.
        }
        rewrite L; first by done.
        intros; specialize (H n0).
        by destruct n0; simpl in *; apply/eqP; rewrite -leqn0.
      }
      { rewrite !max0_cons geq_max; apply/andP; split.
        + rewrite leq_max; apply/orP; left.
          by specialize (H 0); simpl in H.
        + rewrite leq_max; apply/orP; right.
          apply IHlen; try (by rewrite ltnS in LE1, LE2).
          by intros; specialize (H n1.+1); simpl in H.
      }
    }
  Qed.

End Max0.

Additional lemmas about rem for lists.
Section RemList.

We prove that if x lies in xs excluding y, then x also lies in xs.
  Lemma rem_in :
    ∀ {X : eqType} (x y : X) (xs : seq X),
      x \in rem y xs → x \in xs.
  Proof.
    intros; induction xs; simpl in H.
    { by rewrite in_nil in H. }
    { rewrite in_cons; apply/orP.
      destruct (a == y) eqn:EQ.
      { by move: EQ ⇒ /eqP EQ; subst a; right. }
      { move: H; rewrite in_cons; move ⇒ /orP [/eqP H | H].
        - by subst a; left.
        - by right; apply IHxs.
      }
    }
  Qed.

Similarly, we prove that if x ≠ y and x lies in xs, then x lies in xs excluding y.
  Lemma in_neq_impl_rem_in :
    ∀ {X : eqType} (x y : X) (xs : seq X),
      x \in xs →
      x != y →
      x \in rem y xs.
  Proof.
    induction xs.
    { intros ?; by done. }
    { rewrite in_cons ⇒ /orP [/eqP EQ | IN]; intros NEQ.
      { rewrite -EQ //=.
        move: NEQ; rewrite -eqbF_neg ⇒ /eqP →.
        by rewrite in_cons; apply/orP; left.
      }
      { simpl; destruct (a == y) eqn:AD; first by done.
        rewrite in_cons; apply/orP; right.
        by apply IHxs.
      }
    }
  Qed.

We prove that if we remove an element x for which P x from a filter, then the size of the filter decreases by 1.
  Lemma filter_size_rem :
    ∀ {X : eqType} (x : X) (xs : seq X) (P : pred X),
      (x \in xs) →
      P x →
      size [seq y <- xs | P y] = size [seq y <- rem x xs | P y] + 1.
  Proof.
    intros; induction xs; first by inversion H.
    move: H; rewrite in_cons; move ⇒ /orP [/eqP H | H]; subst.
    { by simpl; rewrite H0 -[X in X = _]addn1 eq_refl. }
    { specialize (IHxs H); simpl in ×.
      case EQab: (a == x); simpl.
      { move: EQab ⇒ /eqP EQab; subst.
        by rewrite H0 addn1. }
      { case Pa: (P a); simpl.
        - by rewrite IHxs !addn1.
        - by rewrite IHxs.
      }
    }
  Qed.

End RemList.

Additional lemmas about sequences.
Section AdditionalLemmas.

First, we prove that x::xs = ys is a sufficient condition for x to be in ys.
  Lemma mem_head_impl :
    ∀ {X : eqType} (x : X) (xs ys : seq X),
      x::xs = ys →
      x \in ys.
  Proof.
    intros X x xs [ |y ys] EQ; first by done.
    move: EQ ⇒ /eqP; rewrite eqseq_cons ⇒ /andP [/eqP EQ _].
    by subst y; rewrite in_cons; apply/orP; left.
  Qed.

We show that if n > 0, then nth (x::xs) n = nth xs (n-1).
  Lemma nth0_cons :
    ∀ x xs n,
      n > 0 →
      nth 0 (x :: xs) n = nth 0 xs n.-1.
  Proof. by intros; destruct n. Qed.

We prove that a sequence xs of size n.+1 can be destructed into a sequence xs_l of size n and an element x such that x = xs ++ [::x].
  Lemma seq_elim_last :
    ∀ {X : Type} (n : nat) (xs : seq X),
      size xs = n.+1 →
      ∃ x xs__c, xs = xs__c ++ [:: x] ∧ size xs__c = n.
  Proof.
    intros ? ? ? SIZE.
    revert xs SIZE; induction n; intros.
    - destruct xs; first by done.
      destruct xs; last by done.
      by ∃ x, [::]; split.
    - destruct xs; first by done.
      specialize (IHn xs).
      feed IHn; first by simpl in SIZE; apply eq_add_S in SIZE.
      destruct IHn as [x__n [xs__n [EQ__n SIZE__n]]]; subst xs.
      ∃ x__n, (x :: xs__n); split; first by done.
      simpl in SIZE; apply eq_add_S in SIZE.
      rewrite size_cat //= in SIZE; rewrite addn1 in SIZE; apply eq_add_S in SIZE.
      by apply eq_S.
  Qed.

Next, we prove that x ∈ xs implies that xs can be split into two parts such that xs = xsl ++ [::x] ++ [xsr].
  Lemma in_cat :
    ∀ {X : eqType} (x : X) (xs : list X),
      x \in xs → ∃ xsl xsr, xs = xsl ++ [::x] ++ xsr.
  Proof.
    intros ? ? ? SUB.
    induction xs; first by done.
    move: SUB; rewrite in_cons; move ⇒ /orP [/eqP EQ|IN].
    - by subst; ∃ [::], xs.
    - feed IHxs; first by done.
      clear IN; move: IHxs ⇒ [xsl [xsr EQ]].
      by subst; ∃ (a::xsl), xsr.
  Qed.

We prove that for any two sequences xs and ys the fact that xs is a sub-sequence of ys implies that the size of xs is at most the size of ys.
  Lemma subseq_leq_size :
    ∀ {X : eqType} (xs ys: seq X),
      uniq xs →
      (∀ x, x \in xs → x \in ys) →
      size xs ≤ size ys.
  Proof.
    intros ? ? ? UNIQ SUB.
    have EXm: ∃ m, size ys ≤ m; first by ∃ (size ys).
    move: EXm ⇒ [m SIZEm].
    move: xs ys SIZEm UNIQ SUB.
    induction m; intros.
    { move: SIZEm; rewrite leqn0 size_eq0; move ⇒ /eqP SIZEm; subst ys.
      destruct xs; first by done.
      specialize (SUB s).
      by feed SUB; [rewrite in_cons; apply/orP; left | done].
    }
    { destruct xs as [ | x xs]; first by done.
      move: (@in_cat _ x ys) ⇒ Lem.
      feed Lem; first by apply SUB; rewrite in_cons; apply/orP; left.
      move: Lem ⇒ [ysl [ysr EQ]]; subst ys.
      rewrite !size_cat //= -addnC add1n addSn ltnS -size_cat; eapply IHm.
      - move: SIZEm; rewrite !size_cat //= ⇒ SIZE.
        by rewrite add1n addnS ltnS addnC in SIZE.
      - by move: UNIQ; rewrite cons_uniq ⇒ /andP [_ UNIQ].
      - intros a IN.
        destruct (a == x) eqn: EQ.
        { exfalso.
          move: EQ UNIQ; rewrite cons_uniq; move ⇒ /eqP EQ /andP [NIN UNIQ].
          by subst; move: NIN ⇒ /negP NIN; apply: NIN.
        }
        { specialize (SUB a).
          feed SUB; first by rewrite in_cons; apply/orP; right.
          clear IN; move: SUB; rewrite !mem_cat; move ⇒ /orP [IN| /orP [IN|IN]].
          - by apply/orP; right.
          - by exfalso; move: IN; rewrite in_cons ⇒ /orP [IN|IN]; [rewrite IN in EQ | ].
          - by apply/orP; left.
        }
    }
  Qed.

Given two sequences xs and ys, two elements x and y, and an index idx such that nth xs idx = x, nth ys idx = y, we show that the pair (x, y) is in zip xs ys.
  Lemma in_zip :
    ∀ {X Y : eqType} (xs : seq X) (ys : seq Y) (x x__d : X) (y y__d : Y),
      size xs = size ys →
      (∃ idx, idx < size xs ∧ nth x__d xs idx = x ∧ nth y__d ys idx = y) →
      (x, y) \in zip xs ys.
  Proof.
    induction xs as [ | x1 xs]; intros × EQ [idx [LT [NTHx NTHy]]]; first by done.
    destruct ys as [ | y1 ys]; first by done.
    rewrite //= in_cons; apply/orP.
    destruct idx as [ | idx]; [left | right].
    { by simpl in NTHx, NTHy; subst. }
    { simpl in NTHx, NTHy, LT.
      eapply IHxs.
      - by apply eq_add_S.
      - by ∃ idx; repeat split; eauto.
    }
  Qed.

This lemma allows us to check proposition of the form ∀ x ∈ xs, ∃ y ∈ ys, P x y using a boolean expression all P (zip xs ys).
  Lemma forall_exists_implied_by_forall_in_zip:
    ∀ {X Y : eqType} (P_bool : X × Y → bool) (P_prop : X → Y → Prop) (xs : seq X),
      (∀ x y, P_bool (x, y) ↔ P_prop x y) →
      (∃ ys, size xs = size ys ∧ all P_bool (zip xs ys) == true) →
      (∀ x, x \in xs → ∃ y, P_prop x y).
  Proof.
    intros × EQ TR x IN.
    destruct TR as [ys [SIZE ALL]].
    set (idx := index x xs).
    have x__d : Y by destruct xs, ys.
    have y__d : Y by destruct xs, ys.
    ∃ (nth y__d ys idx); apply EQ; clear EQ.
    move: ALL ⇒ /eqP/allP → //.
    eapply in_zip; first by done.
    ∃ idx; repeat split.
    - by rewrite index_mem.
    - by apply nth_index.
    - Unshelve. by done.
  Qed.

Given two sequences xs and ys of equal size and without duplicates, the fact that xs ⊆ ys implies that ys ⊆ xs.
  Lemma subseq_eq:
    ∀ {X : eqType} (xs ys : seq X),
      uniq xs →
      uniq ys →
      size xs = size ys →
      (∀ x, x \in xs → x \in ys) →
      (∀ x, x \in ys → x \in xs).
  Proof.
    intros ? ? ? UNIQ SUB.
    have EXm: ∃ m, size ys ≤ m; first by ∃ (size ys).
    move: EXm ⇒ [m SIZEm].
    move: SIZEm UNIQ SUB; move: xs ys.
    induction m; intros ? ? SIZEm UNIQx UNIQy EQ SUB a IN.
    { by move: SIZEm; rewrite leqn0 size_eq0; move ⇒ /eqP SIZEm; subst ys. }
    { destruct xs as [ | x xs].
      { by move: EQ; simpl ⇒ /eqP; rewrite eq_sym size_eq0 ⇒ /eqP EQ; subst ys. }
      { destruct (x == a) eqn:XA; first by rewrite in_cons eq_sym; apply/orP; left.
        move: XA ⇒ /negP/negP NEQ.
        rewrite in_cons eq_sym; apply/orP; right.
        specialize (IHm xs (rem x ys)); apply IHm.
        { rewrite size_rem; last by apply SUB; rewrite in_cons; apply/orP; left.
          by rewrite -EQ //=; move: SIZEm; rewrite -EQ //=. }
        { by move: UNIQx; rewrite cons_uniq ⇒ /andP [_ UNIQ]. }
        { by apply rem_uniq. }
        { rewrite size_rem; last by apply SUB; rewrite in_cons; apply/orP; left.
          by rewrite -EQ //=. }
        { intros b INb.
          apply in_neq_impl_rem_in. apply SUB. by rewrite in_cons; apply/orP; right.
          move: UNIQx. rewrite cons_uniq ⇒ /andP [NIN _].
          apply/negP ⇒ /eqP EQbx; subst.
          by move: NIN ⇒ /negP NIN; apply: NIN.
        }
        { by apply in_neq_impl_rem_in; last rewrite eq_sym. }
      }
    }
  Qed.

We prove that if no element of a sequence xs satisfies a predicate P, then filter P xs is equal to an empty sequence.
  Lemma filter_in_pred0 :
    ∀ {X : eqType} (xs : seq X) (P : pred X),
      (∀ x, x \in xs → ~~ P x) →
      filter P xs = [::].
  Proof.
    intros × ALLF.
    induction xs; first by done.
    rewrite //= IHxs; last first.
    + by intros; apply ALLF; rewrite in_cons; apply/orP; right.
    + destruct (P a) eqn:EQ; last by done.
      move: EQ ⇒ /eqP; rewrite eqb_id -[P a]Bool.negb_involutive; move ⇒ /negP T.
      exfalso; apply: T.
      by apply ALLF; apply/orP; left.
  Qed.

We show that any two elements having the same index in a sequence must be equal.
  Lemma eq_ind_in_seq :
    ∀ {X : eqType} (a b : X) (xs : seq X),
      index a xs = index b xs →
      a \in xs →
      b \in xs →
      a = b.
  Proof.
    move⇒ X a b xs EQ IN_a IN_b.
    move: (nth_index a IN_a) ⇒ EQ_a.
    move: (nth_index a IN_b) ⇒ EQ_b.
    by rewrite -EQ_a -EQ_b EQ.
  Qed.

We show that the nth element in a sequence is either in the sequence or is the default element.
  Lemma default_or_in :
    ∀ {X : eqType} (n : nat) (d : X) (xs : seq X),
      nth d xs n = d ∨ nth d xs n \in xs.
  Proof.
    intros; destruct (leqP (size xs) n).
    - by left; apply nth_default.
    - by right; apply mem_nth.
  Qed.

We show that in a unique sequence of size greater than one there exist two unique elements.
  Lemma exists_two :
    ∀ {X : eqType} (xs : seq X),
      size xs > 1 →
      uniq xs →
      ∃ a b, a ≠ b ∧ a \in xs ∧ b \in xs.
  Proof.
    move⇒ T xs GT1 UNIQ.
    (* get an element of T so that we can use nth *)
    have HEAD: ∃ x, ohead xs = Some x by elim: xs GT1 UNIQ ⇒ // a l _ _ _; ∃ a ⇒ //.
    move: (HEAD) ⇒ [x0 _].
    have GT0: 0 < size xs by apply ltn_trans with (n := 1).
    ∃ (nth x0 xs 0).
    ∃ (nth x0 xs 1).
    repeat split; try apply mem_nth ⇒ //.
    apply /eqP; apply contraNneq with (b := (0 == 1)) ⇒ // /eqP.
    by rewrite nth_uniq.
  Qed.

End AdditionalLemmas.

Additional lemmas about sorted.
Section Sorted.

We show that if [x | x ∈ xs : P x] is sorted with respect to values of some function f, then it can be split into two parts: [x | x ∈ xs : P x ∧ f x ≤ t] and [x | x ∈ xs : P x ∧ f x ≤ t].
  Lemma sorted_split :
    ∀ {X : eqType} (xs : seq X) P f t,
      sorted (fun x y ⇒ f x ≤ f y) xs →
      [seq x <- xs | P x] = [seq x <- xs | P x & f x ≤ t] ++ [seq x <- xs | P x & f x > t].
  Proof.
    clear; induction xs; intros × SORT; simpl in *; first by done.
    have TR : transitive (T:=X) (fun x y : X ⇒ f x ≤ f y).
    { intros ? ? ? LE1 LE2; ssrlia. }
    destruct (P a) eqn:Pa, (leqP (f a) t) as [R1 | R1]; simpl.
    { erewrite IHxs; first by reflexivity.
      by eapply path_sorted; eauto. }
    { erewrite (IHxs P f t); last by eapply path_sorted; eauto.
      replace ([seq x <- xs | P x & f x ≤ t]) with (@nil X); first by done.
      symmetry; move: SORT; rewrite path_sortedE // ⇒ /andP [ALL SORT].
      apply filter_in_pred0; intros ? IN; apply/negP; intros H; move: H ⇒ /andP [Px LEx].
      by move: ALL ⇒ /allP ALL; specialize (ALL _ IN); simpl in ALL; ssrlia.
    }
    { by eapply IHxs, path_sorted; eauto. }
    { by eapply IHxs, path_sorted; eauto. }
  Qed.

We show that if a sequence xs1 ++ xs2 is sorted, then both subsequences xs1 and xs2 are sorted as well.
  Lemma sorted_cat:
    ∀ {X : eqType} {R : rel X} (xs1 xs2 : seq X),
      transitive R →
      sorted R (xs1 ++ xs2) → sorted R xs1 ∧ sorted R xs2.
  Proof.
    induction xs1; intros ? TR SORT; split; try by done.
    { simpl in *; move: SORT; rewrite //= path_sortedE // all_cat ⇒ /andP [/andP [ALL1 ALL2] SORT].
      by rewrite //= path_sortedE //; apply/andP; split; last apply IHxs1 with (xs2 := xs2).
    }
    { simpl in *; move: SORT; rewrite //= path_sortedE // all_cat ⇒ /andP [/andP [ALL1 ALL2] SORT].
      by apply IHxs1.
    }
  Qed.

End Sorted.

Additional lemmas about last.
Section Last.

First, we show that the default element does not change the value of last for non-empty sequences.
  Lemma nonnil_last :
    ∀ {X : eqType} (xs : seq X) (d1 d2 : X),
      xs != [::] →
      last d1 xs = last d2 xs.
  Proof.
    induction xs; first by done.
    by intros × _; destruct xs.
  Qed.

We show that if a sequence xs contains an element that satisfies a predicate P, then the last element of filter P xs is in xs.
  Lemma filter_last_mem :
    ∀ {X : eqType} (xs : seq X) (d : X) (P : pred X),
      has P xs →
      last d (filter P xs) \in xs.
  Proof.
    induction xs; first by done.
    move ⇒ d P /hasP [x IN Px]; move: IN; rewrite in_cons ⇒ /orP [/eqP EQ | IN].
    { simpl; subst a; rewrite Px; simpl.
      destruct (has P xs) eqn:HAS.
      { by rewrite in_cons; apply/orP; right; apply IHxs. }
      { replace (filter _ _) with (@nil X).
        - by simpl; rewrite in_cons; apply/orP; left.
        - symmetry; apply filter_in_pred0; intros y IN.
          by move: HAS ⇒ /negP/negP/hasPn ALLN; apply: ALLN.
      }
    }
    { rewrite in_cons; apply/orP; right.
      simpl; destruct (P a); simpl; apply IHxs.
      all: by apply/hasP; ∃ x.
    }
  Qed.

End Last.

Function rem from ssreflect removes only the first occurrence of an element in a sequence. We define function rem_all which removes all occurrences of an element in a sequence.
Fixpoint rem_all {X : eqType} (x : X) (xs : seq X) :=
  match xs with
  | [::] ⇒ [::]
  | a :: xs ⇒
    if a == x then rem_all x xs else a :: rem_all x xs
  end.

Additional lemmas about rem_all for lists.
Section RemAllList.

First we prove that x ∉ rem_all x xs.
  Lemma nin_rem_all :
    ∀ {X : eqType} (x : X) (xs : seq X),
      ¬ (x \in rem_all x xs).
  Proof.
    intros ? ? ? IN.
    induction xs; first by done.
    apply: IHxs.
    simpl in IN; destruct (a == x) eqn:EQ; first by done.
    move: IN; rewrite in_cons; move ⇒ /orP [/eqP EQ2 | IN]; last by done.
    by subst; exfalso; rewrite eq_refl in EQ.
  Qed.

Next we show that rem_all x xs ⊆ xs.
  Lemma in_rem_all :
    ∀ {X : eqType} (a x : X) (xs : seq X),
      a \in rem_all x xs → a \in xs.
  Proof.
    intros X a x xs IN.
    induction xs; first by done.
    simpl in IN.
    destruct (a0 == x) eqn:EQ.
    - by rewrite in_cons; apply/orP; right; eauto.
    - move: IN; rewrite in_cons; move ⇒ /orP [EQ2|IN].
      + by rewrite in_cons; apply/orP; left.
      + by rewrite in_cons; apply/orP; right; auto.
  Qed.

If an element x is smaller than any element of a sequence xs, then rem_all x xs = xs.
  Lemma rem_lt_id :
    ∀ x xs,
      (∀ y, y \in xs → x < y) →
      rem_all x xs = xs.
  Proof.
    intros ? ? MIN; induction xs; first by done.
    simpl; have → : (a == x) = false.
    { apply/eqP/eqP; rewrite neq_ltn; apply/orP; right.
      by apply MIN; rewrite in_cons; apply/orP; left.
    }
    rewrite IHxs //.
    intros; apply: MIN.
    by rewrite in_cons; apply/orP; right.
  Qed.

End RemAllList.

To have a more intuitive naming, we introduce the definition of range a b which is equal to index_iota a b.+1.
Definition range (a b : nat) := index_iota a b.+1.

Additional lemmas about index_iota and range for lists.
Section IotaRange.

First, we show that iota m n can be split into two parts iota m nle and iota (m + nle) (n - nle) for any nle ≤ n.
  Lemma iotaD_impl :
    ∀ n_le m n,
      n_le ≤ n →
      iota m n = iota m n_le ++ iota (m + n_le) (n - n_le).
  Proof.
    intros × LE.
    interval_to_duration n_le n k.
    rewrite iotaD.
      by replace (_ + _ - _) with k; last ssrlia.
  Qed.

Next, we prove that index_iota a b = a :: index_iota a.+1 b for a < b.
  Remark index_iota_lt_step :
    ∀ a b,
      a < b →
      index_iota a b = a :: index_iota a.+1 b.
  Proof.
    intros ? ? LT; unfold index_iota.
    destruct b; first by done.
    rewrite ltnS in LT.
    by rewrite subSn //.
  Qed.

We prove that one can remove duplicating element from the head of a sequence by which range is filtered.
  Lemma range_filter_2cons :
    ∀ x xs k,
      [seq ρ <- range 0 k | ρ \in x :: x :: xs] =
      [seq ρ <- range 0 k | ρ \in x :: xs].
  Proof.
    intros.
    apply eq_filter; intros ?.
    repeat rewrite in_cons.
    by destruct (x0 == x), (x0 \in xs).
  Qed.

Consider a, b, and x s.t. a ≤ x < b, then filter of iota_index a b with predicate (_ == x) yields ::x.
  Lemma index_iota_filter_eqx :
    ∀ x a b,
      a ≤ x < b →
      [seq ρ <- index_iota a b | ρ == x] = [::x].
  Proof.
    intros ? ? ?.
    have EX : ∃ k, b - a ≤ k.
    { ∃ (b-a); by simpl. }
    destruct EX as [k BO].
    revert x a b BO; induction k; move ⇒ x a b BO /andP [GE LT].
    { by exfalso; move: BO; rewrite leqn0 subn_eq0; move ⇒ BO; ssrlia. }
    { destruct a.
      { destruct b; first by done.
        rewrite index_iota_lt_step //; simpl.
        destruct (0 == x) eqn:EQ.
        - move: EQ ⇒ /eqP EQ; subst x.
          rewrite filter_in_pred0 //.
          by intros x; rewrite mem_index_iota -lt0n; move ⇒ /andP [T1 _].
        - by apply IHk; ssrlia.
      }
      rewrite index_iota_lt_step; last by ssrlia.
      simpl; destruct (a.+1 == x) eqn:EQ.
      - move: EQ ⇒ /eqP EQ; subst x.
        rewrite filter_in_pred0 //.
        intros x; rewrite mem_index_iota; move ⇒ /andP [T1 _].
        by rewrite neq_ltn; apply/orP; right.
      - by rewrite IHk //; ssrlia.
    }
  Qed.

As a corollary we prove that filter of iota_index a b with predicate (_ ∈ [::x]) yields ::x.
  Corollary index_iota_filter_singl :
    ∀ x a b,
      a ≤ x < b →
      [seq ρ <- index_iota a b | ρ \in [:: x]] = [::x].
  Proof.
    intros ? ? ? NEQ.
    rewrite -{2}(index_iota_filter_eqx _ a b) //.
    apply eq_filter; intros ?.
    by repeat rewrite in_cons; rewrite in_nil Bool.orb_false_r.
  Qed.

Next we prove that if an element x is not in a sequence xs, then iota_index a b filtered with predicate (_ ∈ xs) is equal to iota_index a b filtered with predicate (_ ∈ rem_all x xs).
  Lemma index_iota_filter_inxs :
    ∀ a b x xs,
      x < a →
      [seq ρ <- index_iota a b | ρ \in xs] =
      [seq ρ <- index_iota a b | ρ \in rem_all x xs].
  Proof.
    intros a b x xs LT.
    apply eq_in_filter.
    intros y; rewrite mem_index_iota; move ⇒ /andP [LE GT].
    induction xs as [ | y' xs]; first by done.
    rewrite in_cons IHxs; simpl; clear IHxs.
    destruct (y == y') eqn:EQ1, (y' == x) eqn:EQ2; auto.
    - by exfalso; move: EQ1 EQ2 ⇒ /eqP EQ1 /eqP EQ2; subst; ssrlia.
    - by move: EQ1 ⇒ /eqP EQ1; subst; rewrite in_cons eq_refl.
    - by rewrite in_cons EQ1.
  Qed.

We prove that if an element x is a min of a sequence xs, then iota_index a b filtered with predicate (_ ∈ x::xs) is equal to x :: iota_index a b filtered with predicate (_ ∈ rem_all x xs).
  Lemma index_iota_filter_step :
    ∀ x xs a b,
      a ≤ x < b →
      (∀ y, y \in xs → x ≤ y) →
      [seq ρ <- index_iota a b | ρ \in x :: xs] =
      x :: [seq ρ <- index_iota a b | ρ \in rem_all x xs].
  Proof.
    intros x xs a b B MIN.
    have EX : ∃ k, b - a ≤ k.
    { ∃ (b-a); by simpl. } destruct EX as [k BO].
    revert x xs a b B MIN BO.
    induction k; move ⇒ x xs a b /andP [LE GT] MIN BO.
    - by move_neq_down BO; ssrlia.
    - move: LE; rewrite leq_eqVlt; move ⇒ /orP [/eqP EQ|LT].
      + subst.
        rewrite index_iota_lt_step //.
        replace ([seq ρ <- x :: index_iota x.+1 b | ρ \in x :: xs])
          with (x :: [seq ρ <- index_iota x.+1 b | ρ \in x :: xs]); last first.
        { simpl; replace (@in_mem nat x (@mem nat (seq_predType nat_eqType) (x::xs))) with true.
          all: by auto; rewrite in_cons eq_refl.
        }
        rewrite (index_iota_filter_inxs _ _ x) //; simpl.
        rewrite eq_refl.
        replace (@in_mem nat x (@mem nat (seq_predType nat_eqType) (@rem_all nat_eqType x xs))) with false; last first.
        apply/eqP; rewrite eq_sym eqbF_neg. apply/negP; apply nin_rem_all.
        reflexivity.
      + rewrite index_iota_lt_step //; last by ssrlia.
        replace ([seq ρ <- a :: index_iota a.+1 b | ρ \in x :: xs])
          with ([seq ρ <- index_iota a.+1 b | ρ \in x :: xs]); last first.
        { simpl; replace (@in_mem nat a (@mem nat (seq_predType nat_eqType) (@cons nat x xs))) with false; first by done.
          apply/eqP; rewrite eq_sym eqbF_neg.
          apply/negP; rewrite in_cons; intros C; move: C ⇒ /orP [/eqP C|C].
          - by subst; rewrite ltnn in LT.
          - by move_neq_down LT; apply MIN.
        }
        replace ([seq ρ <- a :: index_iota a.+1 b | ρ \in rem_all x xs])
          with ([seq ρ <- index_iota a.+1 b | ρ \in rem_all x xs]); last first.
        { simpl; replace (@in_mem nat a (@mem nat (seq_predType nat_eqType) (@rem_all nat_eqType x xs))) with false; first by done.
          apply/eqP; rewrite eq_sym eqbF_neg; apply/negP; intros C.
          apply in_rem_all in C.
          by move_neq_down LT; apply MIN.
        }
        by rewrite IHk //; ssrlia.
  Qed.

For convenience, we define a special case of lemma index_iota_filter_step for a = 0 and b = k.+1.
  Corollary range_iota_filter_step:
    ∀ x xs k,
      x ≤ k →
      (∀ y, y \in xs → x ≤ y) →
      [seq ρ <- range 0 k | ρ \in x :: xs] =
      x :: [seq ρ <- range 0 k | ρ \in rem_all x xs].
  Proof.
    intros x xs k LE MIN.
      by apply index_iota_filter_step; auto.
  Qed.

We prove that if x < a, then x < (filter P (index_iota a b))[idx] for any predicate P.
  Lemma iota_filter_gt:
    ∀ x a b idx P,
      x < a →
      idx < size ([seq x <- index_iota a b | P x]) →
      x < nth 0 [seq x <- index_iota a b | P x] idx.
  Proof.
    clear; intros ? ? ? ? ?.
    have EX : ∃ k, b - a ≤ k.
    { ∃ (b-a); by simpl. } destruct EX as [k BO].
    revert x a b idx P BO; induction k.
    - move ⇒ x a b idx P BO LT1 LT2.
      move: BO; rewrite leqn0; move ⇒ /eqP BO.
        by rewrite /index_iota BO in LT2; simpl in LT2.
    - move ⇒ x a b idx P BO LT1 LT2.
      case: (leqP b a) ⇒ [N|N].
      + move: N; rewrite -subn_eq0; move ⇒ /eqP EQ.
          by rewrite /index_iota EQ //= in LT2.
      + rewrite index_iota_lt_step; last by done.
        simpl in *; destruct (P a) eqn:PA.
        × destruct idx; simpl; first by done.
          apply IHk; try ssrlia.
            by rewrite index_iota_lt_step // //= PA //= in LT2.
        × apply IHk; try ssrlia.
            by rewrite index_iota_lt_step // //= PA //= in LT2.
  Qed.

End IotaRange.

A sequence xs is a prefix of another sequence ys iff there exists a sequence xs_tail such that ys is a concatenation of xs and xs_tail.
Definition prefix (T : eqType) (xs ys : seq T) := ∃ xs_tail, xs ++ xs_tail = ys.

Furthermore, every prefix of a sequence is said to be strict if it is not equal to the sequence itself.
Definition strict_prefix (T : eqType) (xs ys : seq T) :=
  ∃ xs_tail, xs_tail ≠ [::] ∧ xs ++ xs_tail = ys.