Library prosa.util.list

Require Import prosa.util.ssromega prosa.util.tactics.
From mathcomp Require Import ssreflect ssrbool eqtype ssrnat seq fintype bigop.

We define a few simple operations on lists. Namely first, last, and max.

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.

Similarly, let xs_r be a non-empty sequence and xs_l be an 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.

We also prove that last0 xs = xs [| size xs -1 |] .

  Lemma last0_nth:
    ∀ xs,
      last0 xs = nth 0 xs (size xs ).-1.

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.

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.

End Last0.

Additional lemmas about max0.

Section Max0.

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).

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.

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.

We prove that for a non-empty sequence xs, max0 xs ∈ xs.

  Lemma max0_in_seq:
    ∀ xs,
      xs ≠ [::] →
      max0 xs \in xs.

Next we prove 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).

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).

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.

Note that the last element is at most the max element.

Lemma last_of_seq_le_max_of_seq:
∀ xs, last0 xs ≤ max0 xs.

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 max of xs is less-than-or-equal-to max of ys.

  Lemma max_of_dominating_seq:
    ∀ (xs ys : seq nat),
      (∀ n, xs [|n |] ≤ ys [|n |]) →
      max0 xs ≤ max0 ys.

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:
    ∀ (T: eqType) (x y: T) (xs: seq T),
      x \in rem y xs → x \in xs.

  (* 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:
    ∀ (T: eqType) (x:T) (xs: seq T) (P: T → bool),
      (x \in xs ) →
      P x →
      size [seq y <- xs | P y ] = size [seq y <- rem x xs | P y ] + 1.

End RemList.

(* Additional lemmas about sequences. *)
Section AdditionalLemmas.

  (* First, 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.

  (* 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.

  (* 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.

  (* 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.

  (* 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,
      (∀ x, x \in xs → ~~ P x ) →
      filter P xs = [::].

End AdditionalLemmas.

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 ).

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.

If an element x1 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.

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 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.

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 ].

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 ].

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 ].

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 ].

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 ].

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 ].

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.

End IotaRange.