Library prosa.util.sum

From mathcomp Require Import ssreflect ssrbool eqtype ssrnat seq fintype bigop path.

Require Export prosa.util.notation.
Require Export prosa.util.nat.

Section SumsOverSequences.

Consider any type I with a decidable equality ...
  Variable (I : eqType).

... and assume we are given a sequence ...
  Variable (r : seq I).

... and a predicate P.
  Variable (P : pred I).

First, we will show some properties of the sum performed over a single function yielding natural numbers.
  Section SumOfOneFunction.

Consider any function that yields natural numbers...
    Variable (F : I → nat).

We start showing that having every member of r equal to zero is equivalent to having the sum of all the elements of r equal to zero, and vice-versa.
    Lemma sum_nat_eq0_nat :
      all (fun x ⇒ F x == 0) r = (\sum_(i <- r) F i == 0).
In the same way, if at least one element of r is not zero, then the sum of all elements of r must be strictly positive, and vice-versa.
    Lemma sum_seq_gt0P:
      reflect (∃ i, i \in r ∧ 0 < F i) (0 < \sum_(i <- r) F i).
Next, we show that if a number a is not contained in r, then filtering or not filtering a when summing leads to the same result.
    Lemma sum_notin_rem_eqn:
      ∀ a,
        a \notin r →
        \sum_(x <- r | P x && (x != a)) F x = \sum_(x <- r | P x) F x.
We prove that if any element of r is bounded by constant c, then the sum of the whole set is bounded by c × size r.
    Lemma sum_majorant_constant:
      ∀ c,
        (∀ a, a \in r → P a → F a ≤ c) →
        \sum_(j <- r | P j) F j ≤ c × (size [seq j <- r | P j]).
Next, we show that the sum of the elements in r respecting P can be obtained by removing from the total sum over r the sum of the elements in r not respecting P.
    Lemma sum_pred_diff:
      \sum_(r <- r | P r) F r =
      \sum_(r <- r) F r - \sum_(r <- r | ~~ P r) F r.

Summing natural numbers over a superset can only yields a greater sum. Requiring the absence of duplicate in r is a simple way to guarantee that the set inclusion r ≤ rs implies the actually required multiset inclusion.
    Lemma leq_sum_sub_uniq :
      ∀ (rs: seq I),
        uniq r →
        {subset r ≤ rs} →
        \sum_(i <- r) F i ≤ \sum_(i <- rs) F i.

  End SumOfOneFunction.

In this section, we show some properties of the sum performed over two different functions.
  Section SumOfTwoFunctions.

Consider three functions that yield natural numbers.
    Variable (E E1 E2 : I → nat).

Besides earlier introduced predicate P, we add two additional predicates P1 and P2.
    Variable (P1 P2 : pred I).

Assume that E2 dominates E1 in all the points contained in the set r and respecting the predicate P. We prove that, if we sum both function over those points, then the sum of E2 will dominate the sum of E1.
    Lemma leq_sum_seq :
      (∀ i, i \in r → P i → E1 i ≤ E2 i) →
      \sum_(i <- r | P i) E1 i ≤ \sum_(i <- r | P i) E2 i.

In the same way, if E1 equals E2 in all the points considered above, then also the two sums will be identical.
    Lemma eq_sum_seq:
      (∀ i, i \in r → P i → E1 i == E2 i) →
      \sum_(i <- r | P i) E1 i == \sum_(i <- r | P i) E2 i.

Assume that P1 implies P2 in all the points contained in the set r. We prove that, if we sum both functions over those points, then the sum of E conditioned by P2 will dominate the sum of E conditioned by P1.
    Lemma leq_sum_seq_pred:
      (∀ i, i \in r → P1 i → P2 i) →
      \sum_(i <- r | P1 i) E i ≤ \sum_(i <- r | P2 i) E i.

Next, we prove that if for any element x of a set xs the following two statements hold (1) F1 x is less than or equal to F2 x and (2) the sum F1 x_1, ..., F1 x_n is equal to the sum of F2 x_1, ..., F2 x_n, then F1 x is equal to F2 x for any element x of xs.
    Lemma sum_majorant_eqn:
      ∀ xs,
        (∀ x, x \in xs → P x → E1 x ≤ E2 x) →
        \sum_(x <- xs | P x) E1 x = \sum_(x <- xs | P x) E2 x →
        (∀ x, x \in xs → P x → E1 x = E2 x).
Next, we prove that the summing over the difference of E1 and E2 is the same as the difference of the two sums performed separately. Since we are using natural numbers, we have to require that E2 dominates E1 over the summing points given by r.
    Lemma sum_seq_diff:
        (∀ i : I, i \in r → E1 i ≤ E2 i) →
        \sum_(i <- r) (E2 i - E1 i) = \sum_(i <- r) E2 i - \sum_(i <- r) E1 i.

  End SumOfTwoFunctions.

End SumsOverSequences.

In this section, we prove a variety of properties of sums performed over ranges.
Section SumsOverRanges.

First, we show a trivial identity: any sum of zeros is zero.
  Lemma sum0 m n:
    \sum_(m ≤ i < n) 0 = 0.

In a similar way, we prove that the sum of Δ ones is equal to Δ.
  Lemma sum_of_ones:
    ∀ t Δ,
      \sum_(t ≤ x < t + Δ) 1 = Δ.

Next, we show that a sum of natural numbers equals zero if and only if all terms are zero.
  Lemma big_nat_eq0 m n F:
    \sum_(m ≤ i < n) F i = 0 ↔ (∀ i, m ≤ i < n → F i = 0).

Moreover, the fact that the sum is smaller than the range of the summation implies the existence of a zero element.
  Lemma sum_le_summation_range:
    ∀ f t Δ,
      \sum_(t ≤ x < t + Δ) f x < Δ →
      ∃ x, t ≤ x < t + Δ ∧ f x = 0.
Next, we prove that the summing over the difference of two functions is the same as summing over the two functions separately, and then taking the difference of the two sums. Since we are using natural numbers, we have to require that one function dominates the other in the summing range.
  Lemma sum_diff:
    ∀ n F G,
      (∀ i, i < n → F i ≥ G i) →
      \sum_(0 ≤ i < n) (F i - G i) =
      (\sum_(0 ≤ i < n) (F i)) - (\sum_(0 ≤ i < n) (G i)).

Given a sequence r, function F, and a predicate P, we prove that the fact that the sum of F conditioned by P is greater than 0 is equivalent to the fact that there exists an element i \in r such that F i > 0 and P i holds.
  Lemma sum_seq_cond_gt0P:
    ∀ (T : eqType) (r : seq T) (P : T → bool) (F : T → nat),
      reflect (∃ i, i \in r ∧ P i ∧ 0 < F i) (0 < \sum_(i <- r | P i) F i).

End SumsOverRanges.

In this section, we show how it is possible to equate the result of two sums performed on two different functions and on different intervals, provided that the two functions match point-wise.
Section SumOfTwoIntervals.

Consider two equally-sized intervals [t1, t1+d) and [t2, t2+d)...
  Variable (t1 t2 : nat).
  Variable (d : nat).

...and two functions F1 and F2.
  Variable (F1 F2 : nat → nat).

Assume that the two functions match point-wise with each other, with the points taken in their respective interval.
  Hypothesis equal_before_d: ∀ g, g < d → F1 (t1 + g) = F2 (t2 + g).

The then summations of F1 over [t1, t1 + d) and F2 over [t2, t2 + d) are equal.
  Lemma big_sum_eq_in_eq_sized_intervals:
    \sum_(t1 ≤ t < t1 + d) F1 t = \sum_(t2 ≤ t < t2 + d) F2 t.

End SumOfTwoIntervals.

In this section, we relate the sum of items with the sum over partitions of those items.
Section SumOverPartitions.

Consider an item type X and a partition type Y.
  Variable X Y : eqType.

x_to_y is the mapping from an item to the partition it is contained in.
  Variable x_to_y : X → Y.

Consider f, a function from X to nat.
  Variable f : X → nat.

Consider an arbitrary predicate P on X.
  Variable P : pred X.

Consider a sequence of items xs and a sequence of partitions ys.
  Variable xs : seq X.
  Variable ys : seq Y.

We assume that any item in xs has its corresponding partition in the sequence of partitions ys.
  Hypothesis H_no_partition_missing : ∀ x, x \in xs → x_to_y x \in ys.

Consider the sum of f x over all x in a given partition y.
  Let sum_of_partition y := \sum_(x <- xs | P x && (x_to_y x == y)) f x.

We prove that summation of f x over all x is less than or equal to the summation of sum_of_partition over all partitions.
  Lemma sum_over_partitions_le :
    \sum_(x <- xs | P x) f x
    ≤ \sum_(y <- ys) sum_of_partition y.

In this section, we prove a stronger result about the equality between the sum over all items and the sum over all partitions of those items.
  Section Equality.

In order to prove the stronger result of equality, we additionally assume that the sequences xs and ys are sets, i.e., that each element is contained at most once.
    Hypothesis H_xs_unique : uniq xs.
    Hypothesis H_ys_unique : uniq ys.

We prove that summation of f x over all x is equal to the summation of sum_of_partition over all partitions.
    Lemma sum_over_partitions_eq :
      \sum_(x <- xs | P x) f x
      = \sum_(y <- ys) sum_of_partition y.

  End Equality.

End SumOverPartitions.