# 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.

(* TODO: PR MathComp *)
Lemma leq_sum_subseq (I : eqType) (r r' : seq I) (P : pred I) (F : I nat) :
subseq r r' \sum_(i <- r | P i) F i \sum_(i <- r' | P i) F i.

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.
(* TODO: PR MathComp
this should probably be named sum_nat_eq0,
but there is already a sum_nat_eq0 that is less generic? *)

Lemma sum_nat_eq0_nat :
(\sum_(i <- r | P i) F i == 0) = all (fun xF x == 0) [seq x <- r | P x].

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.
(* TODO: PR MathComp *)
Lemma sum_nat_gt0 :
(0 < \sum_(i <- r | P i) F i) = has (fun x ⇒ 0 < F x) [seq x <- r | P x].

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.
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.
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.
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.
(* TODO: PR MathComp
- add a condition P i *)

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.
(* TODO: PR MathComp *)
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.
(* TODO: PR MathComp
- generalize as eq_big_seq_cond (nothing specific to addn here)
- replace == with = ? *)

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.
(* TODO: PR MathComp
- maybe leq_sum_seq above should be leq_sum_seqr and this one leq_sum_seql *)

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.

End SumOfTwoFunctions.

End SumsOverSequences.

We continue establishing properties of sums over sequences, but start a new section here because some of the below proofs depend lemmas in the preceding section in their full generality.
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).

Consider two functions that yield natural numbers.
Variable (E1 E2 : I nat).

First, as an auxiliary lemma, we observe that, if E1 j is less than E2 j for some element j involved in a summation (filtered by P), then the corresponding totals are not equal.
Next, we prove that if for any element i of a set r the following two statements hold (1) E1 i is less than or equal to E2 i and (2) the sum E1 x_1, ..., E1 x_n is equal to the sum of E2 x_1, ..., E2 x_n, then E1 x is equal to E2 x for any element x of xs.
(* TODO: PR MathComp *)
Lemma eq_sum_leq_seq :
( i, i \in r P i E1 i E2 i)
\sum_(x <- r | P x) E1 x == \sum_(x <- r | P x) E2 x
= all (fun xE1 x == E2 x) [seq x <- r | P x].

End SumsOverSequences.

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

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