# Library prosa.util.bigcat

Require Export mathcomp.zify.zify.
Require Export prosa.util.tactics prosa.util.notation.
From mathcomp Require Import ssreflect ssrbool eqtype ssrnat seq fintype bigop.
Require Export prosa.util.tactics prosa.util.list.

In this section, we introduce lemmas about the concatenation operation performed over arbitrary sequences.
Section BigCatNatLemmas.

Consider any type T supporting equality comparisons...
Variable T : eqType.

...and a function f that, given an index, yields a sequence.
Variable f : nat seq T.

In this section, we prove that the concatenation over sequences works as expected: no element is lost during the concatenation, and no new element is introduced.
Section BigCatNatElements.

First, we show that the concatenation comprises all the elements of each sequence; i.e. any element contained in one of the sequences will also be an element of the result of the concatenation.
Conversely, we prove that any element belonging to a concatenation of sequences must come from one of the sequences.
We also restate lemma mem_bigcat_nat in terms of ordinals.
Lemma mem_bigcat_ord :
(x : T) (n : nat) (j : 'I_n) (f : 'I_n seq T),
j < n
x \in (f j)
x \in \cat_(i < n) (f i).

End BigCatNatElements.

In this section, we show how we can preserve uniqueness of the elements (i.e. the absence of a duplicate) over a concatenation of sequences.
Assume that there are no duplicates in each of the possible sequences to concatenate...
Hypothesis H_uniq_seq : i, uniq (f i).

...and that there are no elements in common between the sequences.
Hypothesis H_no_elements_in_common :
x i1 i2, x \in f i1 x \in f i2 i1 = i2.

We prove that the concatenation will yield a sequence with unique elements.
Lemma bigcat_nat_uniq :
n1 n2,
uniq (\cat_(n1 i < n2) (f i)).

Conversely, if the concatenation of the sequences has no duplicates, any element can only belong to a single sequence.
We show that filtering a concatenated sequence by any predicate P is the same as concatenating the elements of the sequence that satisfy P.
We show that the size of a concatenated sequence is the same as summation of sizes of each element of the sequence.
Lemma size_big_nat :
{X : Type} (F : nat seq X) (t1 t2 : nat),
\sum_(t1 t < t2) size (F t) =
size (\cat_(t1 t < t2) F t).

End BigCatNatLemmas.

In this section, we introduce a few lemmas about the concatenation operation performed over arbitrary sequences.
Section BigCatLemmas.

Consider any two types X and Y supporting equality comparisons...
Variable X Y : eqType.

...and a function f that, given an index X, yields a sequence of Y.
Variable f : X seq Y.

First, we show that the concatenation comprises all the elements of each sequence; i.e. any element contained in one of the sequences will also be an element of the result of the concatenation.
Conversely, we prove that any element belonging to a concatenation of sequences must come from one of the sequences.
Next, we show that a map and filter operation can be done either before or after a concatenation, leading to the same result.
In this section, we show how we can preserve uniqueness of the elements (i.e. the absence of a duplicate) over a concatenation of sequences.
Section BigCatDistinctElements.

Assume that there are no duplicates in each of the possible sequences to concatenate...
Hypothesis H_uniq_f : x, uniq (f x).

...and that there are no elements in common between the sequences.
Hypothesis H_no_elements_in_common :
x y z,
x \in f y x \in f z y = z.

We prove that the concatenation will yield a sequence with unique elements.
In this section, we show some properties of the concatenation of sequences in combination with a function g that cancels f.
Consider a function g...
Variable g : Y X.

... and assume that g can cancel f starting from an element of the sequence f x.
Hypothesis H_g_cancels_f : x y, y \in f x g y = x.

First, we show that no element of a sequence f x1 can be fed into g and obtain an element x2 which differs from x1. Hence, filtering by this condition always leads to the empty sequence.
Finally, assume we are given an element y which is contained in a duplicate-free sequence xs. Then, f is applied to each element of xs, but only the elements for which g yields y are kept. In this scenario, concatenating the resulting sequences will always lead to f y.
In this section, we show that the number of elements of the result of a nested mapping and concatenation operation is independent from the order in which the concatenations are performed.
Consider any three types supporting equality comparisons...
Variable X Y Z : eqType.

... a function F that, given two indices, yields a sequence...
Variable F : X Y seq Z.

and a predicate P.
Variable P : pred Z.

Assume that, given two sequences xs and ys, their elements are fed to F in a pair-wise fashion. The resulting lists are then concatenated. The following lemma shows that, when the operation described above is performed, the number of elements respecting P in the resulting list is the same, regardless of the order in which xs and ys are combined.