Library prosa.util.minmax

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

Additional Lemmas about \max.

Given two functions F1 and F2, a predicate P, and sequence xs, we show that if F1 x ≤ F2 x for any x \in xs such that P x, then max of {F1 x | ∀ x ∈ xs, P x} is bounded by the max of {F2 x | ∀ x ∈ xs, P x}.

Lemma leq_big_max :
  ∀ {X : eqType} (F1 F2 : X → nat) (P : pred X) (xs : seq X),
    (∀ x, x \in xs → P x → F1 x ≤ F2 x ) →
    \max_(x <- xs | P x ) F1 x ≤ \max_(x <- xs | P x ) F2 x.

We show that for a positive n, max of {0, …, n-1} is smaller than n.

Lemma bigmax_ord_ltn_identity :
  ∀ (n : nat),
    n > 0 →
    \max_(i < n ) i < n.

We state the next lemma in terms of ordinals. Given a natural number n, a predicate P, and an ordinal i0 ∈ {0, …, n-1} satisfying P, we show that max {i | P i} < n. Note that the element satisfying P is needed to prove that {i | P i} is not empty.

Lemma bigmax_ltn_ord :
  ∀ (n : nat) (P : pred nat) (i0 : 'I_n),
    P i0 →
    \max_(i < n | P i ) i < n.

Next, we show that, given a natural number n, a predicate P, and an ordinal i0 ∈ {0, …, n-1} satisfying P, max {i | P i} < n also satisfies P.

Lemma bigmax_pred :
  ∀ (n : nat) (P : pred nat) (i0 : 'I_n),
    P i0 →
    P (\max_(i < n | P i ) i).

Furthermore, we observe that, if there is any element that satisfies the predicate, then there exists a witness for the computed maximum.

Lemma bigmax_witness {T : eqType} :
  ∀ {xs : seq T} {P} F,
    has P xs →
    ∃ x , x \in xs ∧ P x ∧ (F x = \max_(x <- xs | P x ) F x ).

Additionally, we observe that, if two predicates yield different maxima, then there must exist a witness that satisfies only one of the predicates.

Lemma bigmax_witness_diff {T : eqType} :
  ∀ {xs : seq T} {P1 P2 F},
    \max_(x <- xs | P1 x ) F x < \max_(x <- xs | P2 x ) F x →
    ∃ x , x \in xs ∧ ~~ P1 x ∧ P2 x.