Library rt.util.nat
Require Import rt.util.tactics rt.util.ssromega.
From mathcomp Require Import ssreflect ssrbool eqtype ssrnat seq fintype bigop div.
(* Additional lemmas about natural numbers. *)
Section NatLemmas.
Lemma addnb (b1 b2 : bool) : (b1 + b2) != 0 = b1 || b2.
Proof.
by destruct b1,b2;
rewrite ?addn0 ?add0n ?addn1 ?orTb ?orbT ?orbF ?orFb.
Qed.
Lemma subh1 :
∀ m n p,
m ≥ n →
m - n + p = m + p - n.
Proof. by ins; ssromega. Qed.
Lemma subh2 :
∀ m1 m2 n1 n2,
m1 ≥ m2 →
n1 ≥ n2 →
(m1 + n1) - (m2 + n2) = m1 - m2 + (n1 - n2).
Proof. by ins; ssromega. Qed.
Lemma subh3:
∀ m n p,
m + p ≤ n →
m ≤ n - p.
Proof.
clear.
intros.
rewrite <- leq_add2r with (p := p).
rewrite subh1 //.
- by rewrite -addnBA // subnn addn0.
- by apply leq_trans with (m+p); first rewrite leq_addl.
Qed.
Lemma subh4:
∀ m n p,
m ≤ n →
p ≤ n →
(m == n - p) = (p == n - m).
Proof.
intros; apply/eqP.
destruct (p == n - m) eqn:EQ.
by move: EQ ⇒ /eqP EQ; rewrite EQ subKn.
by apply negbT in EQ; unfold not; intro BUG;
rewrite BUG subKn ?eq_refl in EQ.
Qed.
(* Simplify n + a - b + b - a = n if n >= b. *)
Lemma subn_abba:
∀ n a b,
n ≥ b →
n + a - b + b - a = n.
Proof.
move⇒ n a b le_bn.
rewrite subnK;
first by rewrite -addnBA // subnn addn0 //.
rewrite -[b]addn0.
apply leq_trans with (n := n + 0); rewrite !addn0 //.
apply leq_addr.
Qed.
Lemma addmovr:
∀ m n p,
m ≥ n →
(m - n = p ↔ m = p + n).
Proof.
by split; ins; ssromega.
Qed.
Lemma addmovl:
∀ m n p,
m ≥ n →
(p = m - n ↔ p + n = m).
Proof.
by split; ins; ssromega.
Qed.
Lemma ltSnm : ∀ n m, n.+1 < m → n < m.
Proof.
by ins; apply ltn_trans with (n := n.+1); [by apply ltnSn |by ins].
Qed.
Lemma min_lt_same :
∀ x y z,
minn x z < minn y z → x < y.
Proof.
unfold minn; ins; desf.
{
apply negbT in Heq0; rewrite -ltnNge in Heq0.
by apply leq_trans with (n := z).
}
{
apply negbT in Heq; rewrite -ltnNge in Heq.
by apply (ltn_trans H) in Heq0; rewrite ltnn in Heq0.
}
by rewrite ltnn in H.
Qed.
Lemma add_subC:
∀ a b c,
a ≥ c →
b ≥c →
a + (b - c ) = a - c + b.
Proof.
intros× AgeC BgeC.
induction b;induction c;intros;try ssromega.
Qed.
Lemma ltn_subLR:
∀ a b c,
a - c < b →
a < b + c.
Proof.
intros× AC. ssromega.
Qed.
End NatLemmas.
Section Interval.
(* Trivially, points before the start of an interval, or past the end of an
interval, are not included in the interval. *)
Lemma point_not_in_interval:
∀ t1 t2 t',
t2 ≤ t' ∨ t' < t1 →
∀ t,
t1 ≤ t < t2
→ t ≠ t'.
Proof.
move⇒ t1 t2 t' EXCLUDED t /andP [GEQ_t1 LT_t2] EQ.
subst.
case EXCLUDED ⇒ [INEQ | INEQ];
eapply leq_ltn_trans in INEQ; eauto;
by rewrite ltnn in INEQ.
Qed.
End Interval.
Section NatOrderLemmas.
(* Mimic the way implicit arguments are used in ssreflect. *)
Set Implicit Arguments.
Unset Strict Implicit.
(* ltn_leq_trans: Establish that m < p if m < n and n <= p, to mirror the
lemma leq_ltn_trans in ssrnat.
NB: There is a good reason for this lemma to be "missing" in ssrnat --
since m < n is defined as m.+1 <= n, ltn_leq_trans is just
m.+1 <= n -> n <= p -> m.+1 <= p, that is (@leq_trans n m.+1 p).
Nonetheless we introduce it here because an additional (even though
arguably redundant) lemma doesn't hurt, and for newcomers the apparent
absence of the mirror case of leq_ltn_trans can be somewhat confusing. *)
Lemma ltn_leq_trans n m p : m < n → n ≤ p → m < p.
Proof. exact (@leq_trans n m.+1 p). Qed.
End NatOrderLemmas.
From mathcomp Require Import ssreflect ssrbool eqtype ssrnat seq fintype bigop div.
(* Additional lemmas about natural numbers. *)
Section NatLemmas.
Lemma addnb (b1 b2 : bool) : (b1 + b2) != 0 = b1 || b2.
Proof.
by destruct b1,b2;
rewrite ?addn0 ?add0n ?addn1 ?orTb ?orbT ?orbF ?orFb.
Qed.
Lemma subh1 :
∀ m n p,
m ≥ n →
m - n + p = m + p - n.
Proof. by ins; ssromega. Qed.
Lemma subh2 :
∀ m1 m2 n1 n2,
m1 ≥ m2 →
n1 ≥ n2 →
(m1 + n1) - (m2 + n2) = m1 - m2 + (n1 - n2).
Proof. by ins; ssromega. Qed.
Lemma subh3:
∀ m n p,
m + p ≤ n →
m ≤ n - p.
Proof.
clear.
intros.
rewrite <- leq_add2r with (p := p).
rewrite subh1 //.
- by rewrite -addnBA // subnn addn0.
- by apply leq_trans with (m+p); first rewrite leq_addl.
Qed.
Lemma subh4:
∀ m n p,
m ≤ n →
p ≤ n →
(m == n - p) = (p == n - m).
Proof.
intros; apply/eqP.
destruct (p == n - m) eqn:EQ.
by move: EQ ⇒ /eqP EQ; rewrite EQ subKn.
by apply negbT in EQ; unfold not; intro BUG;
rewrite BUG subKn ?eq_refl in EQ.
Qed.
(* Simplify n + a - b + b - a = n if n >= b. *)
Lemma subn_abba:
∀ n a b,
n ≥ b →
n + a - b + b - a = n.
Proof.
move⇒ n a b le_bn.
rewrite subnK;
first by rewrite -addnBA // subnn addn0 //.
rewrite -[b]addn0.
apply leq_trans with (n := n + 0); rewrite !addn0 //.
apply leq_addr.
Qed.
Lemma addmovr:
∀ m n p,
m ≥ n →
(m - n = p ↔ m = p + n).
Proof.
by split; ins; ssromega.
Qed.
Lemma addmovl:
∀ m n p,
m ≥ n →
(p = m - n ↔ p + n = m).
Proof.
by split; ins; ssromega.
Qed.
Lemma ltSnm : ∀ n m, n.+1 < m → n < m.
Proof.
by ins; apply ltn_trans with (n := n.+1); [by apply ltnSn |by ins].
Qed.
Lemma min_lt_same :
∀ x y z,
minn x z < minn y z → x < y.
Proof.
unfold minn; ins; desf.
{
apply negbT in Heq0; rewrite -ltnNge in Heq0.
by apply leq_trans with (n := z).
}
{
apply negbT in Heq; rewrite -ltnNge in Heq.
by apply (ltn_trans H) in Heq0; rewrite ltnn in Heq0.
}
by rewrite ltnn in H.
Qed.
Lemma add_subC:
∀ a b c,
a ≥ c →
b ≥c →
a + (b - c ) = a - c + b.
Proof.
intros× AgeC BgeC.
induction b;induction c;intros;try ssromega.
Qed.
Lemma ltn_subLR:
∀ a b c,
a - c < b →
a < b + c.
Proof.
intros× AC. ssromega.
Qed.
End NatLemmas.
Section Interval.
(* Trivially, points before the start of an interval, or past the end of an
interval, are not included in the interval. *)
Lemma point_not_in_interval:
∀ t1 t2 t',
t2 ≤ t' ∨ t' < t1 →
∀ t,
t1 ≤ t < t2
→ t ≠ t'.
Proof.
move⇒ t1 t2 t' EXCLUDED t /andP [GEQ_t1 LT_t2] EQ.
subst.
case EXCLUDED ⇒ [INEQ | INEQ];
eapply leq_ltn_trans in INEQ; eauto;
by rewrite ltnn in INEQ.
Qed.
End Interval.
Section NatOrderLemmas.
(* Mimic the way implicit arguments are used in ssreflect. *)
Set Implicit Arguments.
Unset Strict Implicit.
(* ltn_leq_trans: Establish that m < p if m < n and n <= p, to mirror the
lemma leq_ltn_trans in ssrnat.
NB: There is a good reason for this lemma to be "missing" in ssrnat --
since m < n is defined as m.+1 <= n, ltn_leq_trans is just
m.+1 <= n -> n <= p -> m.+1 <= p, that is (@leq_trans n m.+1 p).
Nonetheless we introduce it here because an additional (even though
arguably redundant) lemma doesn't hurt, and for newcomers the apparent
absence of the mirror case of leq_ltn_trans can be somewhat confusing. *)
Lemma ltn_leq_trans n m p : m < n → n ≤ p → m < p.
Proof. exact (@leq_trans n m.+1 p). Qed.
End NatOrderLemmas.