我在完成这个引理的证明时遇到了问题:
Lemma l1 : forall m n : nat, m * m = n * n -> m = n.
任何提示都会非常有用。
我开始这样的证明:
Require Import Arith Omega Nat.
Lemma l1 : forall m n : nat, m * m = n * n -> m = n.
Proof.
intros.
destruct (Nat.eq_dec m n ).
trivial.
induction n.
induction m.
auto.
simpl in H;congruence.
答案 0 :(得分:3)
提示:取你假设两边的平方根,然后立即得出结论。对于平方根函数,请使用Nat.sqrt
模块中的Arith
。
第一个解决方案:
Require Import Coq.Arith.Arith.
Lemma l1 : forall m n : nat, m * m = n * n -> m = n.
Proof.
intros m n H. apply (f_equal Nat.sqrt) in H.
now repeat rewrite Nat.sqrt_square in H.
Qed.
第二个解决方案:
引理也可以使用nia
策略证明,这是一个整数非线性算法的不完整证明程序:
Require Import Psatz.
Lemma l1 m n : m * m = n * n -> m = n.
Proof. nia. Qed.
第三个解决方案:
让我们使用标准Nat.square_le_simpl_nonneg
引理的几倍:
forall n m : nat, 0 <= m -> n * n <= m * m -> n <= m
我们走了:
Require Import Coq.Arith.Arith.
Lemma l1 (m n : nat) :
m * m = n * n -> m = n.
Proof with (auto with arith).
intros H.
pose proof (Nat.eq_le_incl _ _ H) as Hle.
pose proof (Nat.eq_le_incl _ _ (eq_sym H)) as Hge.
apply Nat.square_le_simpl_nonneg in Hle...
apply Nat.square_le_simpl_nonneg in Hge...
Qed.
第四个解决方案:
这是一个基于以下相等的经典证明
m * m - n * n = (m + n) * (m - n)
首先,我们需要一个辅助引理,证明上述相等(令人惊讶的是,似乎标准库缺乏这个引理):
Require Import Coq.Arith.Arith.
(* can be proved using `nia` tactic *)
Lemma sqr_diff (m n : nat) :
m * m - n * n = (m + n) * (m - n).
Proof with (auto with arith).
destruct (Nat.lt_trichotomy m n) as [H | [H | H]].
- pose proof H as H'. (* copy hypothesis *)
apply Nat.square_lt_mono_nonneg in H...
repeat match goal with
h : _ < _ |- _ => apply Nat.lt_le_incl, Nat.sub_0_le in h
end.
rewrite H, H'...
- now rewrite H, !Nat.sub_diag.
- rewrite Nat.mul_add_distr_r, !Nat.mul_sub_distr_l.
rewrite Nat.add_sub_assoc...
replace (n * m) with (m * n) by apply Nat.mul_comm.
rewrite Nat.sub_add...
Qed.
现在我们可以证明主要的引理:
Lemma l1 (m n : nat) :
m * m = n * n -> m = n.
Proof.
intros H.
pose proof (Nat.eq_le_incl _ _ H) as Hle;
pose proof (Nat.eq_le_incl _ _ (eq_sym H)) as Hge; clear H.
rewrite <- Nat.sub_0_le in *.
rewrite sqr_diff in *.
destruct (mult_is_O _ _ Hle) as [H | H].
now destruct (plus_is_O _ _ H); subst.
destruct (mult_is_O _ _ Hge) as [H' | H'].
now destruct (plus_is_O _ _ H'); subst.
rewrite Nat.sub_0_le in *.
apply Nat.le_antisymm; assumption.
Qed.