可否讓電腦來驗證數學證明的正確性?

在數學的學習中我發現似乎所有的命題都可以用幾個最基本的概念組合來表達,對證明的正確性的驗證似乎交給電腦就可以辦得到。那麼這在理論上是否可以做到?是否有人已經寫出類似的程式來驗證標準化表述後的數學證明的正確與否?如果要寫這樣的程式需要學習哪些知識?

如果有這樣的程式,助教批改作業就可以省很多功夫了,一些不自信的學生也可以藉此驗證自己的證明,更可以幫助學生了解一些看似直觀的命題具體的證明細節。


能,Proof assistant 比如 coq 乾的就是這個。舉幾個例子:

  • kik/Four-Color-Theorem-Maintenance,四色定理

  • coq-contribs/goedel,G?del 不完備定理

  • c-corn/corn,代數基本定理
  • coq-contribs/jordan-curve-theorem,Jordan 曲線定理

貼個sqrt{2}是無理數的證明(Contribution: QArithSternBrocot):

Require Export ArithRing.
Require Export Compare_dec.
Require Export Wf_nat.
Require Export Arith.
Require Export Omega.

Theorem minus_minus : forall a b c : nat, a - b - c = a - (b + c).
intros a; elim a; auto.
intros n" Hrec b; case b; auto.
Qed.

Remark expand_mult2 : forall x : nat, 2 * x = x + x.
intros x; ring.
Qed.

Theorem lt_neq : forall x y : nat, x &< y -&> x &<&> y.
unfold not in |- *; intros x y H H1; elim (lt_irrefl x);
pattern x at 2 in |- *; rewrite H1; auto.
Qed.
Hint Resolve lt_neq.

Theorem monotonic_inverse :
forall f : nat -&> nat,
(forall x y : nat, x &< y -&> f x &< f y) -&>
forall x y : nat, f x &< f y -&> x &< y. intros f Hmon x y Hlt; case (le_gt_dec y x); auto. intros Hle; elim (le_lt_or_eq _ _ Hle). intros Hlt"; elim (lt_asym _ _ Hlt); apply Hmon; auto. intros Heq; elim (lt_neq _ _ Hlt); rewrite Heq; auto. Qed. Theorem mult_lt : forall a b c : nat, c &<&> 0 -&> a &< b -&> a * c &< b * c. intros a b c; elim c. intros H; elim H; auto. intros c"; case c". intros; omega. intros c"" Hrec Hneq Hlt; repeat rewrite &<- (fun x : nat =&> mult_n_Sm x (S c"")).
auto with *.
Qed.

Remark add_sub_square_identity :
forall a b : nat,
(b + a - b) * (b + a - b) = (b + a) * (b + a) + b * b - 2 * ((b + a) * b).
intros a b; rewrite minus_plus.
repeat rewrite mult_plus_distr_r || rewrite &<- (mult_comm (b + a)). replace (b * b + a * b + (b * a + a * a) + b * b) with (b * b + a * b + (b * b + a * b + a * a)); try (ring; fail). rewrite expand_mult2; repeat rewrite minus_plus; auto with *. Qed. Theorem sub_square_identity : forall a b : nat, b &<= a -&> (a - b) * (a - b) = a * a + b * b - 2 * (a * b).
intros a b H; rewrite (le_plus_minus b a H); apply add_sub_square_identity.
Qed.

Theorem square_monotonic : forall x y : nat, x &< y -&> x * x &< y * y. intros x; case x. intros y; case y; simpl in |- *; auto with *. intros x" y Hlt; apply lt_trans with (S x" * y). rewrite (mult_comm (S x") y); apply mult_lt; auto. apply mult_lt; omega. Qed. Theorem root_monotonic : forall x y : nat, x * x &< y * y -&> x &< y. exact (monotonic_inverse (fun x : nat =&> x * x) square_monotonic).
Qed.

Remark square_recompose : forall x y : nat, x * y * (x * y) = x * x * (y * y).
intros; ring.
Qed.

Remark mult2_recompose : forall x y : nat, x * (2 * y) = x * 2 * y.
intros; ring.
Qed.
Section sqrt2_decrease.
Variables (p q : nat) (pos_q : 0 &< q) (hyp_sqrt : p * p = 2 * (q * q)). Theorem sqrt_q_non_zero : 0 &<&> q * q.
generalize pos_q; case q.
intros H; elim (lt_n_O 0); auto.
intros n H.
simpl in |- *; discriminate.
Qed.
Hint Resolve sqrt_q_non_zero.

Ltac solve_comparison :=
apply root_monotonic; repeat rewrite square_recompose; rewrite hyp_sqrt;
rewrite mult2_recompose; apply mult_lt; auto with arith.

Theorem comparison1 : q &< p. replace q with (1 * q); try ring. replace p with (1 * p); try ring. solve_comparison. Qed. Theorem comparison2 : 2 * p &< 3 * q. solve_comparison. Qed. Theorem comparison3 : 4 * q &< 3 * p. solve_comparison. Qed. Hint Resolve comparison1 comparison2 comparison3: arith. Theorem comparison4 : 3 * q - 2 * p &< q. apply plus_lt_reg_l with (2 * p). rewrite &<- le_plus_minus; try (simple apply lt_le_weak; auto with arith). replace (3 * q) with (2 * q + q); try ring. apply plus_lt_le_compat; auto. repeat rewrite (mult_comm 2); apply mult_lt; auto with arith. Qed. Remark mult_minus_distr_l : forall a b c : nat, a * (b - c) = a * b - a * c. intros a b c; repeat rewrite (mult_comm a); apply mult_minus_distr_r. Qed. Remark minus_eq_decompose : forall a b c d : nat, a = b -&> c = d -&> a - c = b - d.
intros a b c d H H0; rewrite H; rewrite H0; auto.
Qed.

Theorem new_equality :
(3 * p - 4 * q) * (3 * p - 4 * q) = 2 * ((3 * q - 2 * p) * (3 * q - 2 * p)).
repeat rewrite sub_square_identity; auto with arith.
repeat rewrite square_recompose; rewrite mult_minus_distr_l.
apply minus_eq_decompose; try rewrite hyp_sqrt; ring.
Qed.
End sqrt2_decrease.
Hint Resolve lt_le_weak comparison2: sqrt.

Theorem sqrt2_not_rational :
forall p q : nat, q &<&> 0 -&> p * p = 2 * (q * q) -&> False.
intros p q; generalize p; clear p; elim q using (well_founded_ind lt_wf).
clear q; intros q Hrec p Hneq; generalize (neq_O_lt _ (sym_not_equal Hneq));
intros Hlt_O_q Heq.
apply (Hrec (3 * q - 2 * p) (comparison4 _ _ Hlt_O_q Heq) (3 * p - 4 * q)).
apply sym_not_equal; apply lt_neq; apply plus_lt_reg_l with (2 * p);
rewrite &<- plus_n_O; rewrite &<- le_plus_minus; auto with *. apply new_equality; auto. Qed.

你把它貼到 coqide 裡面就能驗證。


謝邀。作為一隻助教,我覺得批改作業分為兩步:

1. 把作業轉化(或者腦補)成合適的表述方式

2. 確認上面的東西的對錯

而事實上,99%的批改作業的時間花在了第一步……


當然可以

數學建模就行啊

lingo matlab……

就是用電腦來證明數學的呀


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