Prove the Existence of Modular Multiplicative Inverse Using Coq
1
I'm trying to prove the Chinese Remainder Theorem using Coq.
currently, I'm trying to prove the existence of modular multiplicative inverse, this is what I've done so far:
Theorem modulo_inv : forall m n : Z, rel_prime m n -> exists x : Z, (m * x == 1 [ n ]) .
Proof .
intros m0 n0 co_prime0 .
destruct (rel_prime_bezout _ _ co_prime0) as [u v Eq].
cut (u * m0 = 1 - v * n0); auto with zarith .
intros Eq0 .
exists u.
(* the subgoal is Eq0 *)
Abort.
and all there's left now is one subgoal, which means the same as Eq0 as shown below:
1 subgoal
m, n : Z
co_prime : rel_prime m n
m0, n0 : Z
co_prime0 : rel_prime m0 n0
u, v : Z
Eq : u * m0 + v * n0 = 1
Eq0 : u * m0 = 1 - v * n0
______________________________________(1/1)
(m0 * u == 1 [n0])
Does anyone know how to show that the subgoal equals to Eq0?
P.S. The definition of Modulo is given below:
Definition modulo (a b n : Z) : Prop := (n | (a - b)) .
Notation "( a == b [ n ])" := (modulo a b n) .
coq
asked Nov 27 '18 at 12:21
Ang.Ang.
62
add a comment |
1
I'm trying to prove the Chinese Remainder Theorem using Coq.
currently, I'm trying to prove the existence of modular multiplicative inverse, this is what I've done so far:
Theorem modulo_inv : forall m n : Z, rel_prime m n -> exists x : Z, (m * x == 1 [ n ]) .
Proof .
intros m0 n0 co_prime0 .
destruct (rel_prime_bezout _ _ co_prime0) as [u v Eq].
cut (u * m0 = 1 - v * n0); auto with zarith .
intros Eq0 .
exists u.
(* the subgoal is Eq0 *)
Abort.
and all there's left now is one subgoal, which means the same as Eq0 as shown below:
1 subgoal
m, n : Z
co_prime : rel_prime m n
m0, n0 : Z
co_prime0 : rel_prime m0 n0
u, v : Z
Eq : u * m0 + v * n0 = 1
Eq0 : u * m0 = 1 - v * n0
______________________________________(1/1)
(m0 * u == 1 [n0])
Does anyone know how to show that the subgoal equals to Eq0?
P.S. The definition of Modulo is given below:
Definition modulo (a b n : Z) : Prop := (n | (a - b)) .
Notation "( a == b [ n ])" := (modulo a b n) .
coq
asked Nov 27 '18 at 12:21
Ang.Ang.
62
2
Are you already familiar with an informal proof (i.e., not in Coq) of this theorem? Because if you do it is easier to then try to map each step of that proof to a statement in Coq.
– Li-yao Xia
Nov 27 '18 at 12:37
How would you prove the theorem on paper?
– gallais
Nov 27 '18 at 12:39
You can also checkout the proof of this theorem in mathcomp library.
– Anton Trunov
Nov 27 '18 at 12:49
@Li-yaoXia I have updated my informal proof, but how do you convert that to formal proof?
– Ang.
Nov 27 '18 at 13:08
@AntonTrunov Thank you, I'll check it out.
– Ang.
Nov 27 '18 at 13:09
add a comment |
1
1
1
I'm trying to prove the Chinese Remainder Theorem using Coq.
currently, I'm trying to prove the existence of modular multiplicative inverse, this is what I've done so far:
Theorem modulo_inv : forall m n : Z, rel_prime m n -> exists x : Z, (m * x == 1 [ n ]) .
Proof .
intros m0 n0 co_prime0 .
destruct (rel_prime_bezout _ _ co_prime0) as [u v Eq].
cut (u * m0 = 1 - v * n0); auto with zarith .
intros Eq0 .
exists u.
(* the subgoal is Eq0 *)
Abort.
and all there's left now is one subgoal, which means the same as Eq0 as shown below:
1 subgoal
m, n : Z
co_prime : rel_prime m n
m0, n0 : Z
co_prime0 : rel_prime m0 n0
u, v : Z
Eq : u * m0 + v * n0 = 1
Eq0 : u * m0 = 1 - v * n0
______________________________________(1/1)
(m0 * u == 1 [n0])
Does anyone know how to show that the subgoal equals to Eq0?
P.S. The definition of Modulo is given below:
Definition modulo (a b n : Z) : Prop := (n | (a - b)) .
Notation "( a == b [ n ])" := (modulo a b n) .
coq
asked Nov 27 '18 at 12:21
Ang.Ang.
62
I'm trying to prove the Chinese Remainder Theorem using Coq.
currently, I'm trying to prove the existence of modular multiplicative inverse, this is what I've done so far:
Theorem modulo_inv : forall m n : Z, rel_prime m n -> exists x : Z, (m * x == 1 [ n ]) .
Proof .
intros m0 n0 co_prime0 .
destruct (rel_prime_bezout _ _ co_prime0) as [u v Eq].
cut (u * m0 = 1 - v * n0); auto with zarith .
intros Eq0 .
exists u.
(* the subgoal is Eq0 *)
Abort.
and all there's left now is one subgoal, which means the same as Eq0 as shown below:
1 subgoal
m, n : Z
co_prime : rel_prime m n
m0, n0 : Z
co_prime0 : rel_prime m0 n0
u, v : Z
Eq : u * m0 + v * n0 = 1
Eq0 : u * m0 = 1 - v * n0
______________________________________(1/1)
(m0 * u == 1 [n0])
Does anyone know how to show that the subgoal equals to Eq0?
P.S. The definition of Modulo is given below:
Definition modulo (a b n : Z) : Prop := (n | (a - b)) .
Notation "( a == b [ n ])" := (modulo a b n) .
coq
coq
asked Nov 27 '18 at 12:21
Ang.Ang.
62
asked Nov 27 '18 at 12:21
Ang.Ang.
62
edited Jan 10 at 11:17
Ang.
asked Nov 27 '18 at 12:21
Ang.Ang.
62
asked Nov 27 '18 at 12:21
Ang.Ang.
62
asked Nov 27 '18 at 12:21
Ang.Ang.
62
62
2
Are you already familiar with an informal proof (i.e., not in Coq) of this theorem? Because if you do it is easier to then try to map each step of that proof to a statement in Coq.
– Li-yao Xia
Nov 27 '18 at 12:37
How would you prove the theorem on paper?
– gallais
Nov 27 '18 at 12:39
You can also checkout the proof of this theorem in mathcomp library.
– Anton Trunov
Nov 27 '18 at 12:49
@Li-yaoXia I have updated my informal proof, but how do you convert that to formal proof?
– Ang.
Nov 27 '18 at 13:08
@AntonTrunov Thank you, I'll check it out.
– Ang.
Nov 27 '18 at 13:09
add a comment |
2
Are you already familiar with an informal proof (i.e., not in Coq) of this theorem? Because if you do it is easier to then try to map each step of that proof to a statement in Coq.
– Li-yao Xia
Nov 27 '18 at 12:37
How would you prove the theorem on paper?
– gallais
Nov 27 '18 at 12:39
You can also checkout the proof of this theorem in mathcomp library.
– Anton Trunov
Nov 27 '18 at 12:49
@Li-yaoXia I have updated my informal proof, but how do you convert that to formal proof?
– Ang.
Nov 27 '18 at 13:08
@AntonTrunov Thank you, I'll check it out.
– Ang.
Nov 27 '18 at 13:09
2
2
Are you already familiar with an informal proof (i.e., not in Coq) of this theorem? Because if you do it is easier to then try to map each step of that proof to a statement in Coq.
– Li-yao Xia
Nov 27 '18 at 12:37
Are you already familiar with an informal proof (i.e., not in Coq) of this theorem? Because if you do it is easier to then try to map each step of that proof to a statement in Coq.
– Li-yao Xia
Nov 27 '18 at 12:37
How would you prove the theorem on paper?
– gallais
Nov 27 '18 at 12:39
How would you prove the theorem on paper?
– gallais
Nov 27 '18 at 12:39
You can also checkout the proof of this theorem in mathcomp library.
– Anton Trunov
Nov 27 '18 at 12:49
You can also checkout the proof of this theorem in mathcomp library.
– Anton Trunov
Nov 27 '18 at 12:49
@Li-yaoXia I have updated my informal proof, but how do you convert that to formal proof?
– Ang.
Nov 27 '18 at 13:08
@Li-yaoXia I have updated my informal proof, but how do you convert that to formal proof?
– Ang.
Nov 27 '18 at 13:08
@AntonTrunov Thank you, I'll check it out.
– Ang.
Nov 27 '18 at 13:09
@AntonTrunov Thank you, I'll check it out.
– Ang.
Nov 27 '18 at 13:09
add a comment |
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2
Are you already familiar with an informal proof (i.e., not in Coq) of this theorem? Because if you do it is easier to then try to map each step of that proof to a statement in Coq.
– Li-yao Xia
Nov 27 '18 at 12:37
How would you prove the theorem on paper?
– gallais
Nov 27 '18 at 12:39
You can also checkout the proof of this theorem in mathcomp library.
– Anton Trunov
Nov 27 '18 at 12:49
@Li-yaoXia I have updated my informal proof, but how do you convert that to formal proof?
– Ang.
Nov 27 '18 at 13:08
@AntonTrunov Thank you, I'll check it out.
– Ang.
Nov 27 '18 at 13:09