## Three idempotent rings

Show commutativity of three idempotent rings

Authors: Quentin Garchery

Topics: Mathematics

Tools: Why3

see also the index (by topic, by tool, by reference, by year)

# Three idempotent rings are commutative

Author: Quentin Garchery (LRI, Université Paris-Sud)

## Definitions

use int.Int

clone export algebra.Ring with
axiom .

let rec ghost function mul (x : t) (n : int) : t
requires { n >= 0 }
variant { n }
=
if n = 0 then pure{zero} else let r = mul x (n-1) in pure {x + r}

Define multiplication by an integer recursively

clone int.Exponentiation with type t = t,
constant one = zero, function ( * ) = (+), function power = mul,
lemma .

We get lemmas from the why3 library

First results :

lemma simpl_left :
forall x y z. x + y = x + z -> y = z
by (-x) + (x + y) = (-x) + (x + z)

lemma simpl_right :
forall x y z. y + x = z + x -> y = z
by y + x + (-x) = z + x + (-x)

lemma zero_star_l :
forall x. zero * x = zero

lemma zero_star_r :
forall x. x * zero = zero

lemma neg_star_r :
forall x y. x * (-y) = - (x * y)
by x * y + x * (-y) = x * y + (- (x * y))

lemma neg_star_l :
forall x y. (-x) * y = - (x * y)
by x * y + (-x) * y = x * y + (- (x * y))

lemma neg_neg :
forall x. - (- x) = x

predicate null (x : t) (n : int) = mul x n = zero

forall x x' n. 0 <= n -> null x n -> null x' n -> null (x + x') n

let rec lemma mul_star_l (x y : t) (n : int)
requires { 0 <= n }
variant { n }
ensures { mul (x * y) n = (mul x n) * y }
=
if n <> 0 then mul_star_l x y (n-1)

let rec lemma mul_star_r (x y : t) (n : int)
requires { 0 <= n }
variant { n }
ensures { mul (x * y) n = x * (mul y n) }
=
if n <> 0 then mul_star_r x y (n-1)

lemma null_star_l :
forall x y n. 0 <= n -> null x n -> null (x * y) n

lemma null_star_r :
forall x y n. 0 <= n -> null y n -> null (x * y) n

lemma null_mul_congr :
forall x k km. k > 0 -> km > 0 -> null x k -> mul x (Int.(+) km k) = mul x km

## ThreeIdem axiom specific results

We now add the following axiom and want to prove the commutative property :

axiom ThreeIdem : forall x. x * x * x = x

Split the problem in two : one where the ring has characteritic 2 and another where the ring has characteristic 3

lemma all_null6 :
forall x. null x 6
by (x + x) * (x + x) * (x + x) = mul (x * x * x) 8
so mul x 8 = zero + x + x /\ mul x 8 = mul x 6 + x + x

First show that the characteristic of the ring divides 6 ...

lemma all_split :
forall x. (exists y z. x = y + z /\ null y 2 /\ null z 3)
by let y = mul x 3 in
let z = mul (-x) 2 in
x = y + z /\ null y 2 /\ null z 3

... use it to show we can split the problem in two ...

lemma free_split :
forall x. null x 2 -> null x 3 -> x = zero

... and show that the two problems are independent

Show the commutative property in characteristic 2 :

lemma null_2_idem :
forall x. null x 2 -> x * x = x
by (x + x * x) * (x + x * x) = zero
so (x + x * x) * (x + x * x) * (x + x * x) = zero
so x + x * x = zero

lemma null2_comm :
forall x y. null x 2 -> null y 2 -> x * y = y * x
by (x + y) * (x + y) = x * x + y * y + x * y + y * x
so x + y = x + y + x * y + y * x

Show the commutative property in characteristic 3 :

lemma swap_equality :
forall x y. null x 3 -> null y 3 ->
y * y * x + y * x * y + x * y * y = zero
by (forall x y. y * y * x + y * x * y + x * y * y +
x * x * y + x * y * x + y * x * x = zero
by ((x + y) * (x + y) * (x + y) = x * x * x + y * y * y + y * y * x +
y * x * y + x * y * y + x * x * y + x * y * x + y * x * x
so x + y + zero = x + y + (y * y * x +
y * x * y + x * y * y + x * x * y + x * y * x + y * x * x)))
so y * y * x + y * x * y + x * y * y + x * x * y + x * y * x + y * x * x = zero
so (y * y * x + y * x * y + x * y * y +
(- (x * x * y)) + (- (x * y * x)) + (- (y * x * x)) = zero
by (-y) * (-y) * x + (-y) * x * (-y) + x * (-y) * (-y) +
x * x * (-y) + x * (-y) * x + (-y) * x * x = zero)
so mul (y * y * x) 2 + mul (y * x * y) 2 + mul (x * y * y) 2 = zero
so mul (y * y * x) 4 + mul (y * x * y) 4 + mul (x * y * y) 4 = zero

lemma null3_comm :
forall x y. null x 3 -> null y 3 -> x * y = y * x
by y * x + y * y * x * y + y * x * y * y = x * y + y * y * x * y + y * x * y * y

Finally, combine the previous results to show the commutative property.

lemma commutative :
forall x y. x * y = y * x
by exists x2 x3 y2 y3. x = x2 + x3 /\ y = y2 + y3 /\ null x2 2 /\ null y2 2 /\
null x3 3 /\ null y3 3
so x2 * y3 = zero /\ y3 * x2 = zero /\ x3 * y2 = zero /\ y2 * x3 = zero