Commit d36d163b by MARCHE Claude

### Merge branch 'three_idem_example' into 'master'

```Add the example about three-idempotent rings

See merge request !97```
parents 58c2bdbb 38d26d39
 (** {1 Three idempotent rings are commutative} *) (** {2 Definitions} *) module ThreeIdempotentRing use int.Int type t val constant zero : t val function (+) t t : t val function ( *) t t : t clone import algebra.Ring as R with type t = t, constant zero = zero, function (+) = (+), function ( *) = ( *), axiom . (** Define multiplication by an integer recursively *) let rec function mul (x : t) (n : int) : t requires { n >= 0 } variant { n } = if n = 0 then zero else x + mul x (n-1) (** We get lemmas from the why3 library *) clone import int.Exponentiation as Mul with type t = t, constant one = zero, function ( *) = (+), function power = mul, lemma . (** {2 General results about unitary rings} *) (** 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 (** Lemmas about nullable elements : *) predicate null (x : t) (n : int) = mul x n = zero lemma null_add : 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 (** {2 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 *) (** First show that the characteristic of the ring divides 6 ... *) 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 (** ... use it to show we can split the problem in two ...*) 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 (** ... and show that the two problems are independent *) lemma free_split : forall x. null x 2 -> null x 3 -> x = zero (** 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 end