Commit de702e5d authored by MARCHE Claude's avatar MARCHE Claude

euler001 continued

parent e46f0376
......@@ -7,6 +7,29 @@ Find the sum of all the multiples of 3 or 5 below 1000.
*)
theory DivModHints
use import int.Int
use import int.EuclideanDivision
lemma mod_succ_1 :
forall x y:int. x >= 0 /\ y > 0 ->
mod (x+1) y <> 0 -> mod (x+1) y = (mod x y) + 1
lemma mod_succ_2 :
forall x y:int. x >= 0 /\ y > 0 ->
mod (x+1) y = 0 -> mod x y = y-1
lemma div_succ_1 :
forall x y:int. x >= 0 /\ y > 0 ->
mod (x+1) y = 0 -> div (x+1) y = (div x y) + 1
lemma div_succ_2 :
forall x y:int. x >= 0 /\ y > 0 ->
mod (x+1) y <> 0 -> div (x+1) y = (div x y)
end
theory SumMultiple
......@@ -27,14 +50,6 @@ theory SumMultiple
mod n 3 = 0 \/ mod n 5 = 0 ->
sum_multiple_3_5_lt (n+1) = sum_multiple_3_5_lt n + n
lemma div_minus1_1 :
forall x y:int. x >= 0 /\ y > 0 ->
mod (x+1) y = 0 -> div (x+1) y = (div x y) + 1
lemma div_minus1_2 :
forall x y:int. x >= 0 /\ y > 0 ->
mod (x+1) y <> 0 -> div (x+1) y = (div x y)
function closed_formula (n:int) : int =
let n3 = div n 3 in
let n5 = div n 5 in
......@@ -43,16 +58,12 @@ theory SumMultiple
5 * n5 * (n5+1) -
15 * n15 * (n15+1)) 2
(*
lemma mod_15 :
forall n:int. n >= 0 ->
mod n 15 = 0 <-> (mod n 3 = 0 /\ mod n 5 = 0)
*)
predicate p (n:int) = sum_multiple_3_5_lt (n+1) = closed_formula n
lemma Closed_formula_0: p 0
use DivModHints
lemma Closed_formula_n:
forall n:int. n > 0 -> p (n-1) ->
mod n 3 <> 0 /\ mod n 5 <> 0 -> p n
......
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