euler001.mlw 1.98 KB
 MARCHE Claude committed Apr 10, 2011 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 ``````(* Euler Project, problem 1 If we list all the natural numbers below 10 that are multiples of 3 or 5, we get 3, 5, 6 and 9. The sum of these multiples is 23. Find the sum of all the multiples of 3 or 5 below 1000. *) theory SumMultiple use import int.Int use import int.EuclideanDivision (* [sum_multiple_3_5_lt n] is the sum of all the multiples of 3 or 5 below n] *) logic sum_multiple_3_5_lt int : int axiom SumEmpty: sum_multiple_3_5_lt 0 = 0 axiom SumNo : forall n:int. n >= 0 -> mod n 3 <> 0 and mod n 5 <> 0 -> sum_multiple_3_5_lt (n+1) = sum_multiple_3_5_lt n axiom SumYes: forall n:int. n >= 0 -> mod n 3 = 0 or 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 and 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 and y > 0 -> mod (x+1) y <> 0 -> div (x+1) y = (div x y) logic closed_formula (n:int) : int = let n3 = div n 3 in let n5 = div n 5 in let n15 = div n 15 in div (3 * n3 * (n3+1) + 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 and mod n 5 = 0) logic p (n:int) = sum_multiple_3_5_lt (n+1) = closed_formula n lemma Closed_formula_0: p 0 lemma Closed_formula_n_1: forall n:int. n > 0 -> p (n-1) -> mod n 3 <> 0 and mod n 5 <> 0 -> p n lemma Closed_formula_n_2: forall n:int. n > 0 -> p (n-1) -> mod n 3 = 0 or mod n 5 = 0 -> p n clone int.Induction as I with logic p = p lemma Closed_formula: forall n:int. 0 <= n -> p n end module Euler001 use import SumMultiple use import int.Int use import int.EuclideanDivision let solve n = { n >= 1 } let n3 = div (n-1) 3 in let n5 = div (n-1) 5 in let n15 = div (n-1) 15 in div (3 * n3 * (n3+1) + 5 * n5 * (n5+1) - 15 * n15 * (n15+1)) 2 { result = sum_multiple_3_5_lt n } end``````