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(* 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

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  (* [sum_multiple_3_5_lt n] is the sum of all the multiples
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     of 3 or 5 below n] *)
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  function sum_multiple_3_5_lt int : int
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  axiom SumEmpty: sum_multiple_3_5_lt 0 = 0

  axiom SumNo : forall n:int. n >= 0 ->
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    mod n 3 <> 0 /\ mod n 5 <> 0 ->
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    sum_multiple_3_5_lt (n+1) = sum_multiple_3_5_lt n

  axiom SumYes: forall n:int. n >= 0 ->
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    mod n 3 = 0 \/ mod n 5 = 0 ->
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    sum_multiple_3_5_lt (n+1) = sum_multiple_3_5_lt n + n

  lemma div_minus1_1 :
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    forall x y:int. x >= 0 /\ y > 0 ->
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      mod (x+1) y = 0 -> div (x+1) y = (div x y) + 1

  lemma div_minus1_2 :
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    forall x y:int. x >= 0 /\ y > 0 ->
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      mod (x+1) y <> 0 -> div (x+1) y = (div x y)
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  function closed_formula (n:int) : int =
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    let n3 = div n 3 in
    let n5 = div n 5 in
    let n15 = div n 15 in
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    div (3 * n3 * (n3+1) +
         5 * n5 * (n5+1) -
         15 * n15 * (n15+1)) 2
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(*
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  lemma mod_15 :
    forall n:int. n >= 0 ->
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      mod n 15 = 0 <-> (mod n 3 = 0 /\ mod n 5 = 0)
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*)
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  predicate p (n:int) = sum_multiple_3_5_lt (n+1) = closed_formula n
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  lemma Closed_formula_0: p 0

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  lemma Closed_formula_n:
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    forall n:int. n > 0 -> p (n-1) ->
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      mod n 3 <> 0 /\ mod n 5 <> 0 -> p n
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  lemma Closed_formula_n_3:
    forall n:int. n > 0 -> p (n-1) ->
      mod n 3 = 0 /\ mod n 5 <> 0 -> p n

  lemma Closed_formula_n_5:
    forall n:int. n > 0 -> p (n-1) ->
      mod n 3 <> 0 /\ mod n 5 = 0 -> p n

  lemma Closed_formula_n_15:
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    forall n:int. n > 0 -> p (n-1) ->
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      mod n 3 = 0 /\ mod n 5 = 0 -> p n
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  clone int.Induction as I with predicate p = p
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  lemma Closed_formula:
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    forall n:int. 0 <= n -> p n

end

module Euler001

  use import SumMultiple
  use import int.Int
  use import int.EuclideanDivision

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  let solve n =
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    { 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 }

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end

(*
Local Variables:
compile-command: "unset LANG; make -C ../.. examples/programs/euler001.gui"
End:
*)