use vstd::prelude::*;
fn main() {}
verus!{



fn f1_element_wise_multiplication(arr1: &Vec<i32>, arr2: &Vec<i32>) -> (result: Vec<i32>)
    requires
        arr1.len() == arr2.len(),
        forall|i: int|
            (0 <= i < arr1.len()) ==> (i32::MIN <= #[trigger] (arr1[i] * arr2[i]) <= i32::MAX),
{
    let mut output_arr = Vec::with_capacity(arr1.len());
    let mut index = 0;
    while index < arr1.len()
    {
        output_arr.push((arr1[index] * arr2[index]));
        index += 1;
    }
    output_arr
}





fn f2_find_negative_numbers(arr: &Vec<i32>) -> (negative_list: Vec<i32>)
{
    let mut negative_list: Vec<i32> = Vec::new();
    let input_len = arr.len();

    let mut index = 0;
    while index < arr.len()
    {
        if (arr[index] < 0) {
            negative_list.push(arr[index]);
        }
        index += 1;
    }
    negative_list
}





fn f3_has_only_one_distinct_element(arr: &Vec<i32>) -> (result: bool)
{
    if arr.len() <= 1 {
        return true;
    }
    let mut index = 1;
    while index < arr.len()
    {
        if arr[0] != arr[index] {
            return false;
        }
        index += 1;
    }
    true
}



spec fn step1(__hp0: Seq<i32>, __hp1: Seq<i32>, outp: Seq<i32>) -> bool {
    (outp.len() == __hp0.len())
    && (forall|i: int|
            0 <= i < outp.len() ==> #[trigger] outp[i] == #[trigger] (__hp0[i] * __hp1[i]))
}

spec fn step2(inp: Seq<i32>, outp: Seq<i32>) -> bool {
    (outp == inp.filter(|x: i32| x < 0))
}

spec fn step3(inp: Seq<i32>, outp: bool) -> bool {
    (outp == (forall|i: int| 1 <= i < inp.len() ==> inp[0] == #[trigger] inp[i]))
}

fn chain_driver(arr1: &Vec<i32>, arr2: &Vec<i32>) -> (r_final: bool)
    requires
        arr1.len() == arr2.len(),
        forall|i: int|
            (0 <= i < arr1.len()) ==> (i32::MIN <= #[trigger] (arr1[i] * arr2[i]) <= i32::MAX),
    ensures
        exists|v1: Seq<i32>, v2: Seq<i32>| #[trigger] step1(arr1@, arr2@, v1) && #[trigger] step2(v1, v2) && step3(v2, r_final),
{
    let x1 = f1_element_wise_multiplication(arr1, arr2);
    let x2 = f2_find_negative_numbers(&x1);
    let x3 = f3_has_only_one_distinct_element(&x2);
    proof {
        assert(step1(arr1@, arr2@, x1@));
        assert(step2(x1@, x2@));
        assert(step3(x2@, x3));
    }
    x3
}

} // verus!