Test Blog

Hi this is test blog

/// Extract all snowflake IDs surrounded by <@!? and >, called mentions, from a string.
#[must_use]
pub fn extract_mentions(s: &str) -> Vec<u64> {
    static REGEX: OnceLock<Regex> = OnceLock::new();

    let regex = REGEX.get_or_init(|| Regex::new(r"<@!?(\d+)>").unwrap());
    regex
        .captures_iter(s)
        .map(|c| c.get(1).unwrap().as_str().parse().unwrap())
        .collect::<Vec<_>>()
}

Eigenproof

1. Sets $𝐴$ and $𝐵$ are defined as follows:

   $$\begin{array}{cc}𝐴& =\left\{𝑛\in \mathrm{\mathbb{Z}}|𝑛=8𝑟-3\text{ for some integer }𝑟\right\}\\ 𝐵& =\left\{𝑚\in \mathrm{\mathbb{Z}}|𝑚=4𝑠+1\text{ for some integer }𝑠\right\}.\end{array}$$
   1. Prove that $𝐴\subseteq 𝐵$.
   2. Disprove that $𝐵\subseteq 𝐴$.

1. $𝐴\subseteq 𝐵$. Proof.

   Suppose $𝑥$ is an element of set $𝐴$, so $𝑥=8𝑟-3$ for some integer $𝑟$. To prove $𝐴\subseteq 𝐵$, we must show that $𝑥$ is also an element of set $𝐵$, so $𝑥=4𝑠+1$ for some integer $𝑠$.

   We can rewrite $𝑥$ as follows:

   $$𝑥=8𝑟-3=4\left(2𝑟-1\right)+1.$$

   If we set $𝑠=2𝑟-1$, then we have $𝑥=4𝑠+1$. $𝑠$ is an integer since $𝑟$ is an integer, and the products and differences of integers are integers. Thus, $𝑥$ is an element of $𝐵$. Since $𝑥$ was an arbitrary element of $𝐴$, we have shown that every element of $𝐴$ is also an element of $𝐵$, so $𝐴\subseteq 𝐵$. $∎$

2. $𝐵⊈𝐴$. Proof.

   Lemma: Parity Theorem for Even Integers. If $𝑥$ is even, then $𝑥+1$ is odd.
   Proof. If $𝑥$ is even, then $𝑥=2𝑘$ for some integer $𝑘$. Then $𝑥+1=2𝑘+1$, so $𝑥+1$ is odd.

   For the sake of contradiction, suppose $𝐵\subseteq 𝐴$. Then for all elements $𝑥\in 𝐵$, $𝑥\in 𝐴$.

   Suppose that $𝑠$ is an even integer, which means $𝑠$ is an integer. If we let $𝑥=4𝑠+1$, then $𝑥\in 𝐵$ by satisfying the predicate for $𝐵$.

   Since $𝑥\in 𝐵$, by the assumption that $𝐵\subseteq 𝐴$, we have $𝑥\in 𝐴$. Thus, there exists some integer $𝑟$ such that $𝑥=8𝑟-3$. Equating the expressions for $𝑥$ in terms of $𝑟$ and $𝑠$, we have:

   $$\begin{array}{cc}8𝑟-3& =4𝑠+1\\ 8𝑟& =4𝑠+4\\ 2𝑟& =𝑠+1.\end{array}$$

   Since $𝑠+1$ can be written as $2𝑟$ for an integer $𝑟$, $𝑠+1$ is an even integer by the definition of even integers. However, we assumed that $𝑠$ is even. By the Parity Theorem for Even Integers, since $𝑠$ is even, $𝑠+1$ must be odd.

   $𝑠$ cannot be even and odd at the same time. This is a contradiction, so our assumption that $𝐵\subseteq 𝐴$ must be false. Thus, $𝐵\subseteq 𝐴$ is false. $∎$