Ctheorems

Touying 能够与 ctheorems 包一起正常工作，你可以直接使用 ctheorems 包。

其中，你还可以使用 #let s = (s.methods.numbering)(self: s, section: "1.", "1.1") 为 sections 和 subsections 设置 numbering。

#import "@preview/touying:0.4.1": *
#import "@preview/ctheorems:1.1.2": *

// Register university theme
#let s = themes.simple.register(aspect-ratio: "16-9")

// Set the numbering of section and subsection
#let s = (s.methods.numbering)(self: s, section: "1.", "1.1")

// Theorems configuration by ctheorems
#show: thmrules.with(qed-symbol: $square$)
#let theorem = thmbox("theorem", "Theorem", fill: rgb("#eeffee"))
#let corollary = thmplain(
  "corollary",
  "Corollary",
  base: "theorem",
  titlefmt: strong
)
#let definition = thmbox("definition", "Definition", inset: (x: 1.2em, top: 1em))
#let example = thmplain("example", "Example").with(numbering: none)
#let proof = thmproof("proof", "Proof")

// Extract methods
#let (init, slides, touying-outline, alert, speaker-note) = utils.methods(s)
#show: init

#show strong: alert

// Extract slide functions
#let (slide, empty-slide) = utils.slides(s)
#show: slides

= Theorems

== Prime numbers

#definition[
  A natural number is called a #highlight[_prime number_] if it is greater
  than 1 and cannot be written as the product of two smaller natural numbers.
]
#example[
  The numbers $2$, $3$, and $17$ are prime.
  @cor_largest_prime shows that this list is not exhaustive!
]

#theorem("Euclid")[
  There are infinitely many primes.
]
#proof[
  Suppose to the contrary that $p_1, p_2, dots, p_n$ is a finite enumeration
  of all primes. Set $P = p_1 p_2 dots p_n$. Since $P + 1$ is not in our list,
  it cannot be prime. Thus, some prime factor $p_j$ divides $P + 1$.  Since
  $p_j$ also divides $P$, it must divide the difference $(P + 1) - P = 1$, a
  contradiction.
]

#corollary[
  There is no largest prime number.
] <cor_largest_prime>
#corollary[
  There are infinitely many composite numbers.
]

#theorem[
  There are arbitrarily long stretches of composite numbers.
]

#proof[
  For any $n > 2$, consider $
    n! + 2, quad n! + 3, quad ..., quad n! + n #qedhere
  $
]