blog/posts-wip/record.md
2024-09-15 17:57:09 +02:00

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---
title: records
date: 2024-06-02
tags: ['quox (language)']
...
n-ary products to replace pairs _and_ boxes, since the nested `case`
expressions for nested pairs are actually really annoying.
# syntax
$$
\newcommand\CEQ{\operatorname{::=}}
\begin{aligned}
A, B, s, t &\CEQ \dotsb
\mid [Δ]
\mid [𝔇]
& \text{terms} \\
e, f &\operatorname{::=} \dotsb
\mid e\mathord·
& \text{eliminations} \\
𝔅 &\operatorname{::=} \dotsb
\mid [ \overline{=x} ] ⇒ s
& \text{case bodies} \\[1em]
&\operatorname{::=} x \mid n & \text{labels ($n ∈ $)} \\[1em]
Δ &\operatorname{::=} \overline{π· =x : A} & \text{telescopes} \\
𝔇 &\operatorname{::=} \overline{=s} & \text{list of bindings} \\
ϕ &\operatorname{::=} \overline{x=e} & \text{substitutions} \\
\end{aligned}
$$
- this __replaces__ the box syntax. pair syntax __still exists__ but
expands to this. this means that
$A \times B \times C$ and $A \times (B \times C)$ are no longer equivalent.
i guess that's fine but it is a change.
- overlined things are comma separated.
- in $Δ$, $π$ is optional and defaults to $1$ like everywhere else.
- the pair syntax
$(x_0, \dotsc, x_n) : (x_0 : A_0) \times \dotsb \times A_n$
is now an abbreviation for
$[0=x_0, \dotsc, n=x_n] : [0=x_0: A_0, \dotsc, n: A_n]$.
- instead of $\operatorname{\mathsf{fst}} e$ and $\operatorname{\mathsf{snd}} e$, it's now $e\mathord.0$
and $e\mathord.1$ (and $e\mathord.2$, …).
- "`'[0. 0 : A]`" looks bad. i know.
- a telescope $Δ$ can be used as a context by ignoring the outer labels $$.
# typing
## telescope checking
:::rulebox
$$Γ ⊢ Δ$$
:::
$$
\frac{}{Γ ⊢ •}
\qquad
\frac{
Γ ⊢ Δ \qquad
Γ, Δ ⊢_0 A ⇐ ★
}{
Γ ⊢ Δ, π·ℓ = x : A
}
$$
## binding list checking
:::rulebox
$$ Γ ⊢ σ·𝔇 ⇐ Δ ⊳ Σ $$
:::
$$ \frac{}{Γ ⊢ σ·• ⇐ • ⊳ 𝟎} $$
$$
\frac{
Γ ⊢ σ·𝔇 ⇐ Δ ⊳ Σ_𝔇 \qquad
Γ ⊢ ⌊σπ⌋ · s ⇐ A[𝔇:Δ] ⊳ Σ_s
}{
Γ ⊢ σ· (𝔇, = s) ⇐ (Δ, π·ℓ = x : A) ⊳ Σ_𝔇 + πΣ_s
}
$$
:::rulebox
$$ 𝔇 : Δ = ϕ $$
:::
produce a substitution from a binding list and a telescope by distributing the
type annotations
$$
\begin{aligned}
• : • &= • \\
(𝔇, x = s) : (Δ, π·ℓ=x : A) &= (𝔇 : Δ), (x = s ∷ A)
\end{aligned}
$$