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170330.tex
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\subsection{$\sigma \splitsto \sd{k}{\tau}$}
Let's revisit all of the above rules in the Debruijn world.
% TODO: motivation?
\begin{mathpar}
\inferr{\Gamma \vd \bind{m}{s} \of \Gamma', A}
{\Gamma \vd s \of \Gamma' \\ \Gamma \vd m \of A[s]}
\inferr{\Gamma, \Gamma' \vd \lift k \of \Gamma}
{|\Gamma'| = k}
\end{mathpar}
If $\Gamma \vd \sigma \of \sig$ and $\sigma \splitsto \sd{\alpha \of k}{\tau}$
then $\Gamma \vd k \of \kind$, $\Gamma, \alpha \of k \vd \tau \of \type$. \\
Now the Debruijn form: \\
If $\Gamma \vd \sigma \of \sig$ and $\sigma \splitsto \sd{k}{\tau}$
then $\Gamma \vd k \of \kind$ and $\Gamma, k \vd \tau \of \type$.
\begin{mathpar}
\inferr{1 \splitsto \sd{1}{\unit}}{\strut}
\inferr{\satom{k} \splitsto \sd{k}{\unit}}{\strut}
\inferr{\datom{\tau} \splitsto \sd{1}{\tau[\lift]}}{\strut}
% See NOTE 0, 1, 2 for info on how we got this
\inferr{\Pia{\sigma_1}{\sigma_2} \splitsto
\sd{\Pi\bind{k_1}{k_2}}
{\forall\bind{k_1[\lift]}
{\tau_1[\bind{0}{\lift^2}] \arrow
\tau_2[\bind{\ap{1}{0}}{\bind{0}{\lift^2}}]}}}
{\sigma_1 \splitsto \sd{k_1}{\tau_1} \\
\sigma_2 \splitsto \sd{k_2}{\tau_2}}
% NOTE 0 These are the derivations on how we got the case for Pi app
%\inferr{\Gamma, A \vd 0 \of A[\lift]}{\strut}
%\inferr{\Gamma \vd i \of A[\lift^{i + 1}]}{\Gamma(i) = A}
%\inferr{\Gamma, \Pi\bind{k_1}{k_2}, k_1[\lift] \vd \bind{0}{\lift^2} \of \Gamma, k_1}
% {\Gamma, \Pi\bind{k_1}{k_2}, k_1[\lift] \vd 0 \of k_1[\lift^2] \\
% \Gamma, \Pi\bind{k_1}{k_2}, k_1[\lift] \vd \lift^2 \of \Gamma} \\
%\inferr{\Gamma, \Pi\bind{k_1}{k_2}, k_1[\lift] \vd
% \bind{\ap{1}{0}}{\bind{0}{\lift^2}} \of \Gamma, k_1, k_2}
% {\inferr{\Gamma, \Pi\bind{k_1}{k_2}, k_1[\lift] \vd
% \ap{1}{0} \of k_2[\bind{0}{\lift^2}]}
% {\dots \vd 1 \of (\Pi\bind{k_1}{k_2})[\lift^2] \\
% \dots \vd 0 \of k_1[\lift^2]}
% \\ \dots } \\
\inferr{\Pig{\of \sigma_1}{\sigma_2} \splitsto \sd{1}{
\forall\bind{k_1[\lift]}{\tau_1[0.\lift^2] \arrow
\exists\bind{k_2[0.\lift^2]}{\tau_2[0.1.\lift^3]}}}}
{\sigma_1 \splitsto \sd{k_1}{\tau_1} \\
\sigma_2 \splitsto \sd{k_2}{\tau_2}}
% NOTE 3 These is one of the derivations for Pi gen
%\inferr{\Gamma, 1, k_1[\lift], k_2[0.\lift^2] \vd 0.1.\lift^3 \of \Gamma, k_1, k_2}
% {\dots \vd 0 \of k_2[1.\lift^3] \\
% \inferr{\dots \vd 1.\lift^3 \of \Gamma, k_1}
% {\dots \vd 1 \of k_1[\lift^3] \\
% \dots \vd \lift^3 \of \Gamma}}
\inferr{\Sigma\bind{\sigma_1}{\sigma_2} \splitsto
\sd{
\Sigma\bind{k_1}{k_2}
}{
\tau_1[\pi_1 0. \lift] \times
\tau_2[\pi_2 0. \pi_1 0. \lift]
}
}
{\sigma_1 \splitsto \sd{k_1}{\tau_1} \\
\sigma_2 \splitsto \sd{k_2}{\tau_2}}
% NOTE 4 These are the derivations for sigma
%\inferrule{
% \inferr{\Gamma, \Sigma\bind{k_1}{k_2} \vd \pi_2 0 \of k_2[\pi_1 0.\lift]}
% {\Gamma, \Sigma\bind{k_1}{k_2} \vd 0 \of \Sigma\bind{k_1[\lift]}{k_2[0.\lift^2]}}
% \inferr{\Gamma, \Sigma\bind{k_1}{k_2} \vd \pi_1 0. \lift \of \Gamma, k_1}
% {\inferr{\Gamma, \Sigma\bind{k_1}{k_2} \vd \pi_1 0 \of k_1[\lift]}
% {\Gamma, \Sigma\bind{k_1}{k_2} \vd 0 \of
% \Sigma\bind{k_1[\lift]}{k_2[0.\lift^2]}} \\
% \Gamma, \Sigma\bind{k_1}{k_2} \vd \lift \of \Gamma}
%}{
% \Gamma, \Sigma\bind{k_1}{k_2} \vd \pi_2 0. \pi_1 0. \lift \of \Gamma, k_1, k_2
%}
\end{mathpar}
%% NOTE 1
%$\Gamma \vd \Pia{\sigma_1}{\sigma_2} \of \sig$ \\
%$\Gamma \vd \sigma_1 \of \sig$ \\
%$\Gamma, k_1 \vd \sigma_2 \of \sig$ \\
%$\Gamma \vd k_1 \of \kind$ \\
%$\Gamma, k_1 \vd \tau_1 \of \type$ \\
%$\Gamma, k_1 \vd k_2 \of \kind$ \\
%$\Gamma, k_1, k_2 \vd \tau_2 \of \type$ \\
%
%where: \\
%$\sigma_1 \splitsto \sd{k_1}{\tau_1}$ \\
%$\sigma_2 \splitsto \sd{k_2}{\tau_2}$ \\
%
%% NOTE 2
%$(\Pi\bind{k_1}{k_2})[\lift^i] = \Pi\bind{k_1[\lift^i]}{k_2[\bind{0}{\lift^{i + 1}}]}$
\subsection{$\target{\Gamma} \tto \Gamma$}
\begin{flalign*}
\target{\epsilon} &= \epsilon &\\
\target{\Gamma, \alpha \of k} &= \target{\Gamma}, \alpha \of k &\\
\target{\Gamma, x \of \tau} &= \target{\Gamma}, x \of \tau &\\
\target{\Gamma, \alpha/s \of \sigma} &= \target{\Gamma}, \alpha \of k, s \of \tau &\\
\end{flalign*}
*NOTE: in the last one, $\sigma \splitsto \sd{\alpha \of k}{\tau}$
\subsection{$\Gamma \vd_P M \of \sigma \splitsto \sd{c}{e}$}
\begin{mathpar}
\inferrule{
\alpha/s \of \sigma \in \Gamma
}{ % Coding note: algorithmically, we will be inferring \sigma, and it will
% actually be the best signature (eg: \singleton{\alpha : \sigma})
\Gamma \vdp s \of \sigma \splitsto \sd{\alpha}{s}
}
\axiom{\Gamma \vdp \ast \of 1 \splitsto \sd{\ast}{\pair{}}}
\inferrule{
\Gamma \vd c \of k
}{
\Gamma \vdp \satom{c} \of \satom{k} \splitsto \sd{c}{\pair{}}
}
\inferrule{
\Gamma \vd e \of \tau
}{ % note the tau is shifted, but the e is not shifted, care for debruijn
\Gamma \vd \datom{e} \of \datom{\tau} \splitsto \sd{\ast}{e}
}
\inferrule{
\Gamma \vd \sigma_1 \of \sig \\
\Gamma, \alpha/s \of \sigma_1 \vdp M \of \sigma_2 \splitsto \sd{c}{e} \\
\sigma_1 \splitsto \sd{\alpha_1 \of k_1}{\tau_1}
}{
\Gamma \vd \lambdaa{\alpha/s \of \sigma_1}{M} \of \Pi\bind{\alpha \of \sigma_1}{\sigma_2}
\splitsto
\sd{
\lambda\bind{\alpha \of k_1}{c}
}{
\Lambda\bind{\alpha \of k_1}{
\lambda\bind{s \of \subst{\alpha}{\alpha_1}{\tau_1}}{
e
}
}
}
}
\end{mathpar}